Chapter 4
Vector Spaces
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Elementary Linear Algebra
R. Larsen et al. (6 Edition)
Vectors in Rn
Vector Spaces
Subspaces of Vector Spaces
Spanning Sets and Linear Independence
Basis and Dimension
Rank of a Matrix and Systems of Linear Equations
Coordinates and Change of Basis
投影片設計編製者
淡江大學 電機系 翁慶昌 教授
n
4.1 Vectors in R

An ordered n-tuple:
a sequence of n real number ( x1 , x2 ,, xn )
n
 n-space: R
the set of all ordered n-tuple
Elementary Linear Algebra: Section 4.1, p.183
2/107

Ex:
1
n=1
R = 1-space
= set of all real number
n=2
R = 2-space
= set of all ordered pair of real numbers ( x1 , x2 )
n=3
2
3
R = 3-space
= set of all ordered triple of real numbers ( x1 , x2 , x3 )
n=4
4
R = 4-space
= set of all ordered quadruple of real numbers ( x1 , x2 , x3 , x4 )
Elementary Linear Algebra: Section 4.1, p.183
3/107

Notes:
n
(1) An n-tuple ( x1 , x2 ,, xn ) can be viewed as a point in R
with the xi’s as its coordinates.
(2) An n-tuple ( x1 , x2 ,, xn ) can be viewed as a vector
x  ( x1 , x2 ,, xn ) in Rn with the xi’s as its components.

Ex:
x1 , x2 
x1 , x2 
a point
Elementary Linear Algebra: Section 4.1, p.183
0,0
a vector
4/107
u  u1 , u2 ,, un , v  v1 , v2 ,, vn 

Equal:
u  v if and only if
(two vectors in Rn)
u1  v1 , u2  v2 , , un  vn
 Vector
addition (the sum of u and v):
u  v  u1  v1, u2  v2 , , un  vn 

Scalar multiplication (the scalar multiple of u by c):
cu  cu1 , cu2 ,, cun 

Notes:
The sum of two vectors and the scalar multiple of a vector
n
in R are called the standard operations in Rn.
Elementary Linear Algebra: Section 4.1, p.183
5/107

Negative:
 u  (u1 ,u2 ,u3 ,...,un )

Difference:
u  v  (u1  v1 , u2  v2 , u3  v3 ,...,un  vn )

Zero vector:
0  (0, 0, ..., 0)

Notes:
(1) The zero vector 0 in Rn is called the additive identity in Rn.
(2) The vector –v is called the additive inverse of v.
Elementary Linear Algebra: Section 4.1, p.184
6/107

Thm 4.2: (Properties of vector addition and scalar multiplication)
n
Let u, v, and w be vectors in R , and let c and d be scalars.
(1)
(2)
(3)
(4)
(5)
(6)
u+v is a vector in Rn
u+v = v+u
(u+v)+w = u+(v+w)
u+0 = u
u+(–u) = 0
cu is a vector in Rn
(7) c(u+v) = cu+cv
(8) (c+d)u = cu+du
(9) c(du) = (cd)u
(10) 1(u) = u
Elementary Linear Algebra: Section 4.1, p.185
7/107

Ex 5: (Vector operations in R4)
Let u=(2, – 1, 5, 0), v=(4, 3, 1, – 1), and w=(– 6, 2, 0, 3) be
4
vectors in R . Solve x for x in each of the following.
(a) x = 2u – (v + 3w)
(b) 3(x+w) = 2u – v+x
Sol: (a) x  2u  ( v  3w )
 2u  v  3w
 (4,  2, 10, 0)  (4, 3, 1,  1)  (18, 6, 0, 9)
 (4  4  18,  2  3  6, 10  1  0, 0  1  9)
 (18,  11, 9,  8).
Elementary Linear Algebra: Section 4.1, p.185
8/107
(b) 3(x  w )  2u  v  x
3x  3w  2u  v  x
3x  x  2u  v  3w
2x  2u  v  3w
x  u  12 v  32 w
 2,1,5,0   2, 23 , 21 , 12   9,3,0, 29 
 9, 211 , 92 ,4
Elementary Linear Algebra: Section 4.1, p.185
9/107

Thm 4.3: (Properties of additive identity and additive inverse)
n
Let v be a vector in R and c be a scalar. Then the following is true.
(1) The additive identity is unique. That is, if u+v=v, then u = 0
(2) The additive inverse of v is unique. That is, if v+u=0, then u = –v
(3) 0v=0
(4) c0=0
(5) If cv=0, then c=0 or v=0
(6) –(– v) = v
Elementary Linear Algebra: Section 4.1, p.186
10/107

Linear combination:
The vector x is called a linear combination of v1 , v 2 ,...,v n ,
if it can be expressed in the form
x  c1v1  c2 v 2   cn v n
c1 , c2 ,  , cn : scalar
 Ex 6:
Given x = (– 1, – 2, – 2), u = (0,1,4), v = (– 1,1,2), and
3
w = (3,1,2) in R , find a, b, and c such that x = au+bv+cw.
Sol:
 b  3c   1
a 
b  c  2
4a  2b  2c   2
 a  1, b  2, c  1
Thus x  u  2 v  w
Elementary Linear Algebra: Section 4.1, p.187
11/107

Notes:
A vector u  (u1 , u2 ,, un ) in R n can be viewed as:
a 1×n row matrix (row vector): u  [u1 , u2 ,, un ]
or
u1 
u 
2
a n×1 column matrix (column vector): u   
 
 
u n 
(The matrix operations of addition and scalar multiplication
give the same results as the corresponding vector operations)
Elementary Linear Algebra: Section 4.1, p.187
12/107
Vector addition
Scalar multiplication
u  v  (u1 , u2 , , un )  (v1 , v2 , , vn )
cu  c(u1 , u2 ,, un )
 (u1  v1 , u2  v2 ,  , un  vn )
u  v  [u1 , u2 ,  , un ]  [v1 , v2 , , vn ]
 [u1  v1 , u2  v2 ,  , un  vn ]
u1  v1  u1  v1 
u  v  u  v 
u v   2   2   2 2
      
    

u
v
u

v
 n  n  n n
Elementary Linear Algebra: Section 4.1, p.188
 (cu1 , cu2 , , cun )
cu  c[u1 , u2 ,, un ]
 [cu1 , cu2 ,, cun ]
u1  cu1 
u  cu 
cu  c  2    2 
   
   
un  cun 
13/107
Keywords in Section 4.1:










ordered n-tuple：有序的n項
n-space：n維空間
equal：相等
vector addition：向量加法
scalar multiplication：純量乘法
negative：負向量
difference：向量差
zero vector：零向量
additive identity：加法單位元素
additive inverse：加法反元素
14/107
4.2 Vector Spaces

Vector spaces:
Let V be a set on which two operations (vector addition and
scalar multiplication) are defined. If the following axioms are
satisfied for every u, v, and w in V and every scalar (real number)
c and d, then V is called a vector space.
Addition:
(1) u+v is in V
(2) u+v=v+u
(3) u+(v+w)=(u+v)+w
(4) V has a zero vector 0 such that for every u in V, u+0=u
(5) For every u in V, there is a vector in V denoted by –u
such that u+(–u)=0
Elementary Linear Algebra: Section 4.2, p.191
15/107
Scalar multiplication:
(6) cu is in V.
(7) c(u  v)  cu  cv
(8) (c  d )u  cu  du
(9) c(du)  (cd )u
(10) 1(u)  u
Elementary Linear Algebra: Section 4.2, p.191
16/107

Notes:
(1) A vector space consists of four entities:
a set of vectors, a set of scalars, and two operations
V：nonempty set
c：scalar
 (u, v)  u  v： vector addition
 (c, u)  cu： scalar multiplication
V ,
(2) V  0：
, 
is called a vector space
zero vector space
Elementary Linear Algebra: Section 4.2, Addition
17/107

Examples of vector spaces:
(1) n-tuple space: Rn
(u1, u2 ,, un )  (v1, v2 ,, vn )  (u1  v1, u2  v2 ,, un  vn ) vector addition
scalar multiplication
k (u1, u2 ,, un )  (ku1, ku2 ,, kun )
(2) Matrix space: V  M mn (the set of all m×n matrices with real values)
Ex: ：(m = n = 2)
u11 u12  v11 v12   u11  v11 u12  v12 
u u   v v   u  v u  v 
 21 22   21 22   21 21 22 22 
u11 u12   ku11 ku12 
k



u21 u22  ku 21 ku 22 
Elementary Linear Algebra: Section 4.2, Addition
vector addition
scalar multiplication
18/107
(3) n-th degree polynomial space: V  Pn (x)
(the set of all real polynomials of degree n or less)
p( x)  q( x)  (a0  b0 )  (a1  b1 ) x   (an  bn ) xn
kp( x)  ka0  ka1x   kan xn
(4) Function space: V  c(, ) (the set of all real-valued
continuous functions defined on the entire real line.)
( f  g )(x)  f ( x)  g ( x)
(kf )(x)  kf ( x)
Elementary Linear Algebra: Section 4.2, p.193
19/107
 Thm
4.4: (Properties of scalar multiplication)
Let v be any element of a vector space V, and let c be any
scalar. Then the following properties are true.
(1) 0 v  0
(2) c0  0
(3) If cv  0, then c  0 or v  0
(4) (1) v   v
Elementary Linear Algebra: Section 4.2, p.195
20/107

Notes: To show that a set is not a vector space, you need
only find one axiom that is not satisfied.
 Ex 6: The set of all integer is not a vector space.
Pf:
1V , 12  R
( 12 )(1)  12 V (it is not closed under scalar multiplication)
  noninteger

scalar
integer
 Ex 7: The set of all second-degree polynomials is not a vector space.
Pf:
Let p( x)  x 2 and q( x)   x 2  x  1
 p( x)  q( x)  x  1V
(it is not closed under vector addition)
Elementary Linear Algebra: Section 4.2, p.195
21/107

Ex 8:
V=R2=the set of all ordered pairs of real numbers
vector addition: (u1 , u2 )  (v1 , v2 )  (u1  v1 , u2  v2 )
scalar multiplication: c(u1 , u2 )  (cu1 ,0)
Verify V is not a vector space.
Sol:
1(1, 1)  (1, 0)  (1, 1)
 the set (together with the two given operations) is
not a vector space
Elementary Linear Algebra: Section 4.2, p.196
22/107
Keywords in Section 4.2:





vector space：向量空間
n-space：n維空間
matrix space：矩陣空間
polynomial space：多項式空間
function space：函數空間
23/107
4.3 Subspaces of Vector Spaces

Subspace:
(V ,,) : a vector space
W  
 : a nonempty subset
W V
(W ,,) ：a vector space (under the operations of addition and
scalar multiplication defined in V)
 W is a subspace of V
 Trivial subspace:
Every vector space V has at least two subspaces.
(1) Zero vector space {0} is a subspace of V.
(2) V is a subspace of V.
Elementary Linear Algebra: Section 4.3, p.198
24/107

Thm 4.5: (Test for a subspace)
If W is a nonempty subset of a vector space V, then W is
a subspace of V if and only if the following conditions hold.
(1) If u and v are in W, then u+v is in W.
(2) If u is in W and c is any scalar, then cu is in W.
Elementary Linear Algebra: Section 4.3, p.199
25/107

Ex: Subspace of R2
(1) 0
0  0, 0
(2) Lines through t he origin
(3) R2

Ex: Subspace of R3
(1) 0
0  0, 0, 0
(2) Lines through t he origin
(3) Planes through t he origin
(4) R3
Elementary Linear Algebra: Section 4.3, p.199
26/107

Ex 2: (A subspace of M2×2)
Let W be the set of all 2×2 symmetric matrices. Show that
W is a subspace of the vector space M2×2, with the standard
operations of matrix addition and scalar multiplication.
Sol:
W  M 22
M 22 : vector sapces
Let A1, A2 W ( A1T  A1, A2T  A2 )
A1 W, A2 W  ( A1  A2 )T  A1T  A2T  A1  A2 ( A1  A2 W )
k  R, A W  (kA)T  kAT  kA
(kAW )
W is a subspace of M 22
Elementary Linear Algebra: Section 4.3, p.200
27/107

Ex 3: (The set of singular matrices is not a subspace of M2×2)
Let W be the set of singular matrices of order 2. Show that
W is not a subspace of M2×2 with the standard operations.
Sol:
1 0
0 0
A
W , B  
W


0 0
0 1
1 0
A B  
W

0 1
W2 is not a subspace of M 22
Elementary Linear Algebra: Section 4.3, p.200
28/107

Ex 4: (The set of first-quadrant vectors is not a subspace of R2)
Show that W  {(x1 , x2 ) : x1  0 and x2  0} , with the standard
operations, is not a subspace of R2.
Sol:
Let u  (1, 1) W
 1u  11, 1  1, 1W
(not closed under scalar
multiplication)
W is not a subspace of R 2
Elementary Linear Algebra: Section 4.3, p.201
29/107

Ex 6: (Determining subspaces of R2)
Which of the following two subsets is a subspace of R2?
(a) The set of points on the line given by x+2y=0.
(b) The set of points on the line given by x+2y=1.
Sol:
(a) W  ( x, y) x  2 y  0  (2t , t ) t  R
Let v1   2t1 , t1  W
v2   2t2 , t2  W
 v1  v2   2t1  t2 ,t1  t2 W (closed under addition)
kv1   2kt1 , kt1 W (closed under scalar multiplication)
W is a subspace of R 2
Elementary Linear Algebra: Section 4.3, p.202
30/107
(b) W  x, y  x  2 y  1
(Note: the zero vector is not on the line)
Let v  (1,0) W
 1v  1,0W
W is not a subspace of R 2
Elementary Linear Algebra: Section 4.3, p.203
31/107

Ex 8: (Determining subspaces of R3)
Which of the following subsets is a subspace of R 3?
(a) W  ( x1 , x2 ,1) x1 , x2  R
(b) W  ( x1 , x1  x3 , x3 ) x1 , x3  R
Sol:
(a) Let v  (0,0,1) W
 (1) v  (0,0,1) W
W is not a subspace of R 3
(b) Let v  ( v1 , v1  v 3 , v 3 )  W , u  (u1 , u1  u 3 , u 3 )  W
 v  u  v1  u1, v1  u1   v3 u3 , v3  u3  W
kv  kv1 , kv1   kv3 , kv3  W
W is a subspace of R 3
Elementary Linear Algebra: Section 4.3, p.204
32/107
 Thm
4.6: (The intersection of two subspaces is a subspace)
If V and W are both subspaces of a vector space U ,
then the intersecti on of V and W (denoted by V  U )
is also a subspace of U .
Elementary Linear Algebra: Section 4.3, p.202
33/107
Keywords in Section 4.3:

subspace：子空間

trivial subspace：顯然子空間
34/107
4.4 Spanning Sets and Linear Independence

Linear combination:
A vector v in a vector space V is called a linear combinatio n of
the vectors u1,u 2 , ,u k in V if v can be written in the form
v  c1u1  c2u2   ck uk
Elementary Linear Algebra: Section 4.4, p.207
c1,c2 , ,ck : scalars
35/107

Ex 2-3: (Finding a linear combination)
v1  (1,2,3) v 2  (0,1,2) v 3  (1,0,1)
Prove (a) w  (1,1,1) is a linear combinatio n of v1 , v 2 , v 3
(b) w  (1,2,2) is not a linear combinatio n of v1 , v 2 , v 3
Sol:
(a) w  c1v1  c2 v 2  c3 v3
1,1,1  c1 1,2,3  c2 0,1,2  c3 1,0,1
 (c1  c3 , 2c1  c2 , 3c1  2c2  c3 )
c1
 c3
 2c1  c2
3c1  2c2  c3
1
1
1
Elementary Linear Algebra: Section 4.4, pp.208-209
36/107
1 0  1 1
Guass  Jordan Elimination
 2 1 0 1 
3 2 1 1
1 0  1 1 
0 1 2  1


0 0 0 0 
 c1  1  t , c2  1  2t , c3  t
(this system has infinitely many solutions)
t 1
 w  2 v1  3v 2  v 3
Elementary Linear Algebra: Section 4.4, pp.208-209
37/107
(b)
w  c1v1  c2 v 2  c3 v 3
1 0  1 1 
Guass  Jordan Elimination
 2 1 0  2 
3 2 1
2 
1 0  1 1 
0 1 2  4 


0 0 0
7 
 this system has no solution ( 0  7)
 w  c1v1  c2 v 2  c3 v3
Elementary Linear Algebra: Section 4.4, pp.208-209
38/107

the span of a set: span (S)
If S={v1, v2,…, vk} is a set of vectors in a vector space V,
then the span of S is the set of all linear combinations of
the vectors in S,
span(S ) c1 v1  c2 v 2    ck v k
ci  R
(the set of all linear combinatio ns of vectors in S )

a spanning set of a vector space:
If every vector in a given vector space can be written as a
linear combination of vectors in a given set S, then S is
called a spanning set of the vector space.
Elementary Linear Algebra: Section 4.4, p.209
39/107
 Notes:
span ( S )  V
 S spans (generates ) V
V is spanned (generated ) by S
S is a spanning set of V
 Notes:
(1) span( )  0
(2) S  span( S )
(3) S1 , S2  V
S1  S2  span(S1 )  span(S2 )
Elementary Linear Algebra: Section 4.4, p.209
40/107
 Ex 5: (A spanning set for R3)
Show that the set S  (1,2,3), (0,1,2), (2,0,1) sapns R 3
Sol:
We must determine whether an arbitrary vector u  (u1 , u2 , u3 )
in R 3 can be as a linear combinatio n of v1 , v 2 , and v 3 .
u  R3  u  c1v1  c2 v 2  c3 v3
 c1
2c1  c2
 2c3  u1
 u2
3c1  2c2  c3  u3
The problem thus reduces to determinin g whether this system
is consistent for all values of u1 , u2 , and u3 .
Elementary Linear Algebra: Section 4.4, p.210
41/107
1 0 2
A2 1
3 2
0 0
1
 Ax  b has exactly one solution for every u.
 span(S )  R3
Elementary Linear Algebra: Section 4.4, p.210
42/107

Thm 4.7: (Span(S) is a subspace of V)
If S={v1, v2,…, vk} is a set of vectors in a vector space V,
then
(a) span (S) is a subspace of V.
(b) span (S) is the smallest subspace of V that contains S.
(Every other subspace of V that contains S must contain span (S).)
Elementary Linear Algebra: Section 4.4, p.211
43/107

Linear Independent (L.I.) and Linear Dependent (L.D.):
S  v1 , v 2 ,, v k  : a set of vectors in a vector space V
c1v1  c2 v 2    ck v k  0
(1) If the equation has only the trivial solution (c1  c2    ck  0)
then S is called linearly independen t.
(2) If the equation has a nontrivial solution (i.e., not all zeros),
then S is called linearly dependent.
Elementary Linear Algebra: Section 4.4, p.211
44/107

Notes:
(1)  is linearly independen t
(2) 0  S  S is linearly dependent.
(3) v  0  vis linearly independent
(4) S1  S2
S1 is linearly dependent  S2 is linearly dependent
S2 is linearly independent  S1 is linearly independent
Elementary Linear Algebra: Section 4.4, p.211
45/107
 Ex 8: (Testing for linearly independent)
Determine whether the following set of vectors in R3 is L.I. or L.D.
S  1, 2, 3, 0,1, 2,  2, 0,1
v1
v2
v3
c1
 2c3  0
Sol:
0
c1 v1  c2 v 2  c3 v 3  0  2c1  c2 
3c1  2c2  c3  0
1 0  2 0 
1 0 0 0 
- Jordan Eliminatio n
 2 1 0 0 Gauss


 0 1 0 0
3 2 1 0
0 0 1 0
 c1  c2  c3  0 only the trivial solution
 S is linearly independen t
Elementary Linear Algebra: Section 4.4, p.213
46/107

Ex 9: (Testing for linearly independent)
Determine whether the following set of vectors in P2 is L.I. or L.D.
Sol:
S = {1+x – 2x2 , 2+5x – x2 , x+x2}
v1
v2
v3
c1v1+c2v2+c3v3 = 0
i.e. c1(1+x – 2x2) + c2(2+5x – x2) + c3(x+x2) = 0+0x+0x2
2 0 0
1
c1+2c2
=0
G. J.
 c1+5c2+c3 = 0   1 5 1 0 
 2  1 1 0
–2c1 – c2+c3 = 0
1 2 0 0

1 
1
1
0

0 0 03 0


 This system has infinitely many solutions.
(i.e., This system has nontrivial solutions.)
 S is linearly dependent.
Elementary Linear Algebra: Section 4.4, p.214
(Ex: c1=2 , c2= – 1 , c3=3)
47/107

Ex 10: (Testing for linearly independent)
Determine whether the following set of vectors in 2×2
matrix space is L.I. or L.D.
2 1 3 0 1 0 
S  
,
,




0 1 2 1 2 0 
Sol:
v1
v2
v3
c1v1+c2v2+c3v3 = 0
2 1
3 0
1 0 0 0
c1 
 c2 
 c3 





0
1
2
1
2
0
0
0





 

Elementary Linear Algebra: Section 4.4, p.215
48/107

2c1+3c2+ c3 = 0
c1
=0
2c2+2c3 = 0
c1 + c2
=0
  2 3 1 0
1 0 0 0 


0 2 2 0


1 1 0 0 
1
0
- Jordan Eliminatio n
Gauss


 
0

0
0
1
0
0
0
0
1
0
0
0
0

0
 c1 = c2 = c3= 0 (This system has only the trivial solution.)
 S is linearly independent.
Elementary Linear Algebra: Section 4.4, p.215
49/107

Thm 4.8: (A property of linearly dependent sets)
A set S = {v1,v2,…,vk}, k2, is linearly independent if and
only if at least one of the vectors vj in S can be written as
a linear combination of the other vectors in S.
Pf:
() c1v1+c2v2+…+ckvk = 0
 S is linearly dependent
 ci  0 for some i
ci 1
ci 1
ck
c1
 v i  v1   
v i 1 
v i 1    v k
ci
ci
ci
ci
Elementary Linear Algebra: Section 4.4, p.217
50/107
()
Let
vi = d1v1+…+di-1vi-1+di+1vi+1+…+dkvk
 d1v1+…+di-1vi-1-vi+di+1vi+1+…+dkvk = 0
 c1=d1, …,ci-1=di-1, ci=-1,ci+1=di+1,…, ck=dk (nontrivial solution)
 S is linearly dependent

Corollary to Theorem 4.8:
Two vectors u and v in a vector space V are linearly dependent
if and only if one is a scalar multiple of the other.
Elementary Linear Algebra: Section 4.4, p.217
51/107
Keywords in Section 4.4:

linear combination：線性組合

spanning set：生成集合

trivial solution：顯然解

linear independent：線性獨立

linear dependent：線性相依
52/107
4.5 Basis and Dimension

Basis:
V：a vector space
S ={v1, v2, …, vn}V
Generating
Sets
Bases
Linearly
Independent
Sets
(a) S spans V (i.e., span(S) = V )

(b) S is linearly independent
 S is called a basis for V
 Notes:
(1) Ø is a basis for {0}
(2) the standard basis for R3:
{i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)
Elementary Linear Algebra: Section 4.5, p.221
53/107
n
(3) the standard basis for R :
{e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1)
Ex: R4
{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}
(4) the standard basis for mn matrix space:
{ Eij | 1im , 1jn }
Ex: 2  2 matrix space:
1 0 0 1 0 0 0 0 
,
,
,





0
0
0
0
1
0
0
1

 
 


(5) the standard basis for Pn(x):
{1, x, x2, …, xn}
Ex: P3(x)
{1, x, x2, x3}
Elementary Linear Algebra: Section 4.5, p.224
54/107

Thm 4.9: (Uniqueness of basis representation)
If S  v1, v 2 ,, v n  is a basis for a vector space V, then every
vector in V can be written in one and only one way as a linear
combination of vectors in S.
Pf:
 1. span(S) = V
 S is a basis  
 2. S is linearly independent
 span(S) = V Let v = c1v1+c2v2+…+cnvn
v = b1v1+b2v2+…+bnvn
 0 = (c1–b1)v1+(c2 – b2)v2+…+(cn – bn)vn
 S is linearly independen t
 c1= b1 , c2= b2 ,…, cn= bn
Elementary Linear Algebra: Section 4.5, p.224
(i.e., uniqueness)
55/107

Thm 4.10: (Bases and linear dependence)
If S  v1, v 2 ,, v n  is a basis for a vector space V, then every
set containing more than n vectors in V is linearly dependent.
Pf:
Let S1 = {u1, u2, …, um} , m > n
 span(S )  V
u1  c11 v1  c21 v 2    cn1 v n
uiV

u 2  c12 v1  c22 v 2    cn 2 v n

u m  c1m v1  c2 m v 2    cnm v n
Elementary Linear Algebra: Section 4.5, p.225
56/107
Let k1u1+k2u2+…+kmum= 0
 d1v1+d2v2+…+dnvn= 0
(where di = ci1k1+ci2k2+…+cimkm)
 S is L.I.
 di=0
i
i.e.
c11k1  c12 k 2    c1m k m  0
c21k1  c22 k 2    c2 m k m  0

cn1k1  cn 2 k 2    cnm k m  0
 Thm 1.1: If the homogeneous system has fewer equations
than variables, then it must have infinitely many solution.
m > n  k1u1+k2u2+…+kmum = 0 has nontrivial solution
 S1 is linearly dependent
Elementary Linear Algebra: Section 4.5, p.225
57/107

Thm 4.11: (Number of vectors in a basis)
If a vector space V has one basis with n vectors, then every
basis for V has n vectors. (All bases for a finite-dimensional
vector space has the same number of vectors.)
Pf:
S ={v1, v2, …, vn}
S'={u1, u2, …, um}
two bases for a vector space

S is a basis  Thm .4.10
  n  m
S ' is L.I. 

n  m
Thm
.
4
.
10
S is L.I. 
  n  m

S ' is a basis 
Elementary Linear Algebra: Section 4.5, p.226
58/107


Finite dimensional:
A vector space V is called finite dimensional,
if it has a basis consisting of a finite number of elements.
Infinite dimensional:
If a vector space V is not finite dimensional,
then it is called infinite dimensional.

Dimension:
The dimension of a finite dimensional vector space V is
defined to be the number of vectors in a basis for V.
V: a vector space
 dim(V) = #(S)
S: a basis for V
(the number of vectors in S)
Elementary Linear Algebra: Section 4.5, p.227
59/107
dim(V) = n
 Notes:
(1) dim({0}) = 0 = #(Ø)
Generating
Sets
#(S) > n
(2) dim(V) = n , SV
Bases
Linearly
Independent
Sets
#(S) = n
#(S) < n
S：a generating set  #(S)  n
S：a L.I. set
 #(S)  n
S：a basis
 #(S) = n
(3) dim(V) = n , W is a subspace of V  dim(W)  n
Elementary Linear Algebra: Section 4.5, Addition
60/107

Ex:
(1) Vector space Rn
 basis {e1 , e2 ,  , en}
 dim(Rn) = n
(2) Vector space Mmn  basis {Eij | 1im , 1jn}
 dim(Mmn)=mn
(3) Vector space Pn(x)  basis {1, x, x2,  , xn}
 dim(Pn(x)) = n+1
(4) Vector space P(x)  basis {1, x, x2, }
 dim(P(x)) = 
Elementary Linear Algebra: Section 4.5, Addition
61/107

Ex 9: (Finding the dimension of a subspace)
(a) W={(d, c–d, c): c and d are real numbers}
(b) W={(2b, b, 0): b is a real number}
Sol: (Note: Find a set of L.I. vectors that spans the subspace)
(a) (d, c– d, c) = c(0, 1, 1) + d(1, – 1, 0)
 S = {(0, 1, 1) , (1, – 1, 0)} (S is L.I. and S spans W)
 S is a basis for W
 dim(W) = #(S) = 2
(b)  2b, b,0  b2,1,0
 S = {(2, 1, 0)} spans W and S is L.I.
 S is a basis for W
 dim(W) = #(S) = 1
Elementary Linear Algebra: Section 4.5, p.228
62/107

Ex 11: (Finding the dimension of a subspace)
Let W be the subspace of all symmetric matrices in M22.
What is the dimension of W?
Sol:


a b 

W  
a , b, c  R 



 b c 

a b 
1 0 0 1 0 0

 a
 b
 c




b
c
0
0
1
0
0
1



 
 

1 0 0 1 0 0 
 S  
,
,
 spans W and S is L.I.



0 0 1 0 0 1 
 S is a basis for W
 dim(W) = #(S) = 3
Elementary Linear Algebra: Section 4.5, p.229
63/107

Thm 4.12: (Basis tests in an n-dimensional space)
Let V be a vector space of dimension n.
(1) If S  v1 , v 2 ,, v n  is a linearly independent set of
vectors in V, then S is a basis for V.
(2) If S  v1 , v 2 ,, v n  spans V, then S is a basis for V.
dim(V) = n
Generating
Sets
Bases
Linearly
Independent
Sets
#(S) > n
#(S) = n
Elementary Linear Algebra: Section 4.5, p.229
#(S) < n
64/107
Keywords in Section 4.5:

basis：基底

dimension：維度

finite dimension：有限維度

infinite dimension：無限維度
65/107
Let A be an m×n matrix.
 Row space:
The row space of A is the subspace of Rn spanned by
the row vectors of A.
RS( A)  {1 A(1)   2 A( 2)  ...   m A( m) | 1 ,  2 ,..., m  R}
 Column space:
The column space of A is the subspace of Rm spanned by
the column vectors of A.

CS  A  {1 A(1)  2 A(2)   n A( n) 1, 2 ,n  R}
Null space:
The null space of A is the set of all solutions of Ax=0 and
it is a subspace of Rn.
NS ( A)  {x  Rn | Ax  0}
Elementary Linear Algebra: Section 4.6, p.233
66/107
4.6 Rank of a Matrix and Systems of Linear Equations
 row vectors:
 a11
a
A   21
 

am1
a12
a22

am 2
Row vectors of A
 a1n   A1 


 a2 n   A2  

    

 
 amn   Am  
[ a11 , a12 ,  , a1n ]  A(1)
[ a21 , a22 ,  , a2n ]  A(2)

[ am1 , am2 ,  , amn ]  A( n )
 column vectors:
 a11
a
A   21
 

am1
a12
a22

am 2
 a1n 
 a2 n 
 A1  A2   An 
 

 amn 
Elementary Linear Algebra: Section 4.6, p.232

Column vectors of A
 a11   a12   a1n 
a   a  a 
 21   22    2 n 
       
    
am1  am 2  amn 
||
||
||
(1)
(2)
(n)
A
A
A

67/107

Thm 4.13: (Row-equivalent matrices have the same row space)
If an mn matrix A is row equivalent to an mn matrix B,
then the row space of A is equal to the row space of B.

Notes:
(1) The row space of a matrix is not changed by elementary
row operations.
RS(r(A)) = RS(A)
r: elementary row operations
(2) Elementary row operations can change the column space.
Elementary Linear Algebra: Section 4.6, p.233
68/107

Thm 4.14: (Basis for the row space of a matrix)
If a matrix A is row equivalent to a matrix B in row-echelon
form, then the nonzero row vectors of B form a basis for the
row space of A.
Elementary Linear Algebra: Section 4.6, p.234
69/107
 Ex 2: ( Finding a basis for a row space)
 1
 0

Find a basis of row space of A =  3

 3
 2
Sol:
 1
 0

 3
A= 
 3
 2
a1
3
1
1
0
1
6
4 2
0 4
a2 a3
3
0
 1

1
2
a4
Elementary Linear Algebra: Section 4.6, p.234
3
1
1
0
1
6
4 2
0 4
1
0

.E.
0
G

B= 
0
0
b1
3
0
 1

1
2
3 1 3 w 1
1 1 0 w 2
0 0 1 w 3

0 0 0
0 0 0
b2 b3 b4
70/107
a basis for RS(A) = {the nonzero row vectors of B} (Thm 4.14)
= {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}

Notes:
(1) b3  2b1  b2  a3  2a1  a2
(2) {b1 , b2 , b4 } is L.I.  {a1 , a2 , a4 } is L.I.
Elementary Linear Algebra: Section 4.6, p.234
71/107

Ex 3: (Finding a basis for a subspace)
Find a basis for the subspace of R3 spanned by
v1
v3
v2
S  {(1, 2, 5), (3, 0, 3), (5,1, 8)}
Sol:
  1 2 5 v 1
A =  3 0 3 v 2
 5 1 8 v 3
G.E.
 1  2  5 w 1
B  0 1 3 w 2
0 0 0
a basis for span({v1, v2, v3})
= a basis for RS(A)
= {the nonzero row vectors of B}
= {w1, w2}
= {(1, –2, – 5) , (0, 1, 3)}
Elementary Linear Algebra: Section 4.6, p.235
(Thm 4.14)
72/107

Ex 4: (Finding a basis for the column space of a matrix)
Find a basis for the column space of the matrix A given in Ex 2.
 1
 0

A   3

 3
 2
3
1
1
0
1
6
4 2
0 4
3
0
 1

1
 2
Sol. 1:
1
3
AT  
1

3
0 3
3
2
1
0
1 0
4
0
.E.
G
B
0
1 6  2  4


0 1
1  2
0
Elementary Linear Algebra: Section 4.6, p.236
0 3
3
2 w1
1 9  5  6 w 2
0
1  1  1 w3

0
0
0
0
73/107
 CS(A)=RS(AT)
 a basis for CS(A)
= a basis for RS(AT)
= {the nonzero vectors of B}
= {w1, w2, w3}
 1 
 
 0 

  3,
 3 
 
 2 
 0   0 
 1   0 
    
 9 ,  1   (a basis for the column space of A)
   
  5  1 
 6  1 

 Note: This basis is not a subset of {c1, c2, c3, c4}.
Elementary Linear Algebra: Section 4.6, p.236
74/107

Sol. 2:
1
0

A   3

3
 2
c1
Leading 1
3
1
0
4
0
c2
3
1
0
1
0 

G.E .
 B  0
6  1  


2 1 
0
0
 4  2
c3 c4
v1
1
3 1 3
1 1 0
0 0 1

0 0 0
0 0 0
v2 v3 v4
=> {v1, v2, v4} is a basis for CS(B)
{c1, c2, c4} is a basis for CS(A)
 Notes:
(1) This basis is a subset of {c1, c2, c3, c4}.
(2) v3 = –2v1+ v2, thus c3 = – 2c1+ c2 .
Elementary Linear Algebra: Section 4.6, p.236
75/107

Thm 4.16: (Solutions of a homogeneous system)
If A is an mn matrix, then the set of all solutions of the
homogeneous system of linear equations Ax = 0 is a subspace
of Rn called the nullspace of A.
Pf:
NS ( A)  {x  Rn | Ax  0}
NS ( A)  R n
NS ( A)   ( A0  0)
Let x1, x 2  NS ( A) (i.e. Ax1  0, Ax 2  0)
Then
(1) A(x1  x 2)  Ax1  Ax 2  0  0  0
Addition
(2) A(cx1)  c( Ax1)  c(0)  0
Scalar multiplication
Thus NS ( A) is a subspace of R n
 Notes: The nullspace of A is also called the solution space of
the homogeneous system Ax = 0.
Elementary Linear Algebra: Section 4.6, p.239
76/107
 Ex 6: (Finding the solution space of a homogeneous system)
Find the nullspace of the matrix A.
1 2  2 1 
A  3 6  5 4
1 2 0 3
Sol: The nullspace of A is the solution space of Ax = 0.
1 2  2 1 
1 2 0 3
.J .E
A  3 6  5 4 G
0 0 1 1
1 2 0 3
0 0 0 0
 x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t
 x1   2s  3t 
 2  3
x   s 
 1  0
2
  s    t    sv 1  t v 2
x 
 x3    t 
 0   1
  

   
x

 0  1
 4  t
 NS ( A)  {sv1  tv 2 | s, t  R}
Elementary Linear Algebra: Section 4.6, p.240
77/107

Thm 4.15: (Row and column space have equal dimensions)
If A is an mn matrix, then the row space and the column
space of A have the same dimension.
dim(RS(A)) = dim(CS(A))
 Rank:
The dimension of the row (or column) space of a matrix A
is called the rank of A and is denoted by rank(A).
rank(A) = dim(RS(A)) = dim(CS(A))
Elementary Linear Algebra: Section 4.6, pp.237-238
78/107
 Nullity:
The dimension of the nullspace of A is called the nullity of A.
nullity(A) = dim(NS(A))
 Note: rank(AT) = rank(A)
Pf:
rank(AT) = dim(RS(AT)) = dim(CS(A)) = rank(A)
Elementary Linear Algebra: Section 4.6, p.238
79/107

Thm 4.17: (Dimension of the solution space)
If A is an mn matrix of rank r, then the dimension of
the solution space of Ax = 0 is n – r. That is
n = rank(A) + nullity(A)

Notes:
(1) rank(A): The number of leading variables in the solution of Ax=0.
(The number of nonzero rows in the row-echelon form of A)
(2) nullity (A): The number of free variables in the solution of Ax = 0.
Elementary Linear Algebra: Section 4.6, p.241
80/107

Notes:
If A is an mn matrix and rank(A) = r, then
Fundamental Space
Dimension
RS(A)=CS(AT)
r
CS(A)=RS(AT)
r
NS(A)
n–r
NS(AT)
m–r
Elementary Linear Algebra: Section 4.6, Addition
81/107

Ex 7: (Rank and nullity of a matrix)
Let the column vectors of the matrix A be denoted by a1, a2,
a3, a4, and a5.
1
0
 1 0 2
 0 1  3

1
3

A
 2  1
1 1
3


0
3
9
0

12


a1 a2 a3 a4 a5
(a) Find the rank and nullity of A.
(b) Find a subset of the column vectors of A that forms a basis for
the column space of A .
(c) If possible, write the third column of A as a linear combination
of the first two columns.
Elementary Linear Algebra: Section 4.6, p.242
82/107
Sol: Let B be the reduced row-echelon form of A.
0
 1 0 2 1
 0 1  3 1

3

A
 2  1 1  1
3


 0 3 9 0  12
a1 a2 a3 a4 a5
(a) rank(A) = 3
1
0
B
0

0
b1
0 2 0
1
1 3 0  4
0 0 1  1

0 0 0 0
b2 b3 b4 b5
(the number of nonzero rows in B)
nuillity( A)  n  rank( A)  5  3  2
Elementary Linear Algebra: Section 4.6, p.242
83/107
(b) Leading 1
 {b1 , b 2 , b 4 } is a basis for CS ( B)
{a1 , a 2 , a 4 } is a basis for CS ( A)
 1
 0
 1
 0
 1
 1
a1   , a 2   , and a 4   ,
  2
 1
 1
 0
 3
 0
 
 
 
(c) b3  2b1  3b2  a3  2a1  3a2
Elementary Linear Algebra: Section 4.6, p.243
84/107
 Thm 4.18: (Solutions of a nonhomogeneous linear system)
If xp is a particular solution of the nonhomogeneous system
Ax = b, then every solution of this system can be written in
the form x = xp + xh , wher xh is a solution of the corresponding
homogeneous system Ax = 0.
Pf:
Let x be any solution of Ax = b.
 A(x  x p )  Ax  Ax p  b  b  0.
 (x  x p )
is a solution of Ax = 0
Let xh  x  x p
 x  x p  xh
Elementary Linear Algebra: Section 4.6, p.243
85/107

Ex 8: (Finding the solution set of a nonhomogeneous system)
Find the set of all solution vectors of the system of linear equations.
Sol:
x1
3x1 
x2
x1  2 x2
 2 x3
 5x3

x4
 5x4

5

8
 9
1
5
1
5
1 0  2
1 0  2
3 1  5
 G
0 1

.J .E
0
8


1

3

7




 1 2
0 0
0  5  9
0
0
0
s
t
Elementary Linear Algebra: Section 4.6, p.243
86/107
 x1   2 s
 x   s
 x   2  
 x3   s
  
 x4   0 s

t
 3t
 0t

t
 5
 2  1  5
 1  3  7 
 7 
 s   t     
 1  0  0
 0

     
 0
 0  1  0
 su1  tu2  x p
5
 7 
i.e. x p    is a particular solution vector of Ax=b.
0
 
0
xh = su1 + tu2 is a solution of Ax = 0
Elementary Linear Algebra: Section 4.6, p.244
87/107

Thm 4.19: (Solution of a system of linear equations)
The system of linear equations Ax = b is consistent if and only
if b is in the column space of A.
Pf:
Let
 a11
a
A   21
 

 am1
a12
a22

am 2
 a1n 
 a2 n 
,
 

 amn 
 x1 
x 
x   2,

 
 xn 
and
 b1 
b 
b  2

 
bm 
be the coefficient matrix, the column matrix of unknowns,
and the right-hand side, respectively, of the system Ax = b.
Elementary Linear Algebra: Section 4.6, pp.244-245
88/107
Then
 a11 a12  a1n   x1   a11 x1  a12 x2
a
 x  a x  a x
a

a
22
2n   2 
22 2
Ax   21
  21 1
 

    

  
a
a

a
m2
mn   xn 
 m1
am1 x1  am 2 x2
 a11 
 a12 
 a1n 
a 
a 
a 
 x1  21   x2  22     xn  2 n .
  
  
  
 
 
 
am1 
 am 2 
amn 
  
a1n xn 
   a2 n xn 
 

   amn xn 
Hence, Ax = b is consistent if and only if b is a linear combination
of the columns of A. That is, the system is consistent if and only if
b is in the subspace of Rm spanned by the columns of A.
Elementary Linear Algebra: Section 4.6, p.245
89/107
 Note:
If rank([A|b])=rank(A)
Then the system Ax=b is consistent.

Ex 9: (Consistency of a system of linear equations)
x1 
x2
x1
3x1  2 x2
Sol:
 x3
 x3
 x3
 1
 3

1
1
 1 1  1
1 0
. J .E .
A   1 0
1 G
0 1  2
3 2  1
0 0
0
Elementary Linear Algebra: Section 4.6, p.245
90/107
1
3
 1 1  1  1
1 0
. J .E.
[ A  b]   1 0
1 3 G
0 1  2  4
3 2  1 1
0 0
0
0
c1 c2 c3
b
w1 w2 w3 v
 v  3w1  4w 2
 b  3c1  4c 2  0c3
(b is in the column space of A)
 The system of linear equations is consistent.
 Check:
rank( A)  rank([ A b])  2
Elementary Linear Algebra: Section 4.6, p.245
91/107

Summary of equivalent conditions for square matrices:
If A is an n×n matrix, then the following conditions are equivalent.
(1) A is invertible
(2) Ax = b has a unique solution for any n×1 matrix b.
(3) Ax = 0 has only the trivial solution
(4) A is row-equivalent to In
(5) | A | 0
(6) rank(A) = n
(7) The n row vectors of A are linearly independent.
(8) The n column vectors of A are linearly independent.
Elementary Linear Algebra: Section 4.6, p.246
92/107
Keywords in Section 4.6:

row space : 列空間

column space : 行空間

null space: 零空間

solution space : 解空間

rank: 秩

nullity : 核次數
93/107
4.7 Coordinates and Change of Basis

Coordinate representation relative to a basis
Let B = {v1, v2, …, vn} be an ordered basis for a vector space V
and let x be a vector in V such that
x  c1v1  c2 v 2   cn v n .
The scalars c1, c2, …, cn are called the coordinates of x relative
to the basis B. The coordinate matrix (or coordinate vector)
of x relative to B is the column matrix in Rn whose components
are the coordinates of x.
xB
 c1 
c 
  2

 
cn 
Elementary Linear Algebra: Section 4.7, p.249
94/107
n
 Ex 1: (Coordinates and components in R )
Find the coordinate matrix of x = (–2, 1, 3) in R3
relative to the standard basis
S = {(1, 0, 0), ( 0, 1, 0), (0, 0, 1)}
Sol:
 x  (2, 1, 3)  2(1, 0, 0)  1(0, 1, 0)  3(0, 0, 1),
 2
 [ x]S   1.
 3
Elementary Linear Algebra: Section 4.7, p.250
95/107

Ex 3: (Finding a coordinate matrix relative to a nonstandard basis)
Find the coordinate matrix of x=(1, 2, –1) in R3
relative to the (nonstandard) basis
B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)}
Sol: x  c u  c u  c u  (1, 2, 1)  c (1, 0, 1)  c (0, 1, 2)  c (2, 3,  5)
1
1
2
2
c1

c1
 c2
 2c2
3
3
1
 2c3

 3c3
 5c3


2
1
1 0
1
G.J. E.


 0  1
3 2  0
 1 2  5  1
0
Elementary Linear Algebra: Section 4.7, p.251
2
3
2  c1   1
1 0
2
i.e. 0  1
3 c2    2
 1 2  5 c3   1
1
0 0
5
 5
1 0  8  [ x ]    8
  2 
0 1  2
1
B
96/107

Change of basis problem:
You were given the coordinates of a vector relative to one
basis B and were asked to find the coordinates relative to
another basis B'.

Ex: (Change of basis)
Consider two bases for a vector space V
B  {u1 , u2}, B  {u1 , u2}
a 
c 
If [u1 ]B   , [u2 ]B   
b 
d 
i.e., u1  au1  bu2 , u2  cu1  du2
Elementary Linear Algebra: Section 4.7, p.253
97/107
 k1 
Let v V , [ v]B   
k 2 
 v  k1u1  k2u2
 k1 (au1  bu 2 )  k2 (cu1  du 2 )
 (k1a  k2c)u1  (k1b  k2 d )u 2
 k1a  k 2 c  a c   k1 
 [ v ]B  


 k 
k
b

k
d
b
d

 2 
2 
 1
 u1 B u2 B  v B
Elementary Linear Algebra: Section 4.7, p.253
98/107

Transition matrix from B' to B:
Let B  {u1 , u 2 ,..., u n } and B  {u1 , u2 ..., un } be two bases
for a vector space V
If [v]B is the coordinate matrix of v relative to B
[v]B‘ is the coordinate matrix of v relative to B'
then [ v]B  P[ v]B
 u1 B , u2 B ,...,un B  vB
where
P  u1 B , u2 B , ...,un B 
is called the transition matrix from B' to B
Elementary Linear Algebra: Section 4.7, p.255
99/107

Thm 4.20: (The inverse of a transition matrix)
If P is the transition matrix from a basis B' to a basis B in Rn,
then
(1) P is invertible
–1
(2) The transition matrix from B to B' is P

Notes:
B  {u1 , u 2 , ..., u n }, B'  {u1 , u2 , ..., un }
vB  [u1 ]B , [u2 ]B , ..., [un ]B  vB  P vB
vB  [u1 ]B , [u 2 ]B , ..., [u n ]B  vB  P 1 vB
Elementary Linear Algebra: Section 4.7, p.253
100/107

Thm 4.21: (Transition matrix from B to B')
Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases
n
for R . Then the transition matrix P–1 from B to B' can be found
by using Gauss-Jordan elimination on the n×2n matrix B B
as follows.
B B
I
n
 P 1
Elementary Linear Algebra: Section 4.7, p.255

101/107
 Ex 5: (Finding a transition matrix)
B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R2
(a) Find the transition matrix from B' to B.
1 
(b) Let [ v]B '   , find [ v]B
 2
(c) Find the transition matrix from B to B' .
Elementary Linear Algebra: Section 4.7, p.257
102/107
Sol:
4
(a)  3
 2 2

B
(b)
Check :
 1
2
 2  2
B'
3  2
P  

2

1


G.J.E.
 1 0  3  2
0 1  2  1


I
P
(the transition matrix from B' to B)
 1
3  2 1  1
[v ]B     [v ]B  P [v ]B  
 



2
2  1 2  0 
 1
[v ]B     v  (1)(1,2)  (2)(2,2)  (3,2)
 2
 1
[v ]B     v  (1)(3,2)  (0)(4,2)  (3,2)
 0
Elementary Linear Algebra: Section 4.7, p.258
103/107
(c)
2  3
4
 1
 2  2  2  2


B'
B
  1 2
P  


2
3


1

G.J.E.
 1 0   1 2
0 1   2 3


-1
I
P
(the transition matrix from B to B')
Check:
 3  2   1 2  1 0
PP  

 I2




2  1  2 3 0 1
1
Elementary Linear Algebra: Section 4.7, p.259
104/107
 Ex 6: (Coordinate representation in P3(x))
(a) Find the coordinate matrix of p = 3x3-2x2+4 relative to the
standard basis S = {1, x, x2, x3} in P3(x).
(b) Find the coordinate matrix of p = 3x3-2x2+4 relative to the
basis S = {1, 1+x, 1+ x2, 1+ x3} in P3(x).
Sol:
 4
 0
(a) p  (4)(1) (0)(x)  (-2)(x 2 )  (3)(x 3 )  [ p]B   
  2
 
 3
 3
 0
(b) p  (3)(1) (0)(1 x)  (-2)(1 x 2 )  (3)(1 x 3 )  [ p]B   
  2
 
 3
Elementary Linear Algebra: Section 4.7, p.259
105/107
 Ex: (Coordinate representation in M2x2)
Find the coordinate matrix of x = 5 6 relative to
the standard basis in M2x2.
Sol:
7 8 


B = 1 0, 0 1, 0 0, 0 0 

 0 0 1 0 0 1 
0
0





5 6
1 0 0 1 0 0 0 0
x
 5
 6
 7
 8





7 8 
0 0 0 0 1 0 0 1
5 
6 
 x B   
7 
 
8 
Elementary Linear Algebra: Section 5.7, Addition
106/107
Keywords in Section 4.7:

coordinates of x relative to B：x相對於B的座標

coordinate matrix：座標矩陣

coordinate vector：座標向量

change of basis problem：基底變換問題

transition matrix from B' to B：從 B' 到 B的轉移矩陣
107/107