Chapter 3Vector Space

Objective: To introduce the notion of vector space, subspace, linear independence, basis, coordinate, and change of coordinate.

Recall, In

- vector addition

- scalar multiplication

- norm

- triangle inequality

§3-1 Definition and Examples

2 3&

Why Introduces Vector Space?

It provides comprehensive understanding of many mathematical & physical phenomena.

For example, All the solutions of the ODE can be

described as . Why? Controllability and observability

space in linear control theory.

0" yyxcxcy sincos 21

m

Vector Space Axioms

Definition: Let be set and be a field ( in most

practical case, ).

Define two binary operations

V F

o r F F

:

:

V V V

F V V

Then is a vector space if the follow-

ing Conditions hold:

( , , , )V F

Vector Space Axioms (cont.)

For any ,

A1:

A2:

A3:

A4:

, , and ,x y z V F

and (Closed)x y V x V

(Communicative Law)x y y x

( ) ( ) (Associative Law)x y z x y z

0 , 0 (Zero Vector)V x x x V

Vector Space Axioms (cont.)

A5:

A6:

A7:

, ( ) , ( ) 0 (inverse element)x V x V x x

( )

( )

( ) ( )

x y x y

x x x

x x

1 x x x V

Examples

defined by

over is a vector space.

( , , , )n

1 1 1 1

1 1

---------- (1)

------------------ (2)

n n n n

n n

x y x y

x y x y

x x

x x

n

Examples (cont.)

over is also a vector space with defined by (1) and (2).

nC

C

and

over is a vector space.nC

C over is NOT a vector space. (Why?)n

Examples (cont.)

Let [ , ] { : :[ , ] is continuous on [a,b]}C a b f f a b

over defined by

is a vector space.

[ , ]C a b ( )( ) ( ) ( ) -------- (3)

( )( ) ( ) ---------------- (4)

f g x f x g x

f x f x

{ ( ) | ( ) is a polynomial of degree less than }

with "+" and " " defined by (3) and (4) is a vector

space.

nP p x p x n

Examples (cont.)

defined by

is a vector space.

( )

( ) ( )

ij ij

ij

A B a b

A a

over m nR R

is NOT a vector space. (Why?) ( , ) | 2 1x y x y

is NOT a vector space. (Why?) ( ,sin ) |x x x

Theroem3.1.1: Let be a vector space and . Then

PF:

x VV( ) 0 0

( ) 0

( ) (-1)

i x

ii x y y x

iii x x

4

6

3

2 4

( ) 0 0 ( 0 ) (0 0) ( 0 )

0 0 ( 0 ) 0

( ) - 0 ( )

( )

by A

by A

by A

by A by A

i x x x x

x x x x

ii x x x x y

x x y y

﹡

( ) 6

8 4

( ) 0 0 (1 ( 1)) 1 ( 1)

( 1)

i by A

by A by A

iii x x x x

x x x

﹡

﹡

Definition: If is a nonempty subset of a vector space , and satisfies the following conditions:

then is said to be a subspace of .

§3-2 Subspace

VS

S V

( ) whenever for any scalar

( ) whenever and

i x S x S

ii x y S x S y S

S

Remark 1: Thus every subspace is a vector space in its own right.

Remark 2: In a vector space , it can be readily verified that and are subspaces of . All other subspaces are referred to as proper subspaces.

VV

V{0}

Examples of Subspaces

Example 2. (P.135)

1 1

1 2 1 2 2 2 1

3 3

| , | 0

x x

S x x x S x x

x x

1

33 2 1 2

3

| 0 are subspaces of

x

S x x x

x

Examples of Subspaces (cont.)

Example 3. (P.135)

1 2 2

2

| | 1

are subspaces

Neither no

r

of .

xxS x S x

x

V

Example 4. (P.135)

2 21 2

2 2

| , and |

are subspaces of .

Ta bS a b S A A A

b a

Examples of Subspaces (cont.)

Example 5. (P.136)

; (0) 0 is a subspace of .n nS p P p P

Example 8. (P.136)

2

2

[ , ], ''( ) ( ) 0 is a

subspace of [ , ].

S f C a b such f x f x

C a b

Example 6. (P.136)

[ , ] [ , ] | has a continuous derivative on [a,b].

[ , ] is a subspace of [ , ].

n

n

C a b f C a b f nth

C a b C a b

Nullspace and Range-space

m nA Let ,

( ) | 0nN A x Ax

1

( ) (:, ) | for 1,2,...,n

i ii

R A x A i x i n

※ Define that N(A) is called the nullspace of A; R(A) is called the range(column) space of A.

( ) is a subspace of , and ( ) is a subspace of .n mN A R A

Examples of Nullspaces

Example 9. (P.137)

1 3 4

2 3 4

1 1

- 2 2 1 ( )

2 1 0

0 1

x x xN A

x x x

1 1 1 0

2 1 0 1A

Question: Determine N(A) if .

Answer:

2 21 12( 1) (1) ( 2)

1 1 1 0 0 1 0 1 1 0

2 1 0 1 0 0 1 2 1 0E E E

Note

Note that, both the vector spaces and the solution

set of contain infinite number of elements.

2 1 0 1 1| , ( ) | ,

0 1 0 1

1 0 1 0 | ,

0 1 1

x y x y x y y x y

x y x y

Question: Can a vector space be described by a set of vectors

with number being as small as possible?

Example:

2'' 0y y

Spanning set, linear independent, basis

Span and Spanning Sets

Definition: Let be vectors in a vector space ,

a sum of the form , where are scalars, is

called linear combination of .

Definition:

Definition: is said to be a spanning set for

if

1 2, ,..., nv v v

V

1

n

i ii

v

1,..., n

1 2, ,..., nv v v

1 21

{ , ,..., } | for all n

n i i ii

span v v v v F i

1 2{ , ,..., }nv v v

V 1 2{ , ,..., }.nV span v v v

Examples of Span

Example :

1 2

1 0

0 , 1 | ,

0 0 0

x

span y x y x x plane

1 1

0 , 1 | ,

0 0 0

x

span y x y

3

1 0 0

0 , 1 , 0

0 0 1

span

1 2, ,..., nv v v V

V

Theroem3.2.1: If , then is

a subspace of .1 2{ , ,..., }nspan v v v

Question: Given a vector space and a set

, how to determine whether or

not?

V 1 2{ , ,..., }ns v v v V

1 2{ , ,..., }nV span v v v

Example 11. (P.140)

? ?

31 2 3 1 2 3 , 1 2 3 , ,

Tspan e e e span e e e

Yes, 1 2 3 0 1 2 3 .T T

a b c ae be ce

?3 1 1 1 , 1 1 0 , 1 0 0

T T Tspan

Yes, let 1

1 2 3 2

3||

1 1 1 1 1 1

1 1 0 1 1 0

1 0 0 1 0 0

a a

b b

c c

A

∵ A is nonsingular, The system has a unique solution

1

2

3

c

b c

a c

Example 11.(c) (P.141)

?3

1 0

0 , 1

1 0

span

No,

1 0

0 1

1 0

1 1 0

0 0 , 1

2 1 0

span

Example 12. (P.141)

?

2 23 (1 ), ( 2), P span x x x

Yes, let 23

2 2 21 2 3

3 1 1

2 2

1 2 3

(1 ) ( 2)

2

2 2

ax bx c P

ax bx c x x x

a c b

b b

c a c b

Question: How to find a minimal spanning set of a vector space

(i.e. a spanning set that contains the smallest possible number of vectors.)

(i.e. There is no redundancy in a spanning set.)

§3-3 Linear Independence

.V

3 T1 2 3 1 2 3{ , , } { , , , (1, 2, 3) }span e e e span e e e

It’s unnecessary.

Linear Dependency

Definition: is said to be linear independent if “ ”.

Definition: is said to be linear dependent if there exist scalars NOT all zero such that

1{ ,..., }nv v

1

0 0 n

i i ii

c v c i

1 2, ,..., nc c c1{ ,..., }nv v

1

0 .n

i ii

c v

Lemma : 1Suppose { ,..., },nV span v v

1

1 1 1

, for some scalars 's.

{ ,..., , ,..., }

n

k i i iii k

k k n

k v v

V span v v v v

1 2

1 1 2 2 1

... not all zero,

... 0

n

n

c c c

c v c v c v

Note 1: Linear independency means there is no

redundancy on the spanning set .

Note 2: is a minimal spanning set for iff

is linear independent and spans .

1{ ,..., }nv v

V1{ ,..., }nv v

V1{ ,..., }nv v

Definition: A minimal spanning set is called a basis.

Linear Dependency (cont.)

2 31{ ,..., } in or nv v

Question: How to systematically determine the linear dependency of vectors ?

Geometrical interpretation(see Figure 3.3):

•

•

1 2 1 2 and are linear dependent , will lie along the same line.v v v v

1 2 3 1 2 3, , are linear dependent , , will lie on the same plane.v v v v v v

Example 3. (P.149)

Note that

is redundant for the spanning set. On the other hand,

∵ A is singular det(A)=0.

a nontrivial solution is linear dependent.

1 2 3

1 2 1

1 , 3 , 3

2 1 8

v v v

1

1 1 2 2 3 3 2

3

1 2 1 0

1 3 3 0

2 1 8 0

v v v

A

3 1 23 2 ,v v v

1 2 3{ , , }v v v Th 1.4.3

Theroem3.3.1: Let , Then

is linear independent

PF:

1 2{ , ,..., } nnx x x

1 2{ , ,..., }nx x x

1 2[ ... ] is nonsingular.nX x x x 1 2det( ... ) 0nx x x

1.4.3

0 has no nontrivial solution

is nonsingular. det(X) 0

i i

Th

c x XC

X

Example 4. (P.150)

4 2 2

det 2 3 5 0

3 1 3

4 2 2

2 , 3 , 5 is linear dependent.

3 1 3

Theroem3.3.2: Suppose Then

PF:

1

1 21

1

" " Let { , ,..., }, then

and let

( ) 0 0

linear indep

n

n i

en

ii

n

i ii

den

i

tn

i i i ii

v V span v v v v v

v v

v i

i i i

1 2{ , ,..., }.nV span v v v

1 2

11

{ , ,..., } is linear independent

, ! ...

n

n

n i ii

v v v

v V v v

1

1

1 2

" " Let 0

0 0 0

{ , ,..., } is linear independent.

unique

n

i ii

n

i i

ne

i

n

ss

v

v i

v v v

How to determine linear independency

For the Vector Space Pn (P.151)

21 2 3 1 2 3 1 2 3

1

2

3

1 2 3

( ) 0

( 2 ) ( 2 8 ) (3 8 7 ) 0

1 2 1 0

2 1 8 0 ------- (*)

3 8 7 0

det(A)=0 (*) has nontrivial solution.

( ), ( ), ( ) are linear depen

i i

A

c p x

c c c x c c c x c c c

c

c

c

p x p x p x

dent.

2 2 21 2 3( ) 2 3, ( ) 2 8, ( ) 8 7 p x x x p x x x p x x x

Question: Determine the linear dependency of

Sol:

How to determine linear independency

For the Vector Space C(n-1)[a,b] (P.152)

1

1

( )

11 2' ' '

1 2

1 11

( )...... ( ) are linear dependent

... not all zero ( ) 0

( ) 0, 0,1,..., 1

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

j

j

n

n i i

ji i

n

n

n nn

d

dx

n

f x f x

c c c f x

c f x j n

cf x f x f x

f x f x f x

f x f x c

0

,

0

x

( 1)1( )...... ( ) [ , ],n

nf x f x C a bLet

Suppose

Wronskian

1 2' ' '

1 21 2

1 11

( ) ( ) ( )

( ) ( ) ( )[ , ,..., ]( ) [ , ],

( ) ( )

n

nn

n nn

f x f x f x

f x f x f xW f f f x x a b

f x f x

Definition: Let be functions in C (n-1)[a,b], and define

thus, the function is called the Wronskian of

1 2, ,......, nf f f

1 2[ , ,..., ]nW f f f

1 2, ,..., .nf f f

Theroem3.3.3: Let

if are linear dependent on [a,b]

Cor:

1 2( , ,..., )( ) 0 .nW f f f x x

1 2 0 0

1 2

( , ,..., )( ) 0 for some [ , ]

, ,..., are linear independent.n

n

W f f f x x a b

f f f

( 1)1 2, ,..., [ , ],n

nf f f C a b

1 2, ,..., nf f f

Example of Wronskian

Is linear independent in Yes,

Example 6. (P.153)1( , )?C { , }x xe e

( , ) det 2 0x x

x x

x x

e eW e e

e e

Is linear independent in Yes,

Example 8. (P.154)

4 ?P2 3{1, , , }x x x2 3

22 3

1

0 1 2 3(1, , , ) det 12 0

0 0 2 6

0 0 0 6

x x x

x xW x x x

x

2 1

2

and | | are linear independent in ( , )

even though [ , | |] 0.

x x x C

W x x x

Question: Does the converse of Th 3.3.3 hold?

Answer: No, a counterexample is given as follows

Question: Is linear independent in and Why?

2{ & | |}x x x1(0, )?C

§3-4 Basis and Dimension

Definition: Let be a basis for a vector space if

Example: It is easy to show that

1 2

1 2

( ) { , ,..., } is linear independent.

( ) { , ,..., }.n

n

i v v v

ii V span v v v

V1 2, ,..., nv v v

3

1 0 2

1 , 1 , 0 is a basis for .

1 1 1

2 21 0 0 1 0 0 0 0, , , is a basis for .

0 0 0 0 1 0 0 1

Theroem3.4.1: Suppose

PF:1 2

1 1 2 2

{ , ,..., }

= + ... for 1, 2,..., . (*)i n

i i i ni n

u V span v v v

u a v a v a v i m

1 1{ ...... } and { ...... }n mV span v v u u V

1with . Then { ...... } is linear dependent.nm n u u

1

21 2

1

111 12 1

(*)21 22 2 2

1 2 nonsingular

1 2

0

0Consider

0

0 0

m

i i mi

m

m

bym

m

n n nm m

A

c

cc u u u u

c

ca a a

a a a cv v v Ac

a a a c

c

1.2.1

1 21

ˆ ˆ ˆ ˆ 0 has a nontrivial solution ... , 0mThm T

n i ii

m n Ac c c c c u

1{ ...... } is linear dependent.mu u

Cor: If are both bases for a

vector , then

PF:

1 1{ ...... } and { ...... }m nu u v v

V .m n

3.4.11

1

{ ...... } is linear independent .

{ ...... } is a spanning set

Thn

m

v vm n

u u

3.4.11

1

Similarly,

{ ...... } is linear independent .

{ ...... } is a spanning set

Thus, .

Thm

n

u un m

v v

m n

Dimension

Definition: Let be a vector space. If has a basis consisting of n vectors, we say that has dimension n.

{ } is said to have dimension 0.

is said to be finite dimensional if finite set of vectors that spans ;otherwise we say is infinite-dimensional.

V VV

V

0

VV

Example of Dimension

Example

3

dim { , } 2,

if and are linear independent in .

span x y

x y

ndim( ) n

dim( ( ))np x n

dim( [ , ])C a b

dim cos ,sin |span n t n t n

1 0 0 1dim 2

0 1 1 0span

Theroem3.4.3: If , then linear

independent

PF:

1

11

1

( 3.4.1)

" " Let

, ,..., is linear dependent

, ,..., not all zero 0

0, then ,..., linear dependent. contradictory!

Thus 0 &

n

n

n i ii

n

by Th

v V

v v v

c c c cv c v

If c v v

c v

1

.n

ii

i

cv

c

dim( ) 0V n 1,..., nv v

1,..., .nspan v v V

1

1 1 1

1

" " Suppose, on contary, ,..., is linear dependent.

for some ,..., , ,...,

dim( ) contradictory!

Thus, ,..., is linear independe

n

i i i i i nj i

n

v v

v c v i V span v v v v

V n

v v

nt.

Theroem3.4.4:

(i) No set of less than vectors can span .

(ii) Any subset of less than linear independent vectors can

be extended to form a basis for .

(iii) Any spannnig set containing more than vect

n V

n

V

n ors can be

pared down to form a basis for .V

If dim( ) 0, thenV n

Standard Basis

n1{ ,..., } is usually said standard basit so be for .ne e

1( ) 1, ,..., .nnp x x x

2 2 1 0 0 1 0 0 0 0

0 0 0 0 1 0 0 1

§3-5 Chang of Basis

不同場合用不同座標系統有不同的方便性，如質點 運動適合用體座標 (body frame) 來描述，而飛彈攔 截適合用球面座標。

利用某些特定基底表示時，有時更易使系統特性彰 顯出來。

Question: 不同座標系統間如何轉換？

Definition: Let be a vector space and let

be an ordered basis for .

V 1 2, ,..., nE v v v

V

1 21

If V, then for some scalars , ,...., .n

i i ni

v v c v c c c

1

2[ ] is called t coordhe of

inate vectornE

n

c

cv F V

c

with respect to the ordered basis . E

unique expression

Remark 1:

Lemma 2: Every n-dimensional vector space is isomorphic to

1 1

2 2

1 1

1

1 1

2 2

[ ] = , [ ] =

( )

[ ] [ ] [ ]

n n

i i E i i Ei i

n n

n

i i ii

E E E

n n

c d

c dv c v v w d v w

c d

v w c d v

c d

c dv w v w

c d

.nF

Example 4 (P.168)

1 2 1 2

21 2 1 2

1 2

2

5 7 3 1Let , , , ,

2 3 2 1

and let , and , be two ordered bases for .

(a) Find and for any ,

(b)

E F

w w v v

E w w F v v

xX X X

x

1 1

2 2

( . . , )

Find the relation between and .

E F

E F

c di e X X

c d

X X

Question:

Example 4 (cont.)

1 1 2 2 1 1 2 2

1 1 2 1

2 1 2 2

1 2 1

1 2 2

1 2( , )

1 2

Let

5 7 5 7= =

2 3 2 3

3 3 1 = =

2 2 1

,( )

W w w

X c w c w d v d v

x c c c

x c c c

d d d

d d d

V v v

Solution:

Example 4 (cont.)

1 1 11

2 2 2

1 1 11

2 2 2

1 1 11

2 2 2

3 -7 (a)

-2 5

1 -1

-2 3

3 4 (b)

-4 -5

E

F

c x xX W

c x x

d x xX V

d x x

d c cV W

d c c

Solution:

Transition Matrix

Definition: V is called the transition matrix from the ordered basis F to the standard basis .

Remark 1: V-1 is the transition matrix from to F. Remark 2: S=V-1W is the transition matrix from E to F.

1 2,e e

1 2,e e

1 2,e e

1 2,F v v

1 2,E w w

V-1WV-1

W

P.169 figure. 3.5.2 changing coordinates in R2

Theorem (P.171)

1 1

1 11 1 21 1 1

Let ,..., and ,..., be two ordered bases for .

Each vector can then be expressed as a linear combination of the 's,

......

n n

j i

n n

E w w F v v V

w v

w s v s v s v

2 12 1 22 1 2

1 1 2 1

......

......

Let ,

if [ ] and [ ] ,

th

n n

n n n nn n

E F

w s v s v s v

w s v s v s v

v V

x v y v

en , where is referred to as the transition matrix from to .

y Sx S E F

Theorem (cont.)

1 1 2 2 1 1 2 21 1 1

1 1 2 21 1 1

...... ......

( ) ( ) ...... ( )

n n n

n n i i i i n in ii i i

n n n

j j j j nj j nj j j

v x w x w x w x s v x s v x s v

s x v s x v s x v

PF:

11

1

221

1

[ ] = =

n

j jj

n

j jjF

n n

nj jj

s x

y

s xyy v Sx

y

s x

Theorem (cont.)

1 1 2 2 1 1 2 2

1 1 2 2

1 2 ( is linear independent.)

is nonsingular .

...... ...

0 ...... 0

... 0 i

n n n n

n n

n w

S

y Sx x w x w x w y v y v y v

Sx x w x w x w

x x x

Remark :

1 1

11 2 1 2

Let ,..., and ,..., be two ordered bases for .

Then , where = ... and ... .

nn n

FE n n

E w w F v v

S V W V v v v W w w w

Corr. :

Example 6 (P.170)

1 2 3

1 2 3

Let , , (1,1,1) ,(2,3,2) , (1,5,4)

, , (1,1,0) ,(1,2,0) , (1,2,1)

(a) Find the transition matrix from to ;

(b) Find and , if

T T T

T T T

FE

F F

E w w w

F v v v

S E F

X Z

1 2 3

1 2 3

3 2

3 2

x w w w

z w w w

Question:

Example 6 (cont.)

1

(a) Method 1:

1 2 1 1 1 1

1 3 5 , 1 2 2

1 2 4 0 0 1

2 1 0 1 2 1 1 1 -3

1 1 1 1 3 5 1 1 0

0 0 1 1 2 4 1 2 4

FE

W V

S V W

Solution:

(a) Method 2:

1 1 1 1

1 (1) 1 ( 1) 2 (1) 2

1 0 0 1

2 1 1 1

3 (1) 1 ( 1) 2 (2) 2

2 0 0 1

1 1 1 1

5 ( 3) 1 (0) 2 (4) 2

4 0 0 1

Solution:

1 1 -3

1 1 0

1 2 4

FES

Example 6 (cont.)

3 8

(b) [ ] 2 5

1 3

1 8

[ ] 3 2

2 3

FF E

FF E

X S

Z S

Solution:

Example 7 (P.172)

2 2

3

Let 1, , and 1, 2 , 4 2

be two ordered bases for ( ).

Find the transition matrix from to .

E x x F x x

P x

E F

Question:

Solution:

2

2

2 2

1 1 1 0 0

2 1 0 2 0

4 2 2 1 0 4

x x

x x x

x x x

1

1 1 -3

1 1 0

1 2 4

1 0 1/2

and 0 1/ 2 0

0 0 1/4

EF

FE

S

S

Example 7 (cont.)

23

2

[ ]

Thus, if given any ( ) in

1

21 0 1/21

then [ ] 0 1/ 2 02

0 0 1/4 1

4

1 1 1 ( ) ( ) 1 (4 2)

2 2 4

F

EP

p x a bx cx P

a ca

P b b

cc

p x a c b x c x

Solution:

Application 1: Population Migration (P.164)

0

10

0.94 0.02 0.30 Set and x =

0.06 0.98 0.70

the percentages after years will be given by

x x

for =10, 30, and 50, we can get

0.27 x

0.73

nn

A

n

A

n

20 30

0.25 0.25 , x , x

0.75 0.75

Application 1 (cont.)

1 2

1 1

2

Change of Basis

1 -1 Choose u = , u =

3 1

0.94 0.02 1 1 u u

0.06 0.98 3 3

0.94 0.02 -1 -0.92 u

0.06 0.98 1

A

A

2

0 1 2

n 0 1 2

0.92u0.92

0.3 1 1 x 0.25 0.05 0.25u 0.05u

0.7 3 1

x x 0.25u 0.05(0.92) un nA

Markov (P.165)

Application 1 is an example of a type of mathematical model called Markov Process.

The sequence of vectors is called a Markov Chain.

A is called stochastic matrices, which has special struc- ture in that its entries are nonnegative and its columns all add up to 1.

If A is n×n, then we will want to choose basis vectors so that the effect of the matrix A on each basis vector is simply to scale it by some factor λj, that is,

u u 1, 2,...,j j jA j n u j

1 2x , x ,...

§3-6 Row Space and Column Space

Definition: Let

Then,

(:,1) ... (:, )

(1,:)

( ,:)

m n n m

A A A n

A

F

A m

1

1

( ) (:, ) | is called column spa ofce .

( ) ( ,:) | is called ofrow spa .ce

nm

i ii

nn

i ii

col A c A i c F F A

row A c A i c F F A

Example 1 (P.175)

1 0 0Let ,

0 1 0

The row space of is the set of all 3-tuples of the form

(1,0,0) (0,1,0) ( , ,0)

The column space of is the set of all vectors of the form

A

A

A

1 3 2

1 0 0

0 1 0

Thus the row space of is a two-dimensional subspace of

, and the column space of is .

A

A

Theroem3.6.1: Two row equivalent matrices have the same row space.

PF:

(by a finite sequence of row operation)

If is row equivalent to

( ) ( ) (1)

Similarly, If is row equivalent to

B A

B A

R A R B

A

(by a finite sequence of row operation)

( ) ( ) (2)

Thus, from (1) and (2) we can get that ( ) ( ).

B

A B

R B R A

R B R A

Rank

Definition: The rank of a matrix A is the dimension of the row space of A.

Remark 1: The nonzero row of the row echelon mat- rix will form a basis for the row space.

Remark 2: To determine the rank of a matrix, we can reduce the matrix to the echelon matrix.

Example 2 (P.175)

1 2 3 1 2 3

Let 2 5 1 0 1 5 (row echelon form)

1 4 7 0 0 0

Clearly, (1, 2,3) and (0, 1, 5) form a basis for the row sequence

of . Since and are row equivalent, they have tha same row

sp

A U

U U A

ace, and hence the rank of is 2.A

Theroem3.6.2: is consistent

PF:111 12 1

21 22 2 21 2

1 2

1 2

...

has one solution.

is linear combination of , , ... ,

( )

n

nn

m m mmn

n

aa a b

a a a bAx x x x

a a ba

b a a a

b col A

Ax b ( ).b col A

Consistency Theorem for Linear System

Theroem3.6.3: Let , then

PF:

m nA (i) , is consistent col( )= ;

(ii) , has at most one solution

column of are linear independent.

nb Ax b A

b Ax b

A

1

(i) trivial.

(ii) " " 0 has at most one solution

0 is the only solution.

(:, ) 0 implies 0, .

columns of are linear independ

m

i ii

Ax

A i i

A

ent.

1 2

1 2

" " Columns of are linear independent.

0 has only solution 0.

Suppose has two solutions and .

( ) 0.

A

Ax x

Ax b x x

A x x b b

1 2 1 2 0 .x x x x

Corollary 3.6.4:

.3.3.1

.3.4.3

is nonsingular.

the column vectors of are linear independent.

the column vectors of form a basis for .

( )

n n

Th

Thn

n

A

A

A

col A

In general, the rank and the dimension of nullspace

always add up to the number of columns of the matrix.

The dimension of the nullspace of a matrix is called the

of the mnull atity rix.

Definition:

Theroem3.6.5: Let , then

PF:

If ( )

has nonzero rows.

0 has free variables.

( ) number of free variables.

rank A r

U r

Ux n r

Nullity A

m nA ( ) ( ).n Nullity A Rank A

The Rank-Nullity Theorem

Let be the reduced row echelon form of .

and ( ) | 0 | 0 ( )

U A

N A x Ax x Ux N U

Example 3 (P.177)

1 2 1 1 1 2 0 3

Let 2 4 3 0 0 0 1 2 ( ) 2.

1 2 1 5 0 0 0 0

A U rank A

1 2 4 1 2 4

3 4 3 4

2 3 0 2 3( ) ( )

2 0 2

2 3 2 3

1 0( ) | , , .

2 0 2

0 1

x x x x x xN A N U

x x x x

N A span

( ) 2

This agrees with th 3.6.4.

Nullity A

Let ( ) ( ).

but ( ) ( ) in general.

For example,

1 1 1 1

1 1 0 0

( ) 1,1 ( )

but ( )

A U R A R U

col A col U

A U

R A span R U

col A

1 1( ) .

1 0col U

Remark

Theroem3.6.6: Let , then

PF:

m nA F

Let ~ , row echelon form of . ( ) ( ) .

Denote the matrix obtained from by deleting the columns of

free variables.

Similarly, Denote the matrix obtained fr

L

L

A U A rank A rank U r

U U

A

om ' by deleting the

same columns as those of .

~ Columns of are linear independent.

0, implies 0.

L L L

L

row

A

U

A U U

U x x

~ 0, implies 0.

Columns of are linear independent.

dim( ( )) dim( ( )) .

Similarly, dim( ( )) dim( ( )) dim(

L L L

L

T

A U A x x

A

col A col A r

row A col A ro

( )) dim( ( )).

This completes the proof.

Tw A col A

dim( ( )) dim( ( )).R A col A