chapter 3 vector space objective: to introduce the notion of vector space, subspace, linear...
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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