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6-7-2011

Linear Algebra ()

100 (Linearization)

()

(i)

(ii) active and alive SIAM

(SIMAX)

(iii)

1G. Strang, Linear Algebra and Its Applications, 4th edition, Brooks/ Cole,

2006.

2. K. Hoffman and R. Kunze, Linear Algebra, 2nd edition, Prentice-Hall, 1971.

3. S.H. Friedberg, A.J. Insel, L.E Spence, 4th edition, Linear Algebra,

Prentice-Hall, 2003.

Gilbert Strang MIT Lectures

visualized

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appreciate

classical

LUQRSVD least squares

3

1. Webpage Video Lectures

2.

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Ch1 Vector Spaces ()

H.W. 1,9,11-19

(Vector Spaces, VS)

VS

VS (prototype)

Definitions

A vector space V over a field F consists of a set on which two operations (called

addition and scalar multiplication) are defined, so that the following 10 properties

hold.

(VS-1) whenever , . ( )

(VS 0) whenever and . ( )

(VS 1) for all , . ( )

(VS 2) ( ) ( ) for all , , . ( )

(VS 3) so that

x y V x y V

x V F x V

x y y x x y V

x y z x y z x y z V

V x

0 0

for all . ( )

(VS 4) For each , s.t. . ( )

x x

x V y V x y

0

V

(VS 5) For each element in , 1 . Here 1 .

and is the identity element for multiplication.

x V x x F

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4

(VS 6) ( ) ( ) for all , and . ( )

(VS 7) ( ) for all and , . ( )

(VS 8) ( ) for all , and . ( )

ab x a bx a b F x V

a x y ax ay a F x y V

a b x ax bx a b F x V

Remark

(i) The elements of the field F are called scalars.

(ii) The elements of the VS are called vectors.

(iii) (field) 4 ()

()

C

Ex1.

1

1

(i) : , 1, 2,..., , is a field , , VS.

(ii) : R, 1, 2,..., .

Is a VS over the complex numbers ?

No, (VS0) is viola

n n ni

n

i

n

aF a F i n F R

a

aV a i n F

a

V

ted. ( 1.2- Ex15).

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5

Ex2.

(i) ( ) : , 1 , 1 is a VS.

(ii) , , . is a VS.

(iii) , , R. is not a VS.

m n ij ijm n

ij ijm n

ij ij

M F A a a F i m j n

V A a a R F Q V

V A a a Q F V

Ex3.

( , ) : , where is a field is a VS.F S F f S F F

Ex4.

-1-1 0( ) : ( ) ... , , 1 , is a VS.n nn n nP f x f x a x a x a a F i n n N Ex5.

1( ) : is a VS.i i iS a a F VS

Ex6.

1 2 1 2( , ) : , . .S a a a a R F R

1 2 1 2 1 1 2 2

1 2 1 2

: ( , ) ( , ) ( , - ).

Scalar multiplication (SM): ( , ) ( , ).

(VS1), (VS2) and (VS8) fail to hold.

a a b b a b a b

c a a ca ca

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Ex7.

1 2 1 2 1 1

1 2 1

S .

: ( , ) ( , ) ( , 0).

SM: ( , ) ( ,0).

(VS3), (VS5) (X)

F R

a a b b a b

c a a c a

,

Ex8.

1 2 1 21 2 1 2 1 1 2 2

1 2 1 2

( , ) : , .

: ( , ) ( , ) ( 2 , +3 ).

SM: ( , ) ( , ).

(VS1)(VS2)(VS8) (X)

V a a a a R

a a b b a b a b

c a a ca ca

Ex9.

1 2 21

:

SM:

(0,0) if 0, ( , )

, if 0.

(VS8) (X)

V

cc a a aca c

c

.

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(3) (4) (2) (2)

Thm1.1 If , , are in , a VS, such that , then . ( )

Pf: (4) with .

( ) ( ) ( ) ( ) .

Cor1. in (VS3) is unique.

x y z V V x z y z x y

v V z v

x x x z v x z v y z v y z v y

0

0

0

Pf: Let and be such that for all .

.

x V

x x

0 0

0

Thm 1.1

= .

= .

x x

x x

0

0 0 0 0

Cor2. The vector in (VS4) is unique.

Pf: Let and in V be such that . Then .

Remark: Since is unique, we shall denote such by - .

y

y y x y x y y y

y y x

0

Thm1.2 : a vector space. Then

(a) 0 for (0 , ).

(b)( - ) -( ) (- ) for a and .

(c) for a .

V

x x V F V

a x ax a x F x V

a F

0 0

0 0

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Thm1.1

Pf: (a) 0 0 (0 0) 0 0 .

0 .

(b) - ( ) is the unique element of such that -( ) .

Wanted: (- ) . (*)

(Then - ( )

x x x x x

x

ax V ax ax

ax a x

ax

0

0

0

0

=(- ) .)a x

(8) (a)

To see(*), we have that:

(- ) =( (- )) 0 = .

-( ) = (- ) .

-(1 )= (-1) - (-1) .

ax a x a a x x

ax a x

x x x x

0

Hence,

(- )= (-1) = (-1) = (- ) .a x a x a x a x

(c) a +a = a( + )= a = a + a = .0 0 0 0 0 0 0 0 0

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H.W. 1,8,17-21,23,26,30,31

1.3 Subspaces ()

1

subspace 23subspace

Def: Let be a VS over a field . Then is called a subspace of if is a VS

over the under the operations of addition and SM defined on .

Remarks:

(i) and are subspaces of .

V F W V V W

F V

V V

0

(ii) (VS 1-2, 5-8) subspaces

(VS -1,0,3,4).

(iii) (VS -1,0,3) , (VS -1,0).

W W

V

Thm1.3 , where is a VS. Then is a subspace of

(a) 0 . (VS3)

(b) whenever, , . (VS-1)

(c) whenever, , . (VS0)

W V V W V

W

x y W x y W

cx W c F x W

(3) (3)

Thm1.1

Pf: ( )

Let 0 and 0 be zero vectors for and , respectively. Let . Then

0 and 0 . Hence, 0 0 .

0 0 .

V W

W V x W

x x x x x x

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Thm1.2-b

( ) To complete the proof, we need to show that

( ), ( ), ( ) For each , s.t. 0.

To see this, let . Then - ( ) = (-1) .

By (c) (-1) . Hence, -( ) .

Re mark

a b c x W y W x y

x W V x x

x W x W

: Thm1.2-a , ( ), ( ) (Ex.17)

Def: Let ( ) . Define the tramspose of by ( ) .

is symmetric .

Ex1. The set of all symmetric matrices in ( ) is a subspace of

t tij m n ji n m

T

n n

W b c

A a A A A a

A A A

W M F

.

( ).

Ex2. ( ) ( ), a subspace.

Ex3. ( ) ( , ), a subspace.

Ex4. The set of diagonal matrices is a subspace of ( ).

n n

n

n n

M F

P F P F

C R F R R

M F

2

Ex5. ( ) : Trace 0 .

Ex6. ( ) ( ) : 0 is not a subspace of ( ).

Ex7. ( , ) : , is fixed is a subspace of .

( , ) : , and 0 are fixed is not a subspace

n n

ij n n ij n n

W A M F A

W A a M F a M F

W x mx x R m R

W x mx b x R m b

2 of .R

1 2

Thm 1.4 Any intersection of subspace of a vector space is a subspace of .

Is a subspace of ?

V V

W W V

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1 2 1 2 2 1

1 2 1 2

(Ex19. a subspace of or .)

Def. { : and y }.

W W V W W W W

S S x y x S S

1 2

1 2 1 2

1 2

Def. A vector space is called the direct sum of and if (i) , 1, 2,

are subspace, (ii) {0} and (iii) . We denote that

is the direct sum of and by wri

iV W W W i

W W W W V V

W W

1 2ting .V W W

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H.W. 1, 3(a),4(a),5(g),9-12,15,16 1.4 Linear Combinations and Systems of Linear Equations

linear combination ( )

span Span "

"

S S V V

S

S S S

1

Def. : a vector space

, .

A vector is called a linear combination ( ) of vectors of

if , , 1, 2,..., . s.t.

.

In this case, we also say that is a li

i i

n

i ii

V

S V S

v V S

u V a F i n

v a u

v

1 2

1 2

near combination ( .) of , ,..., and

call , ,..., the coefficients of the linear combination.

n

n

l.c u u u

a a a

1

Def: , , : a vector space.

The span of span ( ) = The set of all of the vectors of

: , , 1, 2,..., , .

For convenience, we define span

n

i i i ii

S V S V

S S l.c.

S a u u S a F i n n N

( ) 0 .

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Def: A subset of a vector generates (or spans) if Span ( ) .

In this case, we also say that the vectors of generate (or span) .

S V V S V

S V

2 21 2 1 2Ex1. Let , {(0,1), (1,0)}. Let ( , ) . Then (1,0) (0,1).

Hence, any vector is a of vectors of .

V R S x x x R x x x

l.c. S

Thm1.5

(i) Span ( ) : is a subspace of .S V

(ii) : a subspace of and .W V W S

Then Span ( ) ( pan ( ) ).

Proof:

(i) Span( ) 0 is a subspace.

0 0 span ( ), where .

If and Span ( ) , then , , 1, 2,... . and , , 1,i i j j

W S S S S

S

S z S z S

x y S a F u S i n b F v S j

1 1 1 1

2,... . s.t.

, , and so Span ( ). n m n m

i i j j i i j ji j i j

m

x a u y b u x y a u b u S

1

1

Moreover, for .

( ) Span ( ).

Span ( ) is a subspace of .

(ii) If Span ( ), then , where , .

n

i i ii

n

i i i i i ii

c F

cx c a u S

S V

x S x a u a F u S W a u W

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1 Span ( ) .

n

i ii

a u W x W S W

1 1 2 2 3 3 4 4 5 5

1

Ex2. Can a vector be expressed as a of other vectors ? Such question often reduces to

the problem of solving a system of linear equation.

(2,6,8) .

l.c.

a u a u a u a u a u

u

2 3 4 5(1, 2,1), (-2, -4, -2), (0, 2,3), (2,0, -3) and (-3,8,16).u u u u

1 2 4 5

1 2 3 5

1 2 3 4 5

- 2 2 - 3 2,

2 - 4 2 8 6,

- 2 3 - 3 16 8.

1 0 1 .

0 0 1

a a a a

a a a a

a a a a a

1 2 5

3 5

4 5

( )

- 2 -4,

3 7,

-2 3.

1 -2 0 0 1 -4 0 0 1 0 3 7 .

0 0 0 1 -2 3

a a a

a a

a a

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1 2 5

3 5

4 5

2 5

2 - - 4,

-3 7,

2 3.

,

a a a

a a

a a

a a

.

3

3 2 3 2 3 2

3 2

Ex3. ( )

Is 2 - 2 12 - 6 a of - 2 - 5 - 3, and 3 - 5 -4 -9 ?

How about 3 - 2 7 +8 ?

P R

x x x l.c. x x x x x x

x x x

3 2 3 2 3 2

Sol:

(Case1) 2 - 2 12 - 6 ( - 2 - 5 - 3) (3 - 5 - 4 - 9). x x x a x x x b x x x

3 2,

- 2 - 5 -2,

- 5 - 4 12,

- 3 - 9 -6.

a b

a b

a b

a b

1 3 2 1 3 2 1 0 -4-2 -5 -2 0 1 2 0 1 2

.-5 -4 12 0 11 22 0 0 0-3 -9 -6 0 0 0 0 0 0

(Case2)

1 3 3 1 3 3-2 -5 -2 0 1 4

.-5 -4 7 0 11 22-3 -9 8 0 0 17

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3Ex4. span (1,0,0)(0,1,0) = the -plane in .Re mark.1 0 span ( ).

xy R

S

3

1 1

2 2 1

3 3

Ex5. (1,1,0), (1,0,1) and (0,1,1) spans .

1 1 0 1 1 0 1 0 1 0 -1 1 - .

0 1 1 0 1 1

S R

a aa a aa a

1 2

1 2 1 2

3 3 2 1

2 3 1

3 2 1

3 2 1

1 1 0 1 0 1 0 1 -1 - 0 1 -1 - .

0 1 1 0 0 2 -

11 0 0 ( - + )21 0 1 0 ( - + ) .210 0 1 ( + - )2

a aa a a a

a a a a

a a a

a a a

a a a

2 2 2 2

2 2

Ex6. 3 2,2 5 - 3, - - 4 4 span(S)= ( ).

Ex7.

1 1 1 1 1 0 0 1 (i) , , , span(S) ( ).

1 0 0 1 1 1 1 1

1 1 1 1 1 0 (ii) , and .

1 0 0 1 1 1

T

S x x x x x x P R

S M R

S

2 2hen Span ( ). S M R

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1 1

2 2 1

3 3 1

4 4

1 2 1

2 1 2 1

1 3 1 3

4 1 2 4

1 1 1 1 1 11 0 1 0 1 0

.1 1 0 0 0 10 1 1 0 1 1

1 1 1 1 0 1 20 1 0 0 1 0

.0 0 1 0 0 10 1 1 0 0 1

a aa a aa a aa a

a a aa a a aa a a a

a a a a

1 2 3

2 1

1 3

3 2 4

1 33 2 4 1 3

2 4

2 2

1 0 00 1 0

.0 0 10 0 1

,

1 0 1 1 1 0Ex8. Span , and .

0 1 0 1 1 1

any linear combination of the vectors

a a aa aa a

a a a

a aa a a a a A S

a a

R

of the above set has equal diagonal entries.

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H.W. 1, 3, 8,12,16,17,20

1.5 Linear Dependence and Linear Independence

Goal: Let W be a subspace of a vector space V.

Finding a smallest finite subset S that generates W.

Finding such smallest finite subset leads to the concept of linear

dependence ( l.d.) and linear independence (l.i.).

1

Def (i) A subset of a vector space is called l.d. if distinct vectors , 1,..., in

and scalars a , 1, 2,..., , not all zero, s.t.

0.

i

i

n

i ii

s S V u i n S

i n

a u

In this case, we also say that the vectors of are l.d. (Any set containing the zero vector of l.d.).

l.d. "

S S

trial".

"

0.

. trial"

1 2 1 21

(ii) If is not l.d., then is called l.i. (For any finite number of distinct vectors

u , ,.. in and 0, where , then ... 0).n

n i i i ni

S S

u u S a u a F a a a

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1 2 3 41 2 3 4

Ex1. (1,3, 4,2), (2, 2, 4,0), (1, 3,2, 4), ( 1,0,1,0) , , , .

Then 4 3 2 0 0.

is l.d.

S u u u u

u u u u

S .

Ex2.

1 -3 2 -3 7 4 -2 3 11 , , .

-4 0 5 6 -2 -7 -1 -3 2

1 -3 -2 0 1 -3 -2 0-4 6 -1 0 0 -6 -9 0-3 7 3 0 0 -2 -3 0

.0 -2 -3 0 0 -2 -3 02 4 11 0 0 10 15 05 -7 2 0 0 8 2 0

1 -

S

3 2 1

3 -2 03 50 1 0 1 0 02 23 30 1 0 0 1 02 2

.3 0 0 0 00 1 02 0 0 0 03 0 0 0 00 1 02

0 0 0 030 1 02

-2, 3, 5. is l.du u u S ..

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1 2 1 2

2 1

Thm1.6 Let be a vector space, and let . If is l.d., then is l.d.

Cor. If is l.i , then is l.i.

Thm1.7 : l.i. and - . Then is l.d Span(S), or equivalen

V S S V S S .

S . S .

S v V S S v . v

tly Span ( ) is l.i.( , , l.d. Span( )).

v S S v .

S v S v S

1

1 21

Pf: ( ) is l.d. distinct vectors ,..., and , 1, 2,..., ,

not all zero; such that 0. Moreover, , ..., . Otherwise, = 0

for all 1, 2...... . WLOG, let

n i

n

i i n ii

n

S v u u S v a F i n

a u v u u u a

i n u

. Then , 1, 2,..., -1.iv u S i n

-1

1

-1

1

0.

Hence 0. Otherwise, 0, for all 1, 2,..., -1,

- Span ( ).

( ) Span(S) , , 1, 2,..., , s.t.

n

i i n ni

n i

ni

ii n

i i

a u a v

a a i n

av u v S

a

v u S a F i n

1

1

. Here we may assume that , are distinct.

Hence, , ,..., are distinct.

n

i i ii

n

v a u u

v u u

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1

1

Thm1.6

0 (-1) .

, ,..., is l.d..

is l.d..

n

i ii

n

v a u

v u u

S v

Re mark: 1. If no proper subset of generates the span of , then must be linearly independent.

Otherwise, a proper subset of with Span( ) = Span( ), a contraditoin.

S S S

S S S S

2. If Span( ) , where is a subspace of a vector space , and no proper subset

of is a generating set for , then is a l.i. Such is called a linearly independent

gener

S W W V

S W S S

ating set for .W

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H.W. 1,2(a),3(a),5,10(a),13,14,21,22,30,31

1.6 Bases and Dimension ()

building block linearly independent generating (l.i.g.)

set

(basis)

(dimension)

1

1 2

Def: A basis for a vector space is a l.i. subset of that generates .

Ex1. Span( ) 0 =

Ex2. (0,0, ,...,0).Then e , ,..., for .

Ex3. = that matrix whose only nonzero entry is

in

i n

ij

V V V

e i e e F

E

2

a 1 in the th row and the th column.

Then ,1 ,1 for ( ).

Ex4. = 1, ,..., for ( ).

Ex5. = 1, , ... for ( ).

ijm n

nn

i j

E i m j n M F

x x P F

x x P F

1 2

1

Thm1.8 : a vector space. , ,..., .

Then

is a basis for each .

! , 1 , s.t.

.

n

i

n

i ii

V u u u

V v V

a F i n

v a u

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1 1

Pf: ( ) Let , , 1 , so that

. Then

i i

n n

i i i ii i

a b F i n

v a u b u

1

1 1

1

( ) 0. Hence, for all .

( ) Let 0. Since 0 0,

we have, via the uniqueness of , that =0 for all .

is l.i

n

i i i i ii

n n

i i ii i

i i

ni i

a b u a b i

a u u

a a i

u ..

Remark: 1. .

2.

nV F

.

Thm1.9

Thm1.9 Let Span( ), where #( ) .

Then some subset of is a basis for . Hence has a finite basis.

Pf. Case (i) or 0 Then 0 and is a basis for .

Case (ii) contains a n

V S S

S V V

S S V V

S

1

1

1 2 1 2

onzero vector .

Then is a l.i. set. #( ) ,

, ,..., , - ,

( " " ).

k k

u

u S

u u u u u u S v S v

S

, .... ,

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Claim: is a basis for ( Span( ), Thm1.5-(i,ii)

Span( ) Span( ) ). Let . If ,then Span( ). If ,

then Thm1.7 Span( ). Thu

V S

S V v S v v v

v v

s Span( ).S

1

Remark: Constructive proof,

Step1: Find a finite generating set . That is , span( ) .

Step2: Select a nonzero vector in .

S S V

u S

1 Step3: Find a "largest" possible l.i. set , where and .

Such is a finite basis.

u S

3

Ex: (2, -3,5), (8, -12, 20), (1,0 - 2), (0, 2, -1), (7, 2,0) .

Sol: (2, -3,5), (1,0, -2), (0, 2, -1) is a basis for Span( ).

?

S

R S

Thm1.10 ( Re placement Theorem)

Let Span{ }, where #( ) .

Let is a subset of with #( ) .

V G G n

L l.i. V L m

(i) . ( span vector space

).

m n G

G

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(ii) , #( ) - , s.t. span( ) .

H G H n m L H V

Proof: The proof is by mathmatical induction on .

(i) Let 0 . Then = . 0, (i) and (ii) .

(ii) 0 (i) (ii) , (i) (ii) 1 .

m

m L m n H G m

m m

1 1 1

1 -

1

Let ,..., be a l.i. subset of , L = ,..., l.i. induction ,

,..., , s.t. #( ) - and span( ) . , ,

Span( ) . Span( ) is l.d. ( ).

Hence,

m m

n m

m

L v v V v v

H u u G H n m L H V n m H

L V v L L

, 1.n m n m

1

-

1 11 1

-

1 1 11 21

(ii) :

Span ( '),

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

1WLOG, we may assume that 0, - .

m

m n m

m i i i i i m ii i

m n m

m i i i ii i

v V L H

v a v b u b v v i m

b u v a v b ub

2 - 1,..., , #( ) - -1, span( ). 1-5, span( ) span( ) .

. span( ) .

n mH u u H n m u L H

L H L H V

L H G L H V

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( ) .

Corollarly: (i) Let span ( ) and # ( ) . If is a l.i. subset of , then #( ).

(ii) Let be a vector space having a finite basis. Then every basis for contains

V G G n L G n G

V V

the same number of vectors.

Pf of (i): Let be a l.i. subset of . If #( ) , then a subset of containing 1

vectors. Moreover, by Thm 1.6, is a l.i. set. By Thm1.

L G L L L n

L

10-(i) #( ) 1,

a contraction. Thus #( ) . Again, by Thm1.10-(i) #( ) . Pf of (ii): Thm1.10, . Let and be any two finite

bases for that containin

n L n

L L n

V

V

g and vectors, respertively. 1.10

. .

n m n m

m n n m

Definitions: (i) A vector space, Corollary, well-defined, is called

finite-dimensional if it has a finite basis.

(ii) The number of vectors in a basis is called the dimension of and is denoted by dim( ).

(iii) A vector space that is not finite-dimensional is called infinite-dimentsional.

Ex7. 0 .

V V

V dim( ) 0. ( )V

Ex8. . dim( ) .

Ex9. ( ). dim( ) .

Ex10. ( ). dim( ) 1.

Ex11. , dim( ) 1 1 .

Ex12. , dim( ) 2, 1, .

n

m n

n

V F V n

V M F V m n

V P F V n

F R V R V

F R V C V i

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Corollary2. Let dim( ) .

(a) Let and Span( ) . Then #( ) . Moreover, if # ( ) , then

is a basis for . ( " " genetating set).

V n

G V G V G n G n

G V

(b) Let and is . Then #( ) . Moreover, if #( ) , then is

a basis for . ( " " l.i. set).

(c) Every l.i. subset of can be extended to a basis for .

Pf: (a)

L V L l.i L n L n L

V

V V

Let be a finite basis of .

1.10-(i) #( ) #( ). 1.9, #( ) . #( ) ,

.

(b) Thm1.10-(i). Thm1.10-(ii) .

(

V

n G

G H H V H n G n

G H

c) Thm1.10 .

Ex: If , , is a basis for a vector space , then , , is also a basis for .

Pf: By Cor 2(b), it suffies to show that , , is l.i..

Let ( ) ( ) 0

u v w V u v w v w w V

u v w v w w

a u v w b v w cw au

( ) ( ) 0.

, , is a basis 0, 0, 0 0.

a b u a b c w

u v w a a b a b c a b c

The dimension of subspaces.

Thm1.11 : a subspace of a vector space , where dim( ) .

Then

(i) dim( ) dim( ).

(ii) If dim( ) dim( ), then .

W V V

W V

W V W V

I

28

Pr oof:

(i) Thm1.10-(i) . Let be a finite basis for . Then span( ) . Let be a

basis for . Then and is l.i.. Hence, #( ) #( ). That is, dim( ) dim( ).

(

V V

W V W V

ii) Thm1.10-(ii) , where #( ) dim( ) - dim( ), s.t. span( ) .

#( ) 0 span( ) .

H H V W H V

H V W V

51 2 3 4 5 1 3 5 2 4Ex17. ( , , , , ) : 0, dim( ) 3.Ex18. = The set of diagonal matices dim( ) .

1Ex19. = The set of symmetric matices dim( ) ( 1).2

W a a a a a F a a a a a W

W n n W n

W n n W n n

3

Corollarly: If is a subspace of a finite-dimensional vector space , then any basis for

can be extended to a basis for .

Ex. Describe subspaces of that have dimentsion 3,2,1 and 0, representivel

W V W

V

R

0 1

0

y.

The Lagrange Interpolation Formula.

Lagrange polynomails associated woth , ,... .

- ( ) .

-

n

nk

i ki kk i

c c c

x cf x

c c

0 1

0

0

0 ( ) .

1

Claim: , ,... is a subset of ( ).

Let 0 ( ) for some scalars , 0,1..., .

( ) 0 ( ) 0.

i j

n n

n

i i ii

n

i i j ji

j if c

j i

f f f l.i. P F

a f a F i n

a f c c

I

29

0

But ( ) for any 0,1,..., .

Hence, 0. Since dimen( ( )) 1.

is a basis for ( ).

Every polynomial in ( ) can be uniquely ecpress as a of .

n

i i j ji

j n

n

n

a f c a j n

a P F n

P F

g P F l.c.

0

0

0

Let = .

Then ( ) ( ) .

So ( ) .

This representation is called Lagrange interpolation formula.

( 1) n n

1

n

i ii

n

j i i j ji

n

i ii

g b f

g c b f c b

g g c f

n

n