Texas A&M Universityroquesol/Math_304_Fall_2018_Session… · Vector Spaces Linear Dependence and...

Vector Spaces

Linear Algebra. Session 6

Dr. Marco A Roque Sol

10/02/2018

Dr. Marco A Roque Sol Linear Algebra. Session 6

Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix

Linear Dependence and Independence

Let S0 and S be subsets of a vector space V

More facts on linear independence

If S0 ⊂ S and S is linearly independent, then so is S0.

If S0 ⊂ S and S0 is linearly dependent, then so is S.

If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.

The empty set is linearly independent.

Any set containing 0 is linearly dependent.

S0 and S be subsets of a vector space V

Let S0 and

S be subsets of a vector space V

Let S0 and S

be subsets of a vector space V

Let S0 and S be subsets

of a vector space V

Let S0 and S be subsets of a vector space

If S0 ⊂ S and

S is linearly independent, then so is S0.

If S0 ⊂ S and S

is linearly independent, then so is S0.

If S0 ⊂ S and S is linearly independent,

then so is S0.

If S0 ⊂ S and S is linearly independent, then so is

⊂ S and S0 is linearly dependent, then so is S.

If S0 ⊂ S and

S0 is linearly dependent, then so is S.

If S0 ⊂ S and S0

is linearly dependent, then so is S.

If S0 ⊂ S and S0 is linearly dependent,

then so is S.

If S0 ⊂ S and S0 is linearly dependent, then so is

S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.

is linearly independent in V and V is a subspace of W,then S is linearly independent in W.

If S is linearly independent

in V and V is a subspace of W,then S is linearly independent in W.

If S is linearly independent in V and

V is a subspace of W,then S is linearly independent in W.

If S is linearly independent in V and V

is a subspace of W,then S is linearly independent in W.

If S is linearly independent in V and V is a subspace

of W,then S is linearly independent in W.

If S is linearly independent in V and V is a subspace of W,

then S is linearly independent in W.

If S is linearly independent in V and V is a subspace of W,then S

is linearly independent in W.

If S is linearly independent in V and V is a subspace of W,then S is linearly independent

The empty set

is linearly independent.

Any set

containing 0 is linearly dependent.

Any set containing 0

is linearly dependent.

Vector Spaces

Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.

If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).

Theorem

Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).

Vector Spaces

Theorem

Vector Spaces

Two nonzero vectors

v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.

Theorem

Vector Spaces

Two nonzero vectors v1 and v2

are linearly dependent if andonly if either of them is a scalar multiple the other.

Theorem

Vector Spaces

Two nonzero vectors v1 and v2 are linearly dependent

if andonly if either of them is a scalar multiple the other.

Theorem

Vector Spaces

Two nonzero vectors v1 and v2 are linearly dependent if andonly if

either of them is a scalar multiple the other.

Theorem

Vector Spaces

Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them

is a scalar multiple the other.

Theorem

Vector Spaces

Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple

the other.

Theorem

Vector Spaces

Theorem

Vector Spaces

is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).

Theorem

Vector Spaces

If S0 is linearly independent and

v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).

Theorem

Vector Spaces

If S0 is linearly independent and v0 ∈ V − S0

then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).

Theorem

Vector Spaces

If S0 is linearly independent and v0 ∈ V − S0 then,

S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).

Theorem

Vector Spaces

If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}

is linearly independent if and only if v0 /∈ Span(S0).

Theorem

Vector Spaces

If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent

if and only if v0 /∈ Span(S0).

Theorem

Vector Spaces

If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if

v0 /∈ Span(S0).

Theorem

Vector Spaces

Theorem

Vector Spaces

Theorem

Vector Spaces

Theorem

Vectors

v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).

Vector Spaces

Theorem

Vectors v1, v2, · · · , vm ∈ Rn

are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).

Vector Spaces

Theorem

Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent

wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).

Vector Spaces

Theorem

Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n

(i.e., the number of coordinates is less than the number ofvectors).

Vector Spaces

Theorem

Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates

is less than the number ofvectors).

Vector Spaces

Theorem

Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than

the number ofvectors).

Vector Spaces

Theorem

Vector Spaces

proof.

Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system

a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m

Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then

the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality

t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the systema11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0

is equivalent to the systema11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent

to the systema11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm

are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns

of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij).

The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number

of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries

in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform

is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n.

If m > n then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n

then there are free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are

free variables, thereforethe zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, therefore

the zero solution is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution

is not unique. �

Vector Spaces

proof.

...an1t1 + an2t2 + · · ·+ anmtn = 0

General results on linear independence in Rn

Theorem 1 Given an m × n matrix the following conditions areequivalent:

i) columns of A are linearly independent (as vectors in Rm )

ii) x = 0 is the only solution of the matrix equation Ax = 0

iii) The row echelon form of A has a leading entry in each column.

Theorem 2 Given a square matrix the following conditions areequivalent:

i) det(A) = 0

ii) Columns of A are linearly independent (as vectors in Rn )

iii) Rows of A are linearly independent (as vectors in Rn )

i) det(A) = 0

Theorem 1

Given an m × n matrix the following conditions areequivalent:

i) det(A) = 0

Theorem 1 Given

an m × n matrix the following conditions areequivalent:

i) det(A) = 0

Theorem 1 Given an m × n matrix

the following conditions areequivalent:

i) det(A) = 0

Theorem 1 Given an m × n matrix the following conditions

areequivalent:

i) det(A) = 0

i) columns of A

are linearly independent (as vectors in Rm )

i) det(A) = 0

i) columns of A are linearly independent

(as vectors in Rm )

i) det(A) = 0

ii) x = 0

is the only solution of the matrix equation Ax = 0

i) det(A) = 0

ii) x = 0 is the only solution

of the matrix equation Ax = 0

i) det(A) = 0

ii) x = 0 is the only solution of the matrix equation

Ax = 0

i) det(A) = 0

iii) The row echelon form

of A has a leading entry in each column.

i) det(A) = 0

iii) The row echelon form of A

has a leading entry in each column.

i) det(A) = 0

iii) The row echelon form of A has a leading entry

in each column.

i) det(A) = 0

Theorem 2

Given a square matrix the following conditions areequivalent:

i) det(A) = 0

Theorem 2 Given

a square matrix the following conditions areequivalent:

i) det(A) = 0

Theorem 2 Given a square matrix

the following conditions areequivalent:

i) det(A) = 0

Theorem 2 Given a square matrix the following conditions

areequivalent:

i) det(A) = 0

ii) Columns of A

are linearly independent (as vectors in Rn )

i) det(A) = 0

ii) Columns of A are linearly independent

(as vectors in Rn )

i) det(A) = 0

iii) Rows of A

are linearly independent (as vectors in Rn )

i) det(A) = 0

iii) Rows of A are linearly independent

(as vectors in Rn )

i) det(A) = 0

iii) Rows of A are linearly independent (as vectors in Rn )Dr. Marco A Roque Sol Linear Algebra. Session 6

Example 6.1

Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3

Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.

Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Consider vectors

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Consider vectors v1 = (1,−1, 1),

v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0),

v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), and

v4 = (1, 2, 4) in R3

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4)

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Two vectors

are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Two vectors are linearly dependent

if and only if they are parallel.Hence v1 and v2 are linearly independent.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Two vectors are linearly dependent if and only if

they are parallel.Hence v1 and v2 are linearly independent.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Two vectors are linearly dependent if and only if they are parallel.

Hence v1 and v2 are linearly independent.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Two vectors are linearly dependent if and only if they are parallel.Hence v1 and

v2 are linearly independent.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2

are linearly independent.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors

v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1,

v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1, v2, and

v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1, v2, and v3

are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1, v2, and v3 are linearly independent

if and only if thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1, v2, and v3 are linearly independent if and only if

thematrix A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1, v2, and v3 are linearly independent if and only if thematrix

A = (v1, v2, v3) is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3)

is invertible.

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ =

−∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ =

2 6= 0

Example 6.1

det(A) =

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ = −∣∣∣∣ −1 1

∣∣∣∣ = 2 6= 0

Therefore v1, v2, v3 are linearly independent.

Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.

Example 6.2Let

(−1 1−1 0

Determine whether matrices A,A2,A3 are linearly independent

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Therefore

v1, v2, v3 are linearly independent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Therefore v1, v2, v3

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Four vectors

in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Four vectors in R3

are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Four vectors in R3 are always linearly dependent.

Thus v1, v2, v3and v4 are linearly dependent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Four vectors in R3 are always linearly dependent. Thus

v1, v2, v3and v4 are linearly dependent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Four vectors in R3 are always linearly dependent. Thus v1, v2, v3

and v4 are linearly dependent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4

are linearly dependent.

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Determine

whether matrices A,A2,A3 are linearly independent

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Determine whether matrices

A,A2,A3 are linearly independent

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Determine whether matrices A,A2,A3

are linearly independent

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

)Dr. Marco A Roque Sol Linear Algebra. Session 6

Example 6.2Let

(−1 1−1 0

Solution

We have

(−1 1−1 0

(0 −11 −1

(1 00 1

)Dr. Marco A Roque Sol Linear Algebra. Session 6

We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A

2 + r3A3 = 0. This matrix equation is equivalent to

a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

The augmented matrix is

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

We need

to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

We need to determine

if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

We need to determine if there exist

r1, r2, r3 ∈ R not all zero suchthat r1A + r2A

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

We need to determine if there exist r1, r2, r3 ∈ R

not all zero suchthat r1A + r2A

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

We need to determine if there exist r1, r2, r3 ∈ R not all zero

suchthat r1A + r2A

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat

r1A + r2A2 + r3A

3 = 0. This matrix equation is equivalent toa system

−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

2 + r3A3 =

0. This matrix equation is equivalent toa system

−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

2 + r3A3 = 0.

This matrix equation is equivalent toa system

−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

2 + r3A3 = 0. This matrix equation

is equivalent toa system

−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

2 + r3A3 = 0. This matrix equation is equivalent

toa system

−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

a system

−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

−1 0 1 00 1 −1 00 0 0 00 0 0 0

−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0

⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0

Linear Dependence and Independences

The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Show that functions ex , e2x , and e3x are linearly independent inC∞(R)

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

The row echelon form

of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix

shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is

afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable.

Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable. Hence

the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable. Hence the system

has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero

solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices

are linearly dependent (one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent

(one relation is A + A2 + A3 = 0)

Example 6.3

Solution

The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is

A + A2 + A3 = 0)

Example 6.3

Solution

Example 6.3

Solution

Example 6.3

Solution

Example 6.3

that functions ex , e2x , and e3x are linearly independent inC∞(R)

Solution

Example 6.3

Show that functions

ex , e2x , and e3x are linearly independent inC∞(R)

Solution

Example 6.3

Show that functions ex , e2x , and e3x are

linearly independent inC∞(R)

Solution

Example 6.3

Show that functions ex , e2x , and e3x are linearly independent in

C∞(R)

Solution

Example 6.3

Solution

Example 6.3

Solution

Example 6.3

Solution

Suppose that

aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0

for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all

x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R

where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where

a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, c

areconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants.

We have to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have

to show that a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that

a = b = c = 0. Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0.

Differentiate thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate

thisidentity twice:

Example 6.3

Solution

Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity

twice:

Example 6.3

Solution

aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0

It follows that A(x)v = 0, where

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

aex + be2x + ce3x = 0

aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0

aex + 4be2x + 9ce3x = 0

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

It follows

that A(x)v = 0, where

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

It follows that A(x)v = 0,

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

A(x) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

∣∣∣∣∣∣1 e2x e3x

1 2e2x 3e3x

1 4e2x 9e3x

∣∣∣∣∣∣ =

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x =

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x =

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ =

e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x

∣∣∣∣ 1 13 8

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ =

2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x)

is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible

we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtain

A(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒

v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒

a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence,

the set of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set

of functions {ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions

{ex , e2x , e3x} is linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is

linearly independent.

exe2x = e3x

∣∣∣∣∣∣1 1 e3x

1 2 3e3x

1 4 9e3x

∣∣∣∣∣∣ =

e3xe3x = e6x

∣∣∣∣∣∣1 1 11 2 31 4 9

∣∣∣∣∣∣ = e6x

∣∣∣∣∣∣1 1 10 1 20 3 8

∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1

∣∣∣∣ = 2e6x 6= 0

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

∣∣∣∣∣∣∣∣∣Theorem

If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn

be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions

on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval

[a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval [a, b].

TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian

W[f1, f2, ..., fn] is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn]

is a function on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function

on [a, b] defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b]

defined by

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

∣∣∣∣∣∣∣∣∣

Theorem

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

If W[f1, f2, ..., fn](x0) 6= 0

for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b]

then the functionsf1, f2, ..., fn are linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then

the functionsf1, f2, ..., fn are linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functions

f1, f2, ..., fn are linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are

linearly independent in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent

in C [a, b].

Wronskian

W[f1, f2, ..., fn](x) =

∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space.

Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent

spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set

Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V,

is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis.

That means, that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means,

that any vector v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector

v ∈ V can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V

can berepresented as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented

as a linear combination

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

where v1, v2, · · · , vk are

distinct vectors from S andr1, r2, · · · , rk ∈ R

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

where v1, v2, · · · , vk are distinct vectors

from S andr1, r2, · · · , rk ∈ R

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

where v1, v2, · · · , vk are distinct vectors from

S andr1, r2, · · · , rk ∈ R

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

where v1, v2, · · · , vk are distinct vectors from S and

r1, r2, · · · , rk ∈ R

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk

Basis and Dimension

Definition.

v = r1v1 + r2v2 + · · ·+ rkvk

Basis and Dimension

“Linearly independent”, in the above definition, implies that theabove representation is unique:

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

“Linearly independent”,

in the above definition, implies that theabove representation is unique:

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

“Linearly independent”, in the above definition,

implies that theabove representation is unique:

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

“Linearly independent”, in the above definition, implies that

theabove representation is unique:

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

“Linearly independent”, in the above definition, implies that theabove representation

is unique:

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒

0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒

(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒

r1 = r ′1, r2 = r ′2, · · · rk = r ′k

Basis and Dimension

Example 6.4Standard Basis for Rn

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0),

e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0),

e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0),

· · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · ,

en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 +

x2e2 + ... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 +

... + xnen

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... +

Basis and Dimension

e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),

Solution

Indeed,

(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen

Basis and Dimension

Example 6.5Standard Basis for M2,2

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

aE12 + bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 +

bE12 + cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 +

cE21 + dE22

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 +

Basis and Dimension

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

Solution

Indeed,

(a bc d

)= aE12 + bE12 + cE21 + dE22

Basis and Dimension

Example 6.6Standard Basis for Polinomial Pn

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

Example 6.7Standard Basis for Polinomials P

a0, a1x , a2x2, ..., anx

n, ...

SolutionPOC ... !!!!

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x ,

a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2,

..., an−1xn−1

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ...,

an−1xn−1

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x ,

a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2,

..., anxn, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ...,

anxn, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Solution

POC ... !!!!

Basis and Dimension

a0, a1x , a2x2, ..., an−1x

Solution

POC ... !!!!

a0, a1x , a2x2, ..., anx

n, ...

Basis and Dimension

Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector

v = r1v1 + r2v2 + · · ·+ rkvk

is equivalent to the matrix equation Ax = v where

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

Let v1, v2, · · · vk ∈ Rn and

r1, r2, · · · rn ∈ R. The vector

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R.

The vector

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

is equivalent

to the matrix equation Ax = v where

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

is equivalent to the matrix equation

Ax = v where

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

is equivalent to the matrix equation Ax = v

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk)

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is,

A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the

n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the n × k matrix

such that vectors v1, v2, · · · vk areconsecutive columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the n × k matrix such that

vectors v1, v2, · · · vk areconsecutive columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the n × k matrix such that vectors

v1, v2, · · · vk areconsecutive columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the n × k matrix such that vectors v1, v2, · · · vk

areconsecutive columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive

columns of A.

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors

v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk

span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn

if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row

echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form

of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A

hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.

Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors

v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are

linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent

if the rowechelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the row

echelon form of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form

of A has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A

has a leading entry in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry

in each column (no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column

(no freevariables).

Basis and Dimension

v = r1v1 + r2v2 + · · ·+ rkvk

A = (v1, v2, · · · vk) x =

r1r2...rk

Basis and Dimension

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Basis and Dimension

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Basis and Dimension

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Basis and Dimension

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Basis and Dimension

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Basis and Dimension

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Basis and Dimension

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Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

If k < n then the vectors v1, v2, · · · vk do not span Rn.

Theorem 2

If k > n then the vectors v1, v2, · · · vk are linearly dependent.

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

k < n then the vectors v1, v2, · · · vk do not span Rn.

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

If k < n

then the vectors v1, v2, · · · vk do not span Rn.

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

If k < n then the vectors v1, v2, · · · vk

do not span Rn.

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

If k < n then the vectors v1, v2, · · · vk do not span

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

k > n then the vectors v1, v2, · · · vk are linearly dependent.

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

If k > n

then the vectors v1, v2, · · · vk are linearly dependent.

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

If k > n then the vectors

v1, v2, · · · vk are linearly dependent.

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

If k > n then the vectors v1, v2, · · · vk are

linearly dependent.

Basis and Dimension

Dimension

Let v1, v2, · · · vk ∈ Rn

Theorem 1

Theorem 2

Basis and Dimension

Theorem 3

If k = n then the following conditions are equivalent:

{v1, v2, · · · vn} is a basis for Rn

{v1, v2, · · · vn} is a spanning set for Rn

{v1, v2, · · · vn}is a linearly independent set.

Basis and Dimension

Theorem 3

Basis and Dimension

Theorem 3

k = n then the following conditions are equivalent:

Basis and Dimension

Theorem 3

If k = n then

the following conditions are equivalent:

Basis and Dimension

Theorem 3

If k = n then the following conditions

are equivalent:

Basis and Dimension

Theorem 3

Basis and Dimension

Theorem 3

{v1, v2, · · · vn}

is a basis for Rn

Basis and Dimension

Theorem 3

{v1, v2, · · · vn} is a basis

for Rn

Basis and Dimension

Theorem 3

Basis and Dimension

Theorem 3

{v1, v2, · · · vn}

is a spanning set for Rn

Basis and Dimension

Theorem 3

{v1, v2, · · · vn} is a spanning set

for Rn

Basis and Dimension

Theorem 3

Basis and Dimension

Theorem 3

{v1, v2, · · · vn}

is a linearly independent set.

Basis and Dimension

Theorem 3

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3

Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Therefore, {v1, v2, v3} is a basis for R3

The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors

v1 and v2 are linearly independent (as they are notparallel), but they do not span R3

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1 and v2

are linearly independent (as they are notparallel), but they do not span R3

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1 and v2 are linearly independent

(as they are notparallel), but they do not span R3

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1 and v2 are linearly independent (as they are notparallel),

but they do not span R3

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not

span R3

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors

v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1, v2 and v3 are

linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

Vectors v1, v2 and v3 are linearly independent since

∣∣∣∣∣∣1 1 1−1 0 11 0 1

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ =

2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Therefore,

{v1, v2, v3} is a basis for R3

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Therefore, {v1, v2, v3}

is a basis for R3

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Therefore, {v1, v2, v3} is a basis

for R3

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

The set

{v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

The set {v1, v2, v3, v4}

span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

The set {v1, v2, v3, v4} span R3

( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3}

already spanR3 ), but they are linearly dependent.

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ),

but they are linearly dependent.

Basis and Dimension

Example 6.8

v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),

∣∣∣∣∣∣ =

∣∣∣∣ −1 11 1

∣∣∣∣ = 2 6= 0

Basis and Dimension

Dimension

Theorem 1

Any vector space has a basis.

Theorem 2

If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.

Definition

The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Any vector space

has a basis.

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

If a vector space

V has a finite basis, then all bases for V are finiteand have the same number of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

If a vector space V

has a finite basis, then all bases for V are finiteand have the same number of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

If a vector space V has a finite basis,

then all bases for V are finiteand have the same number of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

If a vector space V has a finite basis, then all bases

for V are finiteand have the same number of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

If a vector space V has a finite basis, then all bases for V are finiteand

have the same number of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

If a vector space V has a finite basis, then all bases for V are finiteand have the same number

of elements.

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension

of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension of a vector space

V, denoted dim(V), is thenumber of elements in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension of a vector space V,

denoted dim(V), is thenumber of elements in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension of a vector space V, denoted dim(V),

is thenumber of elements in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension of a vector space V, denoted dim(V), is thenumber

of elements in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension of a vector space V, denoted dim(V), is thenumber of elements

in any of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

The dimension of a vector space V, denoted dim(V), is thenumber of elements in any

of its bases.

Basis and Dimension

Dimension

Theorem 1

Theorem 2

Definition

Basis and Dimension

Example 6.9

Rn : n-dimensional space. dim(Rn) = n

M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4

Mm×n : the space of m × n matrices. dim(Mm×n) = mn

Pn : the space of polynomials of degree less than n.dim(Pn) = n

P : the space of all polynomials. dim(Pn) =∞

{0} : the trivial space . dim({0}) = 0

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.9

Rn : n-dimensional space.

dim(Rn) = n

Basis and Dimension

Example 6.9

Rn : n-dimensional space. dim(Rn) =

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.9

M2×2 : the space of 2× 2 matrices.

dim(M2×2) = 4

Basis and Dimension

Example 6.9

M2×2 : the space of 2× 2 matrices. dim(M2×2) =

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.9

Mm×n : the space of m × n matrices.

dim(Mm×n) = mn

Basis and Dimension

Example 6.9

Mm×n : the space of m × n matrices. dim(Mm×n) =

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.9

Pn : the space of polynomials of degree less than n.

dim(Pn) = n

Basis and Dimension

Example 6.9

Pn : the space of polynomials of degree less than n.dim(Pn) =

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.9

P : the space of all polynomials.

dim(Pn) =∞

Basis and Dimension

Example 6.9

P : the space of all polynomials. dim(Pn) =

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.9

{0} : the trivial space .

dim({0}) = 0

Basis and Dimension

Example 6.9

{0} : the trivial space . dim({0}) =

Basis and Dimension

Example 6.9

Basis and Dimension

Example 6.10Find the dimension of the plane x + 2z = 0 in R3

Solution

The general solution of the equation isx = 2sy = tz = s

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Example 6.10

Find the dimension of the plane x + 2z = 0 in R3

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Example 6.10Find

the dimension of the plane x + 2z = 0 in R3

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Example 6.10Find the dimension

of the plane x + 2z = 0 in R3

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Example 6.10Find the dimension of the plane

x + 2z = 0 in R3

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Example 6.10Find the dimension of the plane x + 2z = 0 in

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

The general solution

of the equation isx = 2sy = tz = s

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

The general solution of the equation

isx = 2sy = tz = s

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

The general solution of the equation is

x = 2sy = tz = s

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is,

(x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) =

(−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) =

t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) +

s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence,

the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane

is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span

of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors

v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) and

v2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1).

These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors

are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent

as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they are

not parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus,

{v1, v2} is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2}

is a basis so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis

so that the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that

the dimension of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension

of theplane is 2.

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is

Basis and Dimension

Solution

(t, s,∈ R)

That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)

Basis and Dimension

Theorem

Let S be a subset of a vector space V. Then, the followingconditions are equivalent:

i) S is a basis, i.e., is a linearly independent spanning set for V.

ii) S is a minimal spanning set for V.

iii) S is a maximal linearly independent subset of V.

’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.

’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.

Basis and Dimension

Theorem

Basis and Dimension

Theorem

Basis and Dimension

Theorem

S be a subset of a vector space V. Then, the followingconditions are equivalent:

Basis and Dimension

Theorem

be a subset of a vector space V. Then, the followingconditions are equivalent:

Basis and Dimension

Theorem

Let S be a subset

of a vector space V. Then, the followingconditions are equivalent:

Basis and Dimension

Theorem

Let S be a subset of a vector space

V. Then, the followingconditions are equivalent:

Basis and Dimension

Theorem

Let S be a subset of a vector space V.

Then, the followingconditions are equivalent:

Basis and Dimension

Theorem

Let S be a subset of a vector space V. Then,

the followingconditions are equivalent:

Basis and Dimension

Theorem

Let S be a subset of a vector space V. Then, the followingconditions

are equivalent:

Basis and Dimension

Theorem

Basis and Dimension

Theorem

is a basis, i.e., is a linearly independent spanning set for V.

Basis and Dimension

Theorem

i) S is a basis,

i.e., is a linearly independent spanning set for V.

Basis and Dimension

Theorem

i) S is a basis, i.e.,

is a linearly independent spanning set for V.

Basis and Dimension

Theorem

i) S is a basis, i.e., is a linearly independent

spanning set for V.

Basis and Dimension

Theorem

i) S is a basis, i.e., is a linearly independent spanning set

for V.

Basis and Dimension

Theorem

Basis and Dimension

Theorem

Basis and Dimension

Theorem

is a minimal spanning set for V.

Basis and Dimension

Theorem

ii) S is a minimal

spanning set for V.

Basis and Dimension

Theorem

ii) S is a minimal spanning set

for V.

Basis and Dimension

Theorem

Basis and Dimension

Theorem

Basis and Dimension

Theorem

iii) S

is a maximal linearly independent subset of V.

Basis and Dimension

Theorem

iii) S is a maximal

linearly independent subset of V.

Basis and Dimension

Theorem

iii) S is a maximal linearly independent

subset of V.

Basis and Dimension

Theorem

iii) S is a maximal linearly independent subset of

Basis and Dimension

Theorem

Basis and Dimension

Theorem

Basis and Dimension

Theorem

’Minimal spanning set’

means: remove any element from this set,and it is no longer a spanning set.

Basis and Dimension

Theorem

’Minimal spanning set’ means:

remove any element from this set,and it is no longer a spanning set.

Basis and Dimension

Theorem

’Minimal spanning set’ means: remove any element

from this set,and it is no longer a spanning set.

Basis and Dimension

Theorem

’Minimal spanning set’ means: remove any element from this set,and

it is no longer a spanning set.

Basis and Dimension

Theorem

’Minimal spanning set’ means: remove any element from this set,and it is no longer

a spanning set.

Basis and Dimension

Theorem

Basis and Dimension

Theorem

’ Maximal linearly independent subset’

means: add any element ofV to this set, and it will become linearly dependent.

Basis and Dimension

Theorem

’ Maximal linearly independent subset’ means:

add any element ofV to this set, and it will become linearly dependent.

Basis and Dimension

Theorem

’ Maximal linearly independent subset’ means: add any element

ofV to this set, and it will become linearly dependent.

Basis and Dimension

Theorem

’ Maximal linearly independent subset’ means: add any element ofV

to this set, and it will become linearly dependent.

Basis and Dimension

Theorem

’ Maximal linearly independent subset’ means: add any element ofV to this set, and

it will become linearly dependent.

Basis and Dimension

Theorem

’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become

linearly dependent.

Basis and Dimension

Theorem

Basis and Dimension

Theorem

Let V be a vector space. Then

i) Any spanning set for V can be reduced to a minimal spanningset.

ii) any linearly independent subset of V can be extended to amaximal linearly independent set.

Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.

Corollary 2

A vector space is finite-dimensional if and only if it is spanned by afinite set.

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

Theorem

V be a vector space. Then

Corollary 2

Basis and Dimension

Theorem

be a vector space. Then

Corollary 2

Basis and Dimension

Theorem

Let V be a vector space.

Corollary 2

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

Theorem

i) Any

spanning set for V can be reduced to a minimal spanningset.

Corollary 2

Basis and Dimension

Theorem

i) Any spanning set

for V can be reduced to a minimal spanningset.

Corollary 2

Basis and Dimension

Theorem

i) Any spanning set for V

can be reduced to a minimal spanningset.

Corollary 2

Basis and Dimension

Theorem

i) Any spanning set for V can be reduced

to a minimal spanningset.

Corollary 2

Basis and Dimension

Theorem

i) Any spanning set for V can be reduced to a minimal

spanningset.

Corollary 2

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

Theorem

ii) any

linearly independent subset of V can be extended to amaximal linearly independent set.

Corollary 2

Basis and Dimension

Theorem

ii) any linearly independent subset

of V can be extended to amaximal linearly independent set.

Corollary 2

Basis and Dimension

Theorem

ii) any linearly independent subset of V

can be extended to amaximal linearly independent set.

Corollary 2

Basis and Dimension

Theorem

ii) any linearly independent subset of V can be extended

to amaximal linearly independent set.

Corollary 2

Basis and Dimension

Theorem

ii) any linearly independent subset of V can be extended to amaximal

linearly independent set.

Corollary 2

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any

spanning set contains a basis while any linearly independentset is contained in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any spanning set

contains a basis while any linearly independentset is contained in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any spanning set contains

a basis while any linearly independentset is contained in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any spanning set contains a basis

while any linearly independentset is contained in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any spanning set contains a basis while any

linearly independentset is contained in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any spanning set contains a basis while any linearly independent

set is contained in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 1Any spanning set contains a basis while any linearly independentset is contained

in a basis.

Corollary 2

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

Theorem

Corollary 2

A vector space

is finite-dimensional if and only if it is spanned by afinite set.

Basis and Dimension

Theorem

Corollary 2

A vector space is finite-dimensional

if and only if it is spanned by afinite set.

Basis and Dimension

Theorem

Corollary 2

A vector space is finite-dimensional if and only if

it is spanned by afinite set.

Basis and Dimension

Theorem

Corollary 2

A vector space is finite-dimensional if and only if it is spanned

by afinite set.

Basis and Dimension

Theorem

Corollary 2

Basis and Dimension

How to find a basis?

Approach 1.

Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Get a spanning set

for the vector space, then reduce this set to abasis dropping one vector at a time.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Get a spanning set for the vector space,

then reduce this set to abasis dropping one vector at a time.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Get a spanning set for the vector space, then reduce this set

to abasis dropping one vector at a time.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Get a spanning set for the vector space, then reduce this set to abasis

dropping one vector at a time.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Get a spanning set for the vector space, then reduce this set to abasis dropping one vector

at a time.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Theorem

v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk

be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set

for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space

V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V .

If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V . If

v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V . If v0

is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then

v1, v2, ..., vk is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk

is alsoa spanning set for V

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find a

basis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find abasis

for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find abasis for the vector space

V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find abasis for the vector space V

spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),

w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),

w3 = (2, 3, 1) and w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and

w4 = (1, 1, 1)

Basis and Dimension

Approach 1.

Theorem

Example 6.11

Basis and Dimension

Solution

Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

where ri ∈ R are not all equal to zero. Equivalently,

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have

four vector in R3, then they are linearly dependent,and satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have four vector

in R3, then they are linearly dependent,and satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have four vector in R3,

then they are linearly dependent,and satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have four vector in R3, then they are

linearly dependent,and satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have four vector in R3, then they are linearly dependent,and

satisfy an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have four vector in R3, then they are linearly dependent,and satisfy

an equation of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

Since we have four vector in R3, then they are linearly dependent,and satisfy an equation

of the form

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

ri ∈ R are not all equal to zero. Equivalently,

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

where ri ∈ R

are not all equal to zero. Equivalently,

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

where ri ∈ R are not all

equal to zero. Equivalently,

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

where ri ∈ R are not all equal to zero.

Equivalently,

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve

this system of linear equations for r1, r2, r3, r4 we apply rowreduction

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve this system

of linear equations for r1, r2, r3, r4 we apply rowreduction

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve this system of linear equations for

r1, r2, r3, r4 we apply rowreduction

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve this system of linear equations for r1, r2, r3, r4

we apply rowreduction

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve this system of linear equations for r1, r2, r3, r4 we apply

rowreduction

Basis and Dimension

Solution

r1w1 + r2w2 + r3w3 + r4v4 = 0

1 0 2 11 1 3 10 1 1 1

r1r2r3r4

to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction Dr. Marco A Roque Sol Linear Algebra. Session 6

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

→ 1 0 2 1

0 1 1 00 1 1 1

→ 1 0 2 1

0 1 1 00 0 0 1

→ 1 0 2 0

0 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

1 0 2 10 1 1 00 1 1 1

→ 1 0 2 1

0 1 1 00 0 0 1

→ 1 0 2 0

0 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

→ 1 0 2 1

0 1 1 00 1 1 1

1 0 2 10 1 1 00 0 0 1

→ 1 0 2 0

0 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

→ 1 0 2 1

0 1 1 00 1 1 1

→ 1 0 2 1

0 1 1 00 0 0 1

1 0 2 00 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

→ 1 0 2 1

0 1 1 00 1 1 1

→ 1 0 2 1

0 1 1 00 0 0 1

→ 1 0 2 0

0 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

→ 1 0 2 1

0 1 1 00 1 1 1

→ 1 0 2 1

0 1 1 00 0 0 1

→ 1 0 2 0

0 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

1 0 2 11 1 3 10 1 1 1

→ 1 0 2 1

0 1 1 00 1 1 1

→ 1 0 2 1

0 1 1 00 0 0 1

→ 1 0 2 0

0 1 1 00 0 0 1

r1 + 2r3r2 + r3r4 = 0

r1 = −2r3r2 = −r3r4 = 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

and take V = Span(w1,w2,w4)

Let us check whether vectors w1,w3,w4 are linearly independent

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3

Basis and Dimension

the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solution

is (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0),

t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand

a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution

is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0).

We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained

that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒

2w1 + w2 = w3. Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3.

Hence, we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence,

we can dropp w3

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

take V = Span(w1,w2,w4)

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

and take

V = Span(w1,w2,w4)

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Let us check

whether vectors w1,w3,w4 are linearly independent

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Let us check whether vectors

w1,w3,w4 are linearly independent

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Let us check whether vectors w1,w3,w4 are

linearly independent

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ =

1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!.

Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence

{w1,w3,w4} is a basis for V . Besides, itfollows that V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence {w1,w3,w4}

is a basis for V . Besides, itfollows that V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence {w1,w3,w4} is a basis

for V . Besides, itfollows that V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence {w1,w3,w4} is a basis for V .

Besides, itfollows that V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence {w1,w3,w4} is a basis for V . Besides,

itfollows that V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that

V = R3

Basis and Dimension

∣∣∣∣∣∣1 0 11 1 10 1 1

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 10 1 00 1 1

∣∣∣∣∣∣ =

∣∣∣∣ 1 01 1

∣∣∣∣ = 1 6= 0

Basis and Dimension

Approach 2.

Build a maximal linearly independent set adding one vector at atime.

If the vector space V is trivial, it has the empty basis.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Basis and Dimension

Approach 2.

a maximal linearly independent set adding one vector at atime.

Modifications.

Basis and Dimension

Approach 2.

Build a maximal

linearly independent set adding one vector at atime.

Modifications.

Basis and Dimension

Approach 2.

Build a maximal linearly independent set

adding one vector at atime.

Modifications.

Basis and Dimension

Approach 2.

Build a maximal linearly independent set adding one vector

at atime.

Modifications.

Basis and Dimension

Approach 2.

Modifications.

Basis and Dimension

Approach 2.

the vector space V is trivial, it has the empty basis.

Modifications.

Basis and Dimension

Approach 2.

If the vector space

V is trivial, it has the empty basis.

Modifications.

Basis and Dimension

Approach 2.

If the vector space V

is trivial, it has the empty basis.

Modifications.

Basis and Dimension

Approach 2.

If the vector space V is trivial,

it has the empty basis.

Modifications.

Basis and Dimension

Approach 2.

If the vector space V is trivial, it has

the empty basis.

Modifications.

Basis and Dimension

Approach 2.

Modifications.

Basis and Dimension

Approach 2.

V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0,

pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector

v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0,

if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1

spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V,

it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.

Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise

pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector

v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V

that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not

in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of

v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1.

Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. If

v1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2

span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V

they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis.

Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise

pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vector

v3 ∈ V that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V

that is not in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not

in the span of v1, v2. And so on.

Modifications.

Basis and Dimension

Approach 2.

If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2.

And so on.

Modifications.

Basis and Dimension

Approach 2.

Modifications.

Basis and Dimension

Approach 2.

Modifications.

Basis and Dimension

Approach 2.

Modifications.

Instead of

the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set,

we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start

with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any

linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set

(if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one).

If we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If

we are given a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given

a spanningset S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset

S, it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S,

it is enough to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough

to pick new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick

new vectors only in S

Basis and Dimension

Approach 2.

Modifications.

Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors

only in S

Basis and Dimension

Approach 2.

Modifications.

Basis and Dimension

Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Remark.

This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure

works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure works for

finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure works for finite-dimensional

vector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure works for finite-dimensionalvector spaces.

There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure works for finite-dimensionalvector spaces. There is

an analogous procedure forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure

forinfinite-dimensional spaces (transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces

(transfinite induction ).

Example 6.12

Solution

Basis and Dimension

Example 6.12

Solution

Basis and Dimension

Example 6.12

Solution

Basis and Dimension

Example 6.12

Vectors

v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3

Solution

Basis and Dimension

Example 6.12

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are

linearly independent.Extend the set {v1, v2} to a basis for R3

Solution

Basis and Dimension

Example 6.12

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.

Extend the set {v1, v2} to a basis for R3

Solution

Basis and Dimension

Example 6.12

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set

{v1, v2} to a basis for R3

Solution

Basis and Dimension

Example 6.12

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2}

to a basis for R3

Solution

Basis and Dimension

Example 6.12

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis

for R3

Solution

Basis and Dimension

Example 6.12

Solution

Basis and Dimension

Example 6.12

Solution

Basis and Dimension

Example 6.12

Solution

Our task

is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find

a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector

v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3

that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not

a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination

ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2.

Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now,

the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors

v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2,

span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane

x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and

thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1)

does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie

in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane.

Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence,

it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not

alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination

of v1, v2. Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2.

Thus, {v1, v2, v3} is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3}

is a basis for R3.

Basis and Dimension

Example 6.12

Solution

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3

Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors

v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors v1 = (0, 1, 0) and

v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1)

are linearly independent.Extend the set {v1, v2} to a basis for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.

Extend the set {v1, v2} to a basis for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend

the set {v1, v2} to a basis for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2}

to a basis for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis

for R3

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors

e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)

form aspanning set for R3, at least one of them can be chosen as v3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form a

spanning set for R3, at least one of them can be chosen as v3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set

for R3, at least one of them can be chosen as v3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3,

at least one of them can be chosen as v3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them

can be chosen as v3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Let us check

that {v1, v2, e3} and {v1, v2, e2} form a basis for R3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Let us check that {v1, v2, e3} and

{v1, v2, e2} form a basis for R3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Let us check that {v1, v2, e3} and {v1, v2, e2}

form a basis for R3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis

for R3

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ =

1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ =

2 6= 0

Basis and Dimension

Example 6.13

Solution

∣∣∣∣∣∣0 −2 11 0 00 1 0

∣∣∣∣∣∣ = 1 6= 0

∣∣∣∣∣∣0 −2 11 0 00 1 1

∣∣∣∣∣∣ = 2 6= 0

Row space of a matrix

The row space of an m × n matrix A is the subspace of Rn

spanned by the rows of A

The dimension of the row space is called the rank of the matrix A.

Theorem 1

The rank of a matrix A is the maximal number of linearlyindependent rows in A.

Theorem 2

If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.

Theorem 1

Theorem 2

The row space

of an m × n matrix A is the subspace of Rn

Theorem 1

Theorem 2

The row space of an m × n

matrix A is the subspace of Rn

Theorem 1

Theorem 2

The row space of an m × n matrix A

is the subspace of Rn

Theorem 1

Theorem 2

The row space of an m × n matrix A is the subspace

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

The dimension

of the row space is called the rank of the matrix A.

Theorem 1

Theorem 2

The dimension of the row space

is called the rank of the matrix A.

Theorem 1

Theorem 2

The dimension of the row space is called

the rank of the matrix A.

Theorem 1

Theorem 2

The dimension of the row space is called the rank

of the matrix A.

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

The rank

of a matrix A is the maximal number of linearlyindependent rows in A.

Theorem 2

Theorem 1

The rank of a matrix A

is the maximal number of linearlyindependent rows in A.

Theorem 2

Theorem 1

The rank of a matrix A is the maximal

number of linearlyindependent rows in A.

Theorem 2

Theorem 1

The rank of a matrix A is the maximal number

of linearlyindependent rows in A.

Theorem 2

Theorem 1

The rank of a matrix A is the maximal number of linearlyindependent

rows in A.

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

If a matrix A

is in row echelon form, then the nonzero rows of Aare linearly independent.

Theorem 1

Theorem 2

If a matrix A is in

row echelon form, then the nonzero rows of Aare linearly independent.

Theorem 1

Theorem 2

If a matrix A is in row echelon form,

then the nonzero rows of Aare linearly independent.

Theorem 1

Theorem 2

If a matrix A is in row echelon form, then the nonzero rows

of Aare linearly independent.

Theorem 1

Theorem 2

If a matrix A is in row echelon form, then the nonzero rows of A

Theorem 1

Theorem 2

Corollary

The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.

Theorem 3

Elementary row operations do not change the row space of amatrix.

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Corollary

The rank

of a matrix is equal to the number of nonzero rows in itsrow echelon form.

Theorem 3

Corollary

The rank of a matrix

is equal to the number of nonzero rows in itsrow echelon form.

Theorem 3

Corollary

The rank of a matrix is equal

to the number of nonzero rows in itsrow echelon form.

Theorem 3

Corollary

The rank of a matrix is equal to the number

of nonzero rows in itsrow echelon form.

Theorem 3

Corollary

The rank of a matrix is equal to the number of nonzero rows

in itsrow echelon form.

Theorem 3

Corollary

The rank of a matrix is equal to the number of nonzero rows in itsrow

echelon form.

Theorem 3

Corollary

Theorem 3

Corollary

Theorem 3

Corollary

Theorem 3

Elementary

row operations do not change the row space of amatrix.

Corollary

Theorem 3

Elementary row operations

do not change the row space of amatrix.

Corollary

Theorem 3

Elementary row operations do not change

the row space of amatrix.

Corollary

Theorem 3

Elementary row operations do not change the row space

of amatrix.

Corollary

Theorem 3

Corollary

Theorem 3

Corollary

Theorem 3

Suppose

that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B

are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices

such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B

is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtained

from A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A

by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary

row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation.

Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am

be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rows

of A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and

b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm,

be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B.

We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have

to show thatSpan(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show that

Span(a1, ..., am) = Span(b1, ...,bm)

Corollary

Theorem 3

Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) =

Span(b1, ...,bm)

Corollary

Theorem 3

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)

However,

we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe

that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row

bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi

of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B

belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am).

Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either

bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m

(interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or

bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai

for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar

r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0

(multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), or

bi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj

for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some

i 6= j and r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and

r ∈ R ( adding to a row, amultiple of another row )

However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R

( adding to a row, amultiple of another row )

It follows that

Span(b1, ...,bm) ⊂ Span(a1, ..., am)

It follows that Span(b1, ...,bm) ⊂

Span(a1, ..., am)

Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

Find the rank of the matrix

1 1 00 1 12 3 11 1 1

the matrix A can also be obtained from B by an elementaryrow operation. By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Now, the matrix A

can also be obtained from B by an elementaryrow operation. By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Now, the matrix A can also

be obtained from B by an elementaryrow operation. By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Now, the matrix A can also be obtained

from B by an elementaryrow operation. By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Now, the matrix A can also be obtained from B by an elementary

row operation. By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Now, the matrix A can also be obtained from B by an elementaryrow operation.

By the above,

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂

Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

the rank of the matrix

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

Find the rank

of the matrix

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Span(a1, ..., am) ⊂ Span(b1, ...,bm) �

Example 6.14

1 1 00 1 12 3 11 1 1

Solution

Elementary row operations do not change the row space. Let usconvert A to row echelon form:

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

⇒Thus, the rank of A is 3.

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

Elementary

row operations do not change the row space. Let usconvert A to row echelon form:

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

do not change the row space. Let usconvert A to row echelon form:

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

the row space. Let usconvert A to row echelon form:

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

Elementary row operations do not change the row space.

Let usconvert A to row echelon form:

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

Elementary row operations do not change the row space. Let us

convert A to row echelon form:1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

Elementary row operations do not change the row space. Let usconvert A

to row echelon form:1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Thus, the rank of A is 3.

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Thus, the rank of A is 3.

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

⇒Thus,

the rank of A is 3.

Solution

1 1 00 1 12 3 11 1 1

1 1 00 1 10 1 11 1 1

1 1 00 1 10 1 10 0 1

1 1 00 1 10 0 00 0 1

1 1 00 1 10 0 10 0 0

Column space of a matrix

Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A

Theorem 1

The column space of a matrix A coincides with the row space ofthe transpose matrix AT

Theorem 2

Elementary row operations do not change linear relations betweencolumns of a matrix.

Theorem 1

Theorem 2

Definition.The column space

of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A

Theorem 1

Theorem 2

Definition.The column space of an m × n

matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A

Theorem 1

Theorem 2

Definition.The column space of an m × n matrix A

is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A

Theorem 1

Theorem 2

Definition.The column space of an m × n matrix A is the subspace

of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A

Theorem 1

Theorem 2

Definition.The column space of an m × n matrix A is the subspace of Rm,

spanned by columns of A is the subspace of Rm spanned bycolumns of A

Theorem 1

Theorem 2

Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A

is the subspace of Rm spanned bycolumns of A

Theorem 1

Theorem 2

Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm

spanned bycolumns of A

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

The column space

of a matrix A coincides with the row space ofthe transpose matrix AT

Theorem 2

Theorem 1

The column space of a matrix A

coincides with the row space ofthe transpose matrix AT

Theorem 2

Theorem 1

The column space of a matrix A coincides with

the row space ofthe transpose matrix AT

Theorem 2

Theorem 1

The column space of a matrix A coincides with the row space

ofthe transpose matrix AT

Theorem 2

Theorem 1

The column space of a matrix A coincides with the row space ofthe

transpose matrix AT

Theorem 2

Theorem 1

The column space of a matrix A coincides with the row space ofthe transpose matrix

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

Theorem 1

Theorem 2

do not change linear relations betweencolumns of a matrix.

Theorem 1

Theorem 2

linear relations betweencolumns of a matrix.

Theorem 1

Theorem 2

Elementary row operations do not change linear relations

betweencolumns of a matrix.

Theorem 1

Theorem 2

Elementary row operations do not change linear relations betweencolumns

of a matrix.

Theorem 1

Theorem 2

Theorem 5

Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).

Theorem 4

If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.

Corollary

For any matrix, the row space and the column space have thesame dimension.

Theorem 5

Theorem 4

Corollary

Theorem 5

do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).

Theorem 4

Corollary

Theorem 5

the dimension of thecolumn space of a matrix (however they can change the columnspace).

Theorem 4

Corollary

Theorem 5

Elementary row operations do not change the dimension

of thecolumn space of a matrix (however they can change the columnspace).

Theorem 4

Corollary

Theorem 5

Elementary row operations do not change the dimension of thecolumn space

of a matrix (however they can change the columnspace).

Theorem 4

Corollary

Theorem 5

Elementary row operations do not change the dimension of thecolumn space of a matrix

(however they can change the columnspace).

Theorem 4

Corollary

Theorem 5

Elementary row operations do not change the dimension of thecolumn space of a matrix (however

they can change the columnspace).

Theorem 4

Corollary

Theorem 5

Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change

the columnspace).

Theorem 4

Corollary

Theorem 5

Theorem 4

Corollary

Theorem 5

Theorem 4

Corollary

Theorem 5

Theorem 4

If a matrix

is in row echelon form,then the columns with leadingentries form a basis for the column space.

Corollary

Theorem 5

Theorem 4

If a matrix is in row echelon form,

then the columns with leadingentries form a basis for the column space.

Corollary

Theorem 5

Theorem 4

If a matrix is in row echelon form,then the columns

with leadingentries form a basis for the column space.

Corollary

Theorem 5

Theorem 4

If a matrix is in row echelon form,then the columns with leadingentries

form a basis for the column space.

Corollary

Theorem 5

Theorem 4

If a matrix is in row echelon form,then the columns with leadingentries form a basis

for the column space.

Corollary

Theorem 5

Theorem 4

Corollary

Theorem 5

Theorem 4

Corollary

Theorem 5

Theorem 4

Corollary

For any matrix,

the row space and the column space have thesame dimension.

Theorem 5

Theorem 4

Corollary

For any matrix, the row space and

the column space have thesame dimension.

Theorem 5

Theorem 4

Corollary

For any matrix, the row space and the column space

have thesame dimension.

Theorem 5

Theorem 4

Corollary

Example 6.15

Find a basis for the column space of the matrix

1 1 00 1 12 3 11 1 1

Solution

The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

a basis for the column space of the matrix

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

Find a basis

for the column space of the matrix

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

Find a basis for the column space

of the matrix

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space

of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space of A

coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space of A coincides

with the row space of AT . Tofind a basis, we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space of A coincides with the row space

of AT . Tofind a basis, we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space of A coincides with the row space of AT .

Tofind a basis, we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space of A coincides with the row space of AT . Tofind a basis,

we convert AT to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

The column space of A coincides with the row space of AT . Tofind a basis, we convert AT

to row echelon form:

Example 6.15

1 1 00 1 12 3 11 1 1

Solution

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.

Nullspace of a matrix

Let A = (aij) be an m × n

Definition

The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

1 0 2 10 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

1 0 2 00 1 1 00 0 0 1

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

⇒Vectors

(1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1)

form a basis for thecolumn space of A.

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis

for thecolumn space of A.

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Let A = (aij)

be an m × n

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace

of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A,

denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A, denoted by N(A)

is the set of alln− dimensional column vectors x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A, denoted by N(A) is the set of all

n− dimensional column vectors x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional

column vectors x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors

x such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x

such that Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that

Ax = 0

1 0 2 11 1 3 10 1 1 1

⇒ 1 0 2 1

0 1 1 00 1 1 1

⇒ 1 0 2 0

0 1 1 00 0 0 1

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)

Theorem

N(A) is a subspace of the vector space of Rn

Definition

The dimension of the nullspace N(A) is called the nullity of thematrix A

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

OBS The nullspace N(A)

is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

OBS The nullspace N(A) is the solution set

of a system of linearhomogeneous equations (with A as the coefficient matrix)

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

OBS The nullspace N(A) is the solution set of a system of linear

homogeneous equations (with A as the coefficient matrix)

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations

(with A as the coefficient matrix)

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

is a subspace of the vector space of Rn

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

N(A) is a subspace

of the vector space of Rn

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

N(A) is a subspace of the vector space

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

The dimension

of the nullspace N(A) is called the nullity of thematrix A

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

The dimension of the nullspace N(A)

is called the nullity of thematrix A

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

The dimension of the nullspace N(A) is called the nullity

of thematrix A

a11 a12 · · · a1na21 a22 · · · a2n

...am1 am2 · · · amn

x1x2...xn

00...0

Theorem

Definition

Rank + Nullity

TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.

rank(A) + N(A) = n

Rank + Nullity

rank(A) + N(A) = n

Rank + Nullity

TheoremThe rank

of a matrix A plus the nullity of A equals the number ofcolumns in A.

rank(A) + N(A) = n

Rank + Nullity

TheoremThe rank of a matrix A

plus the nullity of A equals the number ofcolumns in A.

rank(A) + N(A) = n

Rank + Nullity

TheoremThe rank of a matrix A plus

the nullity of A equals the number ofcolumns in A.

rank(A) + N(A) = n

Rank + Nullity

TheoremThe rank of a matrix A plus the nullity of A

equals the number ofcolumns in A.

rank(A) + N(A) = n

Rank + Nullity

TheoremThe rank of a matrix A plus the nullity of A equals

the number ofcolumns in A.

rank(A) + N(A) = n

Rank + Nullity

TheoremThe rank of a matrix A plus the nullity of A equals the number of

columns in A.

rank(A) + N(A) = n

Rank + Nullity

rank(A) + N(A) = n

Rank + Nullity

rank(A) + N(A) = n

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Find the rank and the nullity of the matrix B.

Find a basis for the row space of B, then extend this basis toa basis for R4.

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Find the rank and

the nullity of the matrix B.

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Find a basis for the row space of B,

then extend this basis toa basis for R4.

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Find a basis for the row space of B, then extend this basis toa basis

for R4.

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Example 6.16

Let B given by

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

⇒Interchange the 1st row with the 2nd row:

Solution

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank

(= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and

the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity

(=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity (=dimension of the nullspace)

of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are

preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations.

We apply such operations toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations

toconvert the matrix B into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B

into its row echelon form

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Solution

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Interchange the 1st row with the 2nd row:

Solution

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

Interchange the 1st row with the 2nd row:

Solution

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

⇒Add 3 times the 1st row to the 3rd row,

then subtract 2 times the1st row from the 4th row

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1

1 1 2 −10 −1 4 10 3 5 −32 −1 0 1

1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3

Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3

⇒Add the 4th row to the 3rd row:

1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

⇒Add 3 times the 2nd row to the 4th row:

Multiply the 2nd row by−1

1 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3

1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

Add the 4th row to the 3rd row:1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

⇒Add the 4th row

to the 3rd row:1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

Add 3 times the 2nd row to the 4th row:

1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

Add 3 times the 2nd row to the 4th row:

1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

⇒Add 3 times the 2nd row

to the 4th row:

1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

⇒Add 16 times the 3rd row to the 4th row:

1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

Add 16 times the 3rd row to the 4th row:1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

Add 16 times the 3rd row to the 4th row:1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

⇒Add 16 times the 3rd row

to the 4th row:1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0

1 1 2 −10 −1 −4 −10 0 1 00 0 0 0

Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since

(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,

it follows that the nullity of B equals 1.

Now that the matrix

is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since

Now that the matrix is in row echelon form,

its rank equals thenumber of nonzero rows, which is 3. Since

Now that the matrix is in row echelon form, its rank equals

thenumber of nonzero rows, which is 3. Since

Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows,

which is 3. Since

Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3.

(rank of B ) +

(nullity of B ) = (the number of columns of B ) =4,

(rank of B ) + (nullity of B ) =

(the number of columns of B ) =4,

(rank of B ) + (nullity of B ) = (the number of columns of B ) =

it follows that

the nullity of B equals 1.

it follows that the nullity of B

equals 1.

The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.

The row space of a matrix

is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

The row space of a matrix is invariant under elementary rowoperations.

Therefore the row space of the matrix B is thesame as the row space of its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

The row space of a matrix is invariant under elementary rowoperations. Therefore the row space

of the matrix B is thesame as the row space of its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B

is thesame as the row space of its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as

the row space of its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of

its row echelon form:

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

1 1 2 −10 1 −4 −10 0 1 00 0 0 0

v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)

It is known that at least one of the vectors

e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4

In particular, the vectors v1, v2, v3, e4 form a basis for R4.

e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1),

can be chosen as v4

In particular, the vectors

v1, v2, v3, e4 form a basis for R4.

In particular, the vectors v1, v2, v3, e4

form a basis for R4.

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