5-General Vector Sp

8/8/2019 5-General Vector Sp

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Chapter 5 : 1 /21

Chapter 5

GENERAL VECTOR SPACES

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Contents

5.1 Definition of Vector Space ....................................................................................................... 25.2 Subspaces .................................................................................................................................. 25.3 Linear Independence ................................................................................................................. 65.4 Basis and Dimension................................................................................................................. 95. 5 Row Space, Column Space and Nullspace ............................................................................ 145.6 Rank and Nullity ..................................................................................................................... 18


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Chapter 5 : 2 /21

5.1 Definition of Vector Space

Let V be an arbitrary nonempty set on which two operations of vector addition and scalar

multiplication are defined. If the following axiomsare satisfied by all objects u, v, w in V and

all scalars k and l, then V is called a vector space.

1. Ifu and v are objects in V, then u + v is in V2. u + v = v + u3. u + ( v + w ) = (u + v ) + w4. There is an object 0 in V , called a zero vector for V , such that 0 + u = u + 0 = u for all

u in V

5. For each u in V, there isu in V, called a negative of u, such that u + (-u ) = (-u ) + u =0.

6. If k is any scalar and u is any object in V, then ku is in V.7. k ( u + v ) = ku + kv8. (k + l) u = ku + l u9.

K (l u ) = (k l) u10. 1u = u

5.2 Subspaces

Example 1

The subspaces W of V = 2 are(a) The origin {0}(b) The lines passing through the origin { }(c) The Planes passing through the origin {2}

Definition

A subset W of a vector space V is called a subspace of V if W is itself a vector space under theaddition and scalar multiplication defined on V.

Theorem

If W is a set of one or more vectors from a vector space V, then W is a subspace of V if andonly if the following conditions hold:

(1) Ifu and v are vectors in W , then u + v is in W(Closed under addition)

(2) If k is any scalar and u is any vector in W, then ku is in W.(Closed under scalar multiplication)


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Example 2

The subspaces W of V = 3 are

(a) The origin {0}

(b) The lines passing through the origin { }

(c) The Planes passing through the origin {2}

(d) 3

Example 3

Determine which of the following are subspaces:

(a) of 3: all vectors of the form (a, b, 1).(b) of P3: all polynomials 332210 xaxaxaa +++ for which 0 1 2 3 0a a a a+ + + = (c) of Mnn : all n x n matrices A such that AT = - A.

Solution

(a)

1 2

2

{( , ,1) }

( , ,1)

( , ,1)

Let W a b

u Wsuchthat u u u

v Wsuch that v v v

=

=

=

Condition I:

1 1 2 2( , ,1 1)u v u v u v

u v W

+ = + + +

+ Hence Wis not a subspace ofR 3 .

(b)

(a)

2 3

0 1 2 3 0 1 2 3 0 1 2 3

2 3

0 1 2 3 0 1 2 3

2 3

0 1 2 3 0 1 2 3

{ : , , , , 0}

0,

0,

Let W a a x a x a x a a a a a a a a

u u u x u x u x such that u u u u

v v v x v x v x such that v v v v

= + + + + + + =

= + + + + + + =

= + + + + + + =

Condition 1:

(b)

2 3 2 3

0 1 2 3 0 1 2 3

0 1 2 3 0 1 2 3

( ) ( )

( ) ( ) 0 0 0

u v u u x u x u x v v x v x v x

u u u u v v v v

u v W

+ = + + + + + + +

+ + + + + + + = + =

+


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Chapter 5 : 4 /21

Condition 2:2 3 2 3

0 1 2 3 0 1 2 3

0 1 2 3 0 1 2 3

, ( )

( ) ( ) (0) 0

c cu c u u x u x u x cu cu x cu x cu x

cu cu cu cu c u u u u c

cu W

= + + + = + + +

+ + + = + + + = =

Hence Wis a subspace of P3

(c)

(d)

{ : }

,

,

T

nxn

T

T

Let W A M A A

A W suchthat A A

B W suchthat B B

= =

=

=

Condition 1:

1.

( ) ( ) ( )T T TA B A B A B A B

B W+ = + = + = +

+

Condition 2:

, ( ) ( ) ( ) ( )T Tc cA c A c A cA

cu W

= = =

Hence Wis a subspace of M nxn

Linear Combinations of Vectors

TheoremIf AX = 0 is a homogeneous linear system of m equations in n unknown, then the set of

solution vectors is a subspace ofn (Solution Space).

Definition

A vectorw is called a linear combination of the vectorv1, v2, , vr if it can be expressed

in the form

w = k1v1 + k2v2 + + krvrwhere k1, k2, , krare scalars.


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Chapter 5 : 5 /21

Example 4

Every vector v = (a, b, c) in 3 is expressible as a linear combination of the standard basis

vectorsi, j, and k.v = (a,b,c) = a(1,0,0) + b(0,1,0) + c(0,0,1)

= ai+ bj+ kj.

Example 5

(a) Determine whether (11, 1,6) is a linear combination of ( 2,1, 3)u = and ( 5,1, 4)v = .

Solution

Let (11, 1,6) 1( 2,1, 3)k= 2( 5,1, 4)k+

1 2

1 2

1 2

2 5 11

1

3 4 6

k k

k k

k k

=

+ =

=

2 5 11

1 1 1

3 4 6

1 1 1

2 5 11

3 4 6

1 1 1

0 3 9

0 1 3

1 1 1

0 3 9

0 0 0

1 1 1

0 1 3

0 0 0

1 0 2

0 1 3

0 0 0

1 22, 3k k= = The values 1k and 2k also satisfy the third equation.

Therefore (11, 1,6) 2( 2,1, 3)= 3( 5,1, 4)

Spanning

TheoremIfv1, v2 , , vr are vectors in a vector space V, then

(a) The set W of all linear combinations ofv1, v2 , , vr is a subspace of V.(b) W is the smallest subspace of V that contains v1, v2 , , vr in the sense that every

other subspace of V that contains v1, v2, , vr must contain W.

DefinitionIf S = { v1, v2, , vr} is a set of vectors in a vector space V, then the subspace W of V

consisting of all linear combinations of the vectors in S is called the space spanned by v1,

v2, ..,vr and we say that the vectors v1, v2, ..,vr spanW.

To indicate that W is the space spanned by the vectors in the set S ={ v1, v2, , vr }, we

writeW = span (S) or W = span { v1, v2, , vr }


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Example 6

Determine whether2 31

{ , , }S v v v= spans 3, where

1(1,6,4)v = ,

2(2,4, 1)v = , and

3( 1,2,5)v = .

SolutionChoose any arbitrary vector in

2 31( , , )b b b b= in 3 to represent all the linear combinations of the

vectors in S such that

2 31( , , )b b b 1(1,6,4)k= 2 (2,4, 1)k+ 3( 1,2,5)k+

or

1 2 3 1

1 2 3 2

1 2 3 3

2

6 4 2

4 5

k k k b

k k k b

k k k b

+ =

+ + =

+ =

The problem is to determine whether the system AX = b is consistent for all valuesof

2 31, ,b b and b . Based on the theorem of Equivalent Statement, since A is a square matrix, then

AX = b is consistent if and only if

1 2 1

det( ) 6 4 2

4 1 5

A

=

is nonzero. However, 0)det( =A (verify). Thus, Sdoes not span 3.

Remark1. IfA is NOT a square matrix, then Gaussian elimination method is used. The system

AX = b is consistent if there is no zeros row in the row echelon form.

2. Spanning sets are not unique.

5.3 Linear Independence

Theorem

If S = {v1, v2, , vr} and S = { w1, w2, , wk} are two sets of vectors in a vector space V,then span {v1, v2, , vr} = span { w1, w2, , wk}if and only if each vector in S is a linear combination of those in S and each vector in S is a

linear combination of those in S.

Definition

If S = { v1 , v2 , . . . , vr} is a nonempty set of vectors then the vector equationk1v1+ k2v2+ . . . + krvr = 0

has at least one solution namely

k1= 0, k2= 0, . . ., kr= 0

If this is the only solution, then S is called a linearly independent set. If there are othersolutions, then S is called a linearly dependentset.


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Chapter 5 : 7 /21

Example 7

Consider the vectors i= (1,0,0), j= (0,1,0) and k= (0,0,1) in 3.

The vector equation k1i + k2j + k3k= 0 becomes k1(1,0,0)+ k2(0,1,0) + k3(0,0,1) = (0,0,0)or, equivalently, (k1, k2, k3) = (0 ,0, 0) .

This implies that k1 = k2 = k3 = 0 (trivial solution). Thus, S = {i, j, k} is linearly independent.

Example 8Determine whether the vectors are linearly independent or linearly dependent.

v1 = (-1, 1, 1, 2), v2 = (0, 3, 1, 1) and v3 = (-2, 4, 2, 1)

Solution

Let1 2 3( 1,1,1, 2) (0,3,1,1) ( 2, 4, 2,1) (0, 0, 0, 0)k k k + + = .

02 31 = kk

043 321 =++ kkk

02 321 =++ kkk

02 321 =++ kkk

Solving the system, we obtained 0321 === kkk , i.e. the system has only the trivial solutionand this implies that the set {( 1,1,1, 2), (0,3,1,1),( 2,4,2,1)}S= is linearly independent.

Example 9Ifv1 = (2, -1, 0, 3), v2 = (1, 2, 5, -1 ) and v3 = ( 7, -1, 5,8 ) , then the set S = { v1, v2, v3 } is

linearly dependentsince3v1 + v2 v3 = 0. In this example each vector is expressible as a linear

combination of the other two, say, v3 = 3v1 + v2.

Theorem

A set S with two or more vectors is

(a)Linearly dependent if and only if at least one of the vectors in S is expressible as alinearly combination of the other vectors in S.(b)Linearly independent if and only if no vectors in S is expressible as a linearly

combination of the other vectors in S.

Theorem(a) A finite set of vectors that contains the zero vector is linearly dependent.(b) A set with exactly two vectors is linearly independent if and only if neither vector

is a scalar multiple of the other.


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Example 10

Explain why Sis linearly dependent for3.

(a) { }(1, 3, 4), (0, 0, 0), ( 3, 4, 5)S=

(b) )4,3,2(),8,6,4( =S

Solution

(a) Vector Equation becomes 0(1, 3, 4) (0, 0, 0) 0( 3, 4, 5) 0k + + = This implies that kis are not all zeros.

(b) Scalar multiple of one another.

Geometric Interpretation of Linear Independence

Linear independence has some useful geometric interpretations in 2 and 3:

1. In 2 or3,a set of 2 vectors is linearly independent if and only if the vectors do not lieon the same line when they are placed with their initial points at the origin.2. In 3, a set of 3 vectors is linearly independent if and only if the vectors do not lie in the

same plane when they are placed with their initial points at the origin.

Example 11

Determine whether the set { }xxxS 5,4,2,1 2 =in P2 is linearly dependent.

Solution

Since the set S in P2 has more than three vectors, S is linearly dependent.

Remark

A homogeneous system of linear equations with more unknowns, r than equations, n has

infinitely many solutions.

Nontrivial solutions imply linear dependence.

(A linearly independent set in n can contain at most n vectors.)

Theorem

Let S = { v1 , v2 , . . . , vr} be a set of vectors in Rn. If r > n, then S is linearly dependent.


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Chapter 5 : 9 /21

5.4 Basis and Dimension

Example 12Show that the following set of vectors is a basis for M22.

3 6

3 6

0 1

1 0

0 8

12 4

1 0

1 2

Solution

Let k13 6

3 6

+ k20 1

1 0

+ k30 8

12 4

+ k41 0

1 2

=0 0

0 0

Or considerAX = 0 as given by

1 43 0k k+ =

1 2 36 8 0k k k =

1 2 3 43 12 0k k k k = or equivalently,

1 3 46 4 2 0k k k + =

3 0 0 1

6 1 8 0det( ) det 0

3 1 12 1

6 0 4 2

A

=

(Verify). Based on the theorem of Equivalent

Statements, this implies thatA is invertible andAX = 0 has only trivial solutions. Thus, the set ofvectors is linearly independent.

Then, choose, any arbitrary vector in 2 3 41( , , , )b b b b b= in Mnn to represent all the linearcombinations of the vectors such that

Let k13 6

3 6

+ k20 1

1 0

+ k30 8

12 4

+ k41 0

1 2

=1 2

3 4

b b

b b

Definition

If V is any vector space and S = { v1 , v2 , . . . , vn} is a set of vectors in V, then S is called a

basis for V if the following two conditions hold:

(a) S is linearly independent(b) S spans V


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Chapter 5 : 10 /21

Similarly, det( ) 0A which implies thatAX = b is consistent for all values ofb. Thus, the set ofvectors span Mnn.

Hence, the set of vectors is a basis for Mnn.

Coordinates Relative to a Basis

If S = {v1, v2, , vn} is a basis for a vector space V and

v = c1v1 + c2v2 + + cnvn

is the expression forv in terms of S, then the scalars c1, c2, , cn are called the coordinates ofv

relative to the basis S.

And, the vector (c1, c2, ,cn) in n

is called the coodinate vector ofv relative to S and denoted by (v)s= (c1, c2, ,cn).

Example 13

Find the coordinate vector ofwrelative to S = {u1, u2} for2.

(a)1 2

(3, 7), (1,0), (0,1)w u u= = =

(b) 1 2(1,1), (2, 4), (3,8)w u u= = =

Solution

(a)1 2

3 7w u u=

( )(3, 7)

sw =

(b)1 1 2 2

w k u k u= + yields

1 22 3 1k k+ =

1 24 8 1k k + =

( )

5 3( , )28 14

sw =

Theorem - Uniqueness of Basis Representation

If S = {v1, v2, , vn} is a basis for a vector space V, then every vector v in V can be expressed inthe form v = c1v1 + c2v2 + + cnvn in exactly one way.


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Chapter 5 : 11 /21

Example 14: Standard Basis forn

From the previous section, it was shown that if e1= (1,0,,0), e2= (0,1,..,0) , en =

(0,0,..,1), then S = {e1, e2,,en} is linearly independent set in n

. Moreover, this set S also

spans n since any vectorv = (v1, v2, ,vn) in n

can be written as

v = v1e1 + v2e2++ vnen.Thus S is a basis forn and it is called the standard basis for n .

Remark

From v = v1e1 + v2e2++ vn en, the coordinates ofv = (v1, v2, ,vn) relative to the standard

basis are v1, v2, ,vn , so (v)s= (v1, v2, ,vn) = v.

Example 14

The vector spaces n, Pn and Mmn are finite-dimensional because they have standard bases.

Example 15

Determine whether the following S is a basis.

(a) 1 2{ , , }S o v v= for3

(b)

1 2{ , }S v v= for 2 where

1 22v v=

Solution

(a) Even though r = n, S is not a basis since zero vector exists in the set S.(b) Even though r = n, S is not a basis since a vector is a scalar multiple of the other.

Definition

A nonzero vector space V is called finite-dimensional if it contains a finite set of vectors

{v1 , v2 , . . . , vn } that forms a basis. If no such set exists, V is called infinite-dimensional .

Theorem

Let V be a finite-dimensional vector space, and let {v1, v2, . . . , vn} be any basis.(a) If a set has more than n vectors, then it is linearly dependent

(b) If a set has fewer than n vectors, then it does not span V

TheoremAll bases for a finite-dimensional vector space have the same number of vectors.

(This implies that if S is a basis, then r = n)


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Chapter 5 : 12 /21

Example 16dim (R

n) = n dim (Pn) = n+1 dim(Mmn) = mn

Their standard bases are {e1, e2,, en},{1, x,,xn}, and {M1, M2,,Mn, ,Mmn} respectively.

Example 17

Determine a basis for and the dimension of the solution space of the homogeneous system

1 2 33 0x x x + =

1 2 32 6 2 0x x x + =

1 2 33 9 3 0x x x + =

Solution

1 3 1 0

2 6 2 0

3 9 3 0

...

1 3 1 0

0 0 0 0

0 0 0 0

(verify)

Thus, the general solution of the given system is1 2 3

3 0x x x + = , 2 3,x r x s= = , 1 3x r s=

Therefore, the solution vectors can be written as

1

2

3

x

x

x

=

3r s

r

s

=

3

1

0

r

+

1

0

1

s

which shows that the vectors,1

3

1

0

v

=

and2

1

0

1

v

=

span the solution space. Since they are

also linearly independent,1 2

{ , }v v is a basis, and the solution space is two-dimensional.

Definition

The dimension of a finite-dimensional vector space V, denoted by dim (V) is defined to bethe number of vectors in a basis for V.

TheoremIf V is an n-dimensional vector space, and if S is a set in V with exactly n vectors, then S is a

basis for V if either S spans V or S is linearly independent.


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Chapter 5 : 13 /21

Example 18

Determine whether the following polynomials form a basis for 2P (the set of all

polynomials of degree 2 ).

2

1 84)( xxxp += ,2

2 2)( xxxp += and xxp 71)(3 = .

Solution

r = 3 vectors and n = dim (P2) = 3 . Since r = n, we form only Ax=0.

=

028

714

100

A . det (A) = 0.

Based on the Theorem of Equivalent Statements, det(A) = 0 implies that Ax = 0 has nontrivial

solutions. Therefore the polynomials are linearly dependent. Thus the polynomials do not form abasis for P2.

Example 19

Find the dimension of a subspace {( , , ) : , }W d c d c c d = .

Solution

d

c d

c

=

0

1

1

c

+

1

1

0

d

.

W is spanned by {(0,1,1),(1, 1,0)}S= and S is also linearly independent. Thus, dim (W) = 2since S forms a basis for W.

Theorem

If W is a subspace of a finite-dimensional vector space V, then dim (W) dim (V);Moreover, if dim (W) = dim (V), then W = V.


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Chapter 5 : 14 /21

5. 5 Row Space, Column Space and Nullspace

DefinitionFor an m x n matrix

11 12 1

21 22 2

1 2

...

...

. . .

...

n

n

m m mn

a a a

a a aA

a a a

=

the vectors

[ ]

[ ]

[ ]

1 11 12 1

2 21 22 2

1 2

.. .

.. .

. . .

.. .

n

n

m m m mn

r a a a

r a a a

r a a a

=

=

=

in Rn

formed from the rows of A are called the row vectors of A, and the vectors

c1 =

11

21

1

:

:

n

a

a

a

, c2 =

12

22

2

:

:

n

a

a

a

, .. , cn =

1

2

:

:

n

n

nn

a

a

a

in Rm

formed the columns of A are called the

column vectors of A.

DefinitionIfA is an m x n matrix, then the subspace of R

nspanned by the row vectors of A is called the

row space of A , and the subspace of Rm

spanned by the column vectors of A is called the

column space of A. The solution space of the homogeneous system of equations Ax = 0,which is a subspace of R

n, is called the nullspace of A.

Theorem

A system of linear equationsAx = b is consistent if and only ifb is in the column space ofA.


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Chapter 5 : 15 /21

Example 20

LetAx = b be the linear system

1

2

3

1 2 3 1

2 9 3 10

1 0 4 2

x

x

x

=

Express b as a linear combination of the column vectors of A.

Solution

Solving the system by Gaussian elimination yieldsx1 =2, x2 = 1, x3 = -1. It follows that

1 2 3 1

2 2 1 9 1 3 10

1 0 4 2

+ = .

Thus, the system is consistent since b is in the column space ofA.

General and Particular Solutions

The vectorxo is called a particular solution ofAx = b.

The expression xo + c1v1+ c2v2+ . . . + ckvkis called the general solution ofAx = b.

The expression c1v1+ c2v2+ . . . + ckvkis called the general solution ofAx = 0.

Example 21

Find the vector form of the general solution of the given system Ax = b; then use that result tofind the vector form of the general solution ofAx = 0.

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

2 2 1

2 4 2 4 2

2 2 1

3 6 3 6 3

x x x x

x x x x

x x x x

x x x x

+ + =

+ + =

+ =

+ + =

TheoremIfxo denotes any single solution of a consistent linear system Ax = b, and ifv1 , v2 , . . . , vk

form a basis for the nullspace ofA then every solution ofAx = b can be expressed in the form

x = xo + c1v1+ c2v2+ . . . + ckvkfor all choices of scalars c1 , c2 , . . . , ck, the vectorx in this formula is a solution ofAx = b.


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Chapter 5 : 16 /21

Solution

By Gaussian method, we obtained

1 2 3 41 2 , , , 0x s t x s x t x= + = = = Its vector form is

1 2 1

0 1 0

0 0 1

0 0 0

s t

+ +

. Thus the vector form of the general solution to Ax = 0 is

2 1

1 0

0 1

0 0

s t

+

.

Bases for Row Spaces, Column Spaces, and Nullspaces.

Example 22

The reduced row-echelon form of the matrix

A =

34021

32732

41823

43021

is R =

00000

11000

10110

10201

Theorem

(a) Elementary row operations do not change the nullspace of a matrix

(b) Elementary row operations do not change the row space of a matrix

TheoremIf A and B are row equivalentmatrices then

(a) A given set of column vectors of A is linearly independent if and only if the

corresponding column vectors of B are linearly independent.

(b) A given set of column vectors of A form a basis for the column space of A if andonly if the corresponding column vectors of B form a basis for the column space of B.

Theorem

If a matrix R is in row-echelon form, then the row vectors with the leading 1 s (the

nonzero row vectors) form a basis for the row space ofR, and the column vectors with theleading 1s of the row vectors form a basis for the column space ofR.


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Chapter 5 : 17 /21

Find

(a) a basis for the null space of A(b)a basis for the row space of A(c) a basis for the column space of A

Solution(a) Let rx =

3and sx =

5 sxx ==

54

0532

=++ xxx srx =2

02531

=++ xxx srx = 21

=

s

s

r

sr

sr

x

2

=

1

1

0

1

1

s +

0

0

1

1

2

r

Therefore, basis for the null space ofA is }0,0,1,1,2,1,1,0,1,1{TT ><

(b) Basis for the row space ofA is }1,1,0,0,0,1,0,1,1,0,1,0,2,0,1{ ><

(c) Basis for the column space ofA is }4,2,1,3,2,3,2,2,1,2,3,1{TTT ><

Example 23: Basis and Linear Combinations

Consider the column vectorsin the matrixA given in Example 22 as v1, v2, v3, v4, v5.(a) Find a subset of these vectors that forms a basis for the space spanned by the vectors.

(b) Express each vector that is not in the basis as a linear combination of the basis vectors.

Solution

(a) The leading 1s inR occur in columns 1, 2 and 4. Thus the corresponding column vectors

ofA are the basis vectors for span {v1, v2, v3, v4, v5}, i.e.

{v1, v2, v4}

(c)Denoting the column vectors ofR by w1,w2, w3, w4, w5.3 1 2 4

w aw bw cw= + + 5 1 2 4

w pw qw rw= + +

By inspection, a = 2, b = 1, c = 0 and p = q = 1,r = -1.

Thus,213

2 vvv += and5 1 2 4

v v v v= +


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Chapter 5 : 18 /21

5.6 Rank and Nullity

Four Fundamental matrix spaces associated with matrixA are:

1. Row space ofA = Column Space ofAT2. Column space of A = Row Space ofAT3.

Nullspace of A4. Nullspace of AT

Proof:Rank (A) = dim (row space of A)

= dim (column space of AT) = rank (AT)

TheoremIfA is any matrix, then the row space and column space ofA have the same dimension.

Definition

The common dimension of the row space and column space of a matrixA is called the

rank ofA and is denoted by rank (A); the dimension of the nullspace ofA is called the

nullity ofAand is denoted by nullity (A).

Theorem

IfA is any matrix, then rank (A) = rank (AT)

Dimension Theorem for Matrices

IfA is a matrix with n columns, then

rank (A) + nullity (A) = n.

Theorem

IfA is an m x n matrix, then

(a) rank (A)= The number of leading variables in the solution ofAx =0.

(b) nullity (A) = The number of parameters in the solution ofAx =0.


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Chapter 5 : 19 /21

Example 24

The row-echelon form of the matrix

1 2 2 -1

3 -6 -6 3

4 9 9 -4

2 -1 -1 2

5 8 9 -5

4 2 7 -4

A

=

1 2 2 1

0 1 1 0

0 0 1 0is

0 0 0 0

0 0 0 0

0 0 0 0

R

=

(a) rank (A) (b) nullity (A)(c) rank (A

T) (d) nullity(A

T)

Solution

(a) rank (A) = 3 (b) nullity (A) = 4 3 = 1(c) rank (A

T) = 3 (d) nullity(A

T) = 6 3 = 3

Remark

SupposethatA is an m x n matrix of rank r, thenAT

is an n x m matrix of rank r.

So, nullity (A) = n rnullity (A

T) = m r

THE DIMENSION OF THE FOUR FUNDAMENTAL SPACES OF AN n x m MATRIXA

OF RANK r

Fundamental Space Dimension

Row space ofA r

Column space ofA r

Nullspace ofA n r

Nullspaces ofAT

m r


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Chapter 5 : 20 /21

THE CONSISTENT THEOREM

IfAx = 0 is a linear system of m equations in n unknowns, then the following are

equivalent.

(a) Ax = bis consistent

(b) b is in the column space ofA

(c) The coefficient matrixA and augmented matrix [A | B] have the same rank.

IfAx = b is a linear system of m equations in n unknowns, then the following are

equivalent.

(a) Ax = bis consistent for every m x 1 matrix.

(b) The column vectors of A spans Rm

(c) rank (A) = m

Theorem: EQUIVALENT STATEMENTS

IfA is an n x n matrix, and if TA: Rn R

nis multiplication byA, then the following are

equivalent

(a) A is invertible(b) Ax = 0 has only the trivial solution(c) The reduced row-echelon form ofA is I n(d) A is expressible as a product of elementary matrix(e) Ax = b is consistent for every n x1 matrix(f) Ax = b has exactly one solution for every n x1 matrix b(g) det (A) 0(h) The range of TA is Rn(i) TA is one-to-one(j) The column vectors ofA are linearly independent(k) The row vectors ofA are linearly independent(l) The column vectors ofA spans Rn(m) The row vectors ofA spans Rn(n) The column vectors ofA form a basis for Rn(o) The row vectors ofA form a basis for Rn(p) A has rank n(q) A has nullity 0


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Exercise

Section 5.2

1. Let v1 = < 2, 1, 0, 3 >, v2= < 3, -1, 5, 2 > and v3= < -1, 0, 2,1 > . Determinewhether the following vectors are in span { v1 , v2 , v3}

(a) < 2, 3, -7, 3 > (b) < -4, 6, -13, 4 > (c) < 0, 0, 0, 0 >

5-General Vector Sp

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