Linear Algebra. Session 6 - math.tamu.eduroquesol/Math_304_Spring_2018_Sessi… · Session 6....

Vector Spaces

Linear Algebra. Session 6

Dr. Marco A Roque Sol

08/01/2017

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.



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

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. �




General results on linear independence in Rn

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

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

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 )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

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 6.2

Let

A =

(−1 1−1 0

).

Determine whether matrices A,A2,A3 are linearly independent

Solution

We have

A =

(−1 1−1 0

)A2 =

(0 −11 −1

)A3 =

(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



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




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

, v =

abc

det(A(x)) =

ex e2x e3x

ex 2e2x 3e3x

ex 4e2x 9e3x

= ex

∣∣∣∣∣∣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

3 8

∣∣∣∣ = 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.




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].



Basis and Dimension

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

OBS

“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

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

Example 6.5Standard Basis for M2,2

E11 =

(1 00 0

),E12 =

(0 10 0

),E21 =

(0 01 0

),E22 =

(0 00 1

)Solution

Indeed,

A =

(a bc d

)= aE12 + bE12 + cE21 + dE22



Basis and Dimension

Example 6.6Standard Basis for Polinomial Pn

a0, a1x , a2x2, ..., an−1x

n−1Solution

POC ... !!!!

Example 6.7Standard Basis for Polinomials P

a0, a1x , a2x2, ..., anx

n, ...SolutionPOC ... !!!!



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

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� ∗ ∗ ∗

� ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗� ∗ ∗

� ∗

Spanning/Lin. Indep. No spanning/Lin. Indep.

� ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗

� � ∗ ∗� ∗

�

� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗

� ∗ ∗ ∗ ∗� � ∗ ∗ ∗

� ∗ ∗� �

Spanning/ No Lin. Indep. No spanning/ No Lin. Indep.



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.



Vector Spaces

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.



Vector Spaces

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.



Vector Spaces

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.



Vector Spaces

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



Vector Spaces

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

Basis

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

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

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 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

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

=

0000

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



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 solution is(r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ R and 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

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

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

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

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



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

Elementary row operations do not change the row space of amatrix.




Theorem 3

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

Corollary

The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.

Theorem 4

Elementary row operations do not change the row space of amatrix.




proofSuppose 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)

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 sclar 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)




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




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.




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 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.




Example 6.15

Find a basis for the column space of the matrix

A =

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:




AT =

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




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




Rank + Nullity

TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.

rank(A) + N(A) = n




Example 6.16

Let B given by

B =

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.




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

B =

0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1

⇒Interchange the 1st row with the 2nd row:




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

⇒




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:




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

⇒




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.




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:

B =

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.




It is known that at least one of the vectorse1 = (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


Linear Algebra. Session 6 - math.tamu.eduroquesol/Math_304_Spring_2018_Sessi… · Session 6....

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