I
1
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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6
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 2
1 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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7
(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 4
1 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