Transcript Linear Algebra Chapter 4 General Vector Spaces 大葉大學 資訊工程系 黃鈴玲 4.1 General Vector Spaces and Subspaces Our aim in this section will be to focus.

Linear Algebra
Chapter 4
General Vector Spaces
大葉大學 資訊工程系
黃鈴玲
4.1 General Vector Spaces and Subspaces
Our aim in this section will be to focus on the algebraic properties
of Rn. We draw up a set of axioms (公理) based on the properties
of Rn. Any set that satisfies these axioms will have similar
algebraic properties to the vector space Rn.
Definition
A vector space is a set V of elements called vectors, having
operations of addition and scalar multiplication defined on it
that satisfy the following conditions. (u, v, and w are arbitrary
elements of V, and c and d are scalars.)
Closure Axioms (最重要！)
1. The sum u + v exists and is an element of V. (V is closed
under addition.)
2. cu is an element of V. (V is closed under scalar multiplication.)
Ch04_2
補充範例
(1) V={ …, -3, -1, 1, 3, 5, 7, …}
(所有奇數構成的集合)
V is not closed under addition because
1+3=4  V.
(2)
Z={ …, -2, -1, 0, 1, 2, 3, 4, …}
(所有整數構成的集合)
Z is closed under addition because
for any a, b  Z, a + b Z.
Z is not closed under scalar multiplication
because ½ is a scalar, for any odd a  Z, (½)a  Z.
隨堂作業：14
Ch04_3
Definition of Vector Space (continued)
Addition Axioms
3. u + v = v + u (commutative property)
4. u + (v + w) = (u + v) + w (associative property)
5. There exists an element of V, called the zero vector, denoted 0,
such that u + 0 = u.
6. For every element u of V there exists an element called the
negative of u, denoted -u, such that u + (-u) = 0.
Scalar Multiplication Axioms
7. c(u + v) = cu + cv
8. (c + d)u = cu + du
9. c(du) = (cd)u
10.1u = u
Ch04_4
A Vector Space in R3
補充：
Let W  { a(1, 0, 1) | a  R }. Prove that W is a vector space.
Proof
Let u  a(1,0,1) and v  b(1,0,1) W , for some a, bR.
Axiom 1: u  v  a(1,0,1)  b(1,0,1)  (a  b)(1,0,1)
u + v W. Thus W is closed under addition.
Axiom 2: cu  ca(1,0,1) W.
Thus W is closed under scalar multiplication.
Axiom 5: Let 0 = (0, 0, 0) = 0(1,0,1),
then 0 W and 0+u = u+0 = u for any u  W.
Axiom 6: For any u = a(1,0,1)  W. Let -u = -a(1,0,1),
then -u W and (-u)+u = 0.
隨堂作業：5
Axiom 3,4 and 7~10: trivial
Ch04_5
Vector Spaces of Matrices
 p q
Let M 22  { 
| p, q, r , s  R }. Prove that M22 is a vector space.

 r s
Proof
a b 
e
Let u  
and v  

c d 
g
f
 M 22 .

h
Axiom 1:
a b  e f  a  e b  f 

uv 


 c d   g h  c  g d  h 
u + v is a 2  2 matrix. Thus M22 is closed under addition.
Axiom 3 and 4:
From our previous discussions we know that 2  2 matrices are
communicative and associative under addition (Theorem 2.2).
Ch04_6
Axiom 5:
0 0

The 2  2 zero matrix is 0  0 0 , since


a b  0 0   a b 

u0 


u
 c d  0 0  c d 
Axiom 6:
a b
- a - b


If u 
, then - u 
, since
 c d 
 - c - d 
a b   - a - b   a - a b - b  0 0 
u  (-u)  



0
 c d   - c - d   c - c d - d  0 0
推廣：The set of m  n matrices, Mmn, is a vector space.
 p q
隨堂作業：7 W  { 
| p, q, r , s  0}.

 r s
Ch04_7
Vector Spaces of Functions
(跳過)
Prove that V={ f | f: R R } is a vector space.
Let f, g  V, c  R.
For example: f: R R, f(x)=2x,
g: R R, g(x)=x2+1.
Axiom 1:
f + g is defined by (f + g)(x) = f(x) + g(x).
 f + g : R R
 f + g  V. Thus V is closed under addition.
Axiom 2:
cf is defined by (cf)(x) = c f(x).
 cf : R R
 cf  V. Thus V is closed under scalar multiplication.
Ch04_8
Vector Spaces of Functions
(跳過)
Axiom 5:
Let 0 be the function such that 0(x) = 0 for every xR.
0 is called the zero function.
We get (f + 0)(x) = f(x) + 0(x) = f(x) + 0 = f(x) for every xR.
Thus f + 0 = f.
0 is the zero vector.
Axiom 6:
Let the function –f defined by (-f )(x) = -f (x).
[ f  (- f )](x)  f ( x)  (- f )(x)
 f ( x) - [ f ( x)]
0
 0( x)
Thus [f + (-f )] = 0, -f is the negative of f.
隨堂作業：13 V={ f | f(x)=ax2+bx+c for some a,b,c R}
Ch04_9
The Complex Vector Space Cn
(跳過)
Let (u1 , ..., un ) be a sequence of n complexnumbers.
T heset of all such sequences is denotedC n .
Let operat ionsof addition and scalar multiplication
(by a complexscalar c) be defined on C n as follows:
(u1 , ..., un )  (v1 , ..., vn )  (u1  v1 , ..., un  vn )
c(u1 , ..., un )  (cu1 , ..., cun )
It can be shown that Cn with these two operations is a
complex vector space.
Ch04_10
Theorem 4.1
Let V be a vector space, v a vector in V, 0 the zero vector of V,
c a scalar, and 0 the zero scalar. Then
(a) 0v = 0
(b) c0 = 0
(c) (-1)v = -v
(d) If cv = 0, then either c = 0 or v = 0.
Proof
(a) 0v + 0v = (0 + 0)v = 0v
(0v + 0v) + (-0v) = 0v + (-0v)
0v + [0v + (-0v)] = 0, 0v + 0 = 0, 0v = 0
(c) (-1)v + v = (-1)v + 1v
= [(-1) + 1]v
= 0v = 0
Ch04_11
Subspaces
In general, a subset of a vector space may or may not satisfy the
closure axioms. However, any subset that is closed under both
of these operations satisfies all the other vector space properties.
Definition
Let V be a vector space and U be a nonempty subset of V. If U is
a vector space under the operations of addition and scalar
multiplication of V it is called a subspace of V.
Note. U只要有加法與純量乘法的封閉性，其他axiom都會滿足。
Ch04_12
Example 1
Let W be the subset of R3 consisting of all vectors of the form
(a, a, b). Show that W is a subspace of R3.
Solution
Let (a, a, b), (c, c, d)  W, and let k R.
We get
(a, a, b) + (c, c, d) = (a+c, a+c, b+d)  W
k(a, a, b) = (ka, ka, kb)  W
Thus W is a subspace of R3.
Ch04_13
Example 1’
Let W be the set of vectors of the form (a, a2, b).
Show that W is not a subspace of R3.
Solution
Let (a, a2, b), (c, c2, d)  W.
(a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)
 (a + c, (a + c)2, b + d)
Thus (a, a2, b) + (c, c2, d)  W.
W is not closed under addition.
W is not a subspace.
隨堂作業：18(a), 20(b)
Ch04_14
Example 2
Prove that the set U of 2  2 diagonal matrices is a subspace of
the vector space M22 of 2  2 matrices.
Solution
(+) Let u  a 0 and v   p 0  U.
0 b
 0 q 
a 0  p 0   a  p
0 



We get u  v  



0
b
0
q
b  q 

 
  0
 u + v  U.
 U is closed under addition.
a 0 ca 0 
() Let c R. We get cu  c 

0 b  0 cb 
 cu  U.
 U is closed under scalar multiplication.
 U is a subspace of M22.
隨堂作業：27(a)
Ch04_15
Example 3
(跳過)
Let Pn denoted the set of real polynomial functions of degree  n.
Prove that Pn is a vector space if addition and scalar
multiplication are defined on polynomials in a pointwise manner.
Solution
Let f and g  Pn, where
f ( x)  an x n  an-1 x n-1  ...  a1 x  a0 and
g ( x)  bn x n  bn-1 x n -1  ...  b1 x  b0
(+) ( f  g )(x)
 f ( x)  g ( x)
 [an x n  an -1 x n -1  ...  a1 x  a0 ]  [bn x n  bn -1 x n -1  ...  b1 x  b0 ]
 (an  bn ) x n  (an -1  bn -1 ) x n -1  ...  (a1  b1 ) x  (a0  b0 )
(f + g)(x) is a polynomial of degree  n.
Thus f + g  Pn.
Pn is closed under addition.
Ch04_16
(跳過)
() Let c R, (cf )( x)  c[ f ( x)]
 c[an x n  an -1 x n -1  ...  a1 x  a0 ]
 can x n  can -1 x n -1  ...  ca1 x  ca0
(cf)(x) is a polynomial of degree  n.
Thus cf  Pn.
Pn is closed under scalar multiplication.
By (+) and (), Pn is a subspace of the vector space V of
functions. Therefore Pn is a vector space.
Ch04_17
Theorem 4.2
Let U be a subspace of a vector space V.
U contains the zero vector of V.
Proof
Let u be an arbitrary vector in U and 0 be the zero vector of V.
Let 0 be the zero scalar.
By Theorem 4.5(a) we know that 0u = 0.
Since U is closed under scalar multiplication,
this means that 0 is in U.
Note. Let 0 be the zero vector of V, U is a subset of V.
If 0 U  U is not a subspace of V.
If 0 U  check (+)() to determine if U is a
subspace of V.
Ch04_18
Example 4
Let W be the set of vectors of the form (a, a, a+2).
Show that W is not a subspace of R3.
Solution
If (a, a, a+2) = (0, 0, 0),
then a = 0 and a + 2 = 0. 
 (0, 0, 0)  W.
 W is not a subspace of R3.
隨堂作業：24(a,b), 27(b)
Ch04_19
Homework
Exercise 4.1:
5, 7, 14, 18, 19, 20, 24, 27
Ch04_20
4.2 Linear Combinations
W={(a, a, b) | a,b R}  R3
(a, a, b) = a (1,1,0) + b (0,0,1)
W中的任何向量都可以用 (1,1,0) 及 (0,0,1)來表示
e.g., (2, 2, 3) = 2 (1,1,0) + 3 (0,0,1)
(-1, -1, 7) = -1 (1,1,0) + 7 (0,0,1)
Definition
Let v1, v2, …, vm be vectors in a vector space V. The vector v in V
is a linear combination (線性組合) of v1, v2, …, vm if there exist
scalars c1, c2, …, cm such that v can be written
v = c1v1 + c2v2 + … + cmvm
W中的任何向量都是 (1,1,0) 及 (0,0,1) 的 linear combination.
Ch04_21
Example
The vector (7, 3, 2) is a linear combination of the vector (1, 3, 0)
and (2, -3, 1) since
(7, 3, 2) = 3(1, 3, 0) + 2(2, -3, 1)
The vector (3, 4, 2) is not a linear combination of (1, 1, 0) and
(2, 3, 0) because there are no values of c1 and c2 for which
(3, 4, 2) = c1(1, 1, 0) + c2(2, 3, 0)
is true.
Ch04_22
Example 1
Determine whether or not the vector (8, 0, 5) is a linear
combination of the vectors (1, 2, 3), (0, 1, 4), and (2, -1, 1).
Solution
Suppose c1(1, 2, 3)+c2(0, 1, 4)+c3(2, -1, 1)=(8, 0, 5).
 2c3  8
 c1

 2c1  c2 - c3  0
3c1  4c2  c3  5
 c1  2, c2  -1, c3  3
Thus (8, 0, 5) is a linear combination of (1, 2, 3), (0, 1, 4), and
(2, -1, 1).
隨堂作業：2(a)
Ch04_23
決定一個向量是否是某些向量的linear combination
 求解聯立方程式
 唯一解表示 linear combination 的係數唯一
無限多解表示 linear combination 的係數不唯一
無解表示不是 linear combination
Ch04_24
Example 2
Determine whether the vector (4, 5, 5) is a linear combination of
the vectors (1, 2, 3), (-1, 1, 4), and (3, 3, 2).
Solution
Suppose c1 (1, 2, 3)  c2 (-1, 1, 4)  c3 (3, 3, 2)  (4, 5, 5)
 c1 - c2  3c3  4

 2c1  c2  3c3  5  c1  -2r  3, c2  r - 1, c3  r
3c  4c  2c  5
2
3
 1
Thus (4, 5, 5) can be expressed in many ways as a linear
combination of (1, 2, 3), (-1, 1, 4), and (3, 3, 2):
(4, 5, 5)  (-2r  3)(1, 2, 3)  (r - 1)(-1, 1, 4)  r (2, 3, 6)
Ch04_25
Example 2’
Show that the vector (3, -4, -6) cannot be expressed as a linear
combination of the vectors (1, 2, 3), (-1, -1, -2), and (1, 4, 5).
Solution
Suppose
c1 (1, 2, 3)  c2 (-1, - 1, - 2)  c3 (1, 4, 5)  (3, - 4, - 6)

 c1 - c 2  c3  3

 2c1 - c 2  4c3  -4
3c - 2c  5c  - 6
2
3
 1
This system has no solution.
Thus (3, -4, -6) is not a linear combination of (1, 2, 3),
(-1, -1, -2), and (1, 4, 5).
隨堂作業：2(c)
Ch04_26
Spanning a Vector Space
Definition
Let v1, v2, …, vm be vectors in a vector space V. These vectors
span V if every vector in V can be expressed as a linear
combination of them.
{v1, v2, …, vm} is called a spanning set of V.
Ch04_27
Example 3
Show that the vectors (1, 2, 0), (0, 1, -1), and (1, 1, 2) span R3.
Solution
Let (x, y, z) be an arbitrary element of R3.
Suppose ( x, y, z )  c1 (1, 2, 0)  c2 (0, 1, - 1)  c3 (1, 1, 2)
 ( x, y, z)  (c1  c3 , 2c1  c2  c3 , - c2  2c3 )
 c3  x
 c1

 2c1  c 2  c3  y

- c 2  2c 3  z


c1  3 x - y - z

c 2  -4 x  2 y  z
c  -2 x  y  z
 3
 ( x, y, z )  (3x - y - z )(1,2,0)  (-4 x  2 y  z )(0, 1, - 1)  (-2 x  y  z )(1, 1, 2)
The vectors (1, 2, 0), (0, 1, -1), and (1, 1, 2) span R3.
隨堂作業：4(a)
Ch04_28
Theorem 4.3
Let v1, …, vm be vectors in a vector space V. Let U be the set
consisting of all linear combinations of v1, …, vm . Then U is a
subspace of V spanned the vectors v1, …, vm .
U is said to be the vector space generated by v1, …, vm.
It is denoted Span{v1, …, vm}.
Proof
(+) Let u1 = a1v1 + … + amvm and u2 = b1v1 + … + bmvm  U.
Then u1 + u2 = (a1v1 + … + amvm) + (b1v1 + … + bmvm)
= (a1 + b1) v1 + … + (am + bm) vm
 u1 + u2 is a linear combination of v1, …, vm .
 u1 + u2  U.
 U is closed under vector addition.
Ch04_29
()Let c R. Then
cu1 = c(a1v1 + … + amvm)
= ca1v1 + … + camvm)
 cu1 is a linear combination of v1, …, vm .
 cu1  U.
 U is closed under scalar multiplication.
Thus U is a subspace of V.
By the definition of U, every vector in U can be written as a
linear combination of v1, …, vm . Thus v1, …, vm span U.
Ch04_30
Example 4
Consider the vectors (-1, 5, 3) and (2, -3, 4) in R3.
Let U =Span{(-1, 5, 3), (2, -3, 4)}. U will be a subspace of R3
consisting of all vectors of the form
c1(-1, 5, 3) + c2(2, -3, 4).
The following are examples of some of the vectors in U, which
are obtained by given c1 and c2 various values.
c1  1, c2  1; vector(1, 2, 7)
c1  2, c2  3; vector(4, 1, 18)
We can visualize U. U is made up of all vectors in the plane
defined by the vectors (-1, 5, 3) and (2, -3, 4).
Ch04_31
Figure 4.1
Ch04_32
We can generalize this result.
Let v1 and v2 be vectors in the
space R3.
The subspace U generated by
v1 and v2 is the set of all
vectors of the form c1v1 + c2v2.
If v1 and v2 are not colinear,
then U is the plane defined by
v1 and v2 .
Figure 4.2
Ch04_33
Example 5
Let v1 and v2 span a subspace U of a vector V. Let k1 and k2 be
nonzero scalars. Show that k1v1 and k2v2 also span U.
Solution
Choose any vector v  U.
Since v1 and v2 span U,
 There exist a, b  R such that v = av1 + bv2
a
b
We can write v  (k1v1 )  (k2 v 2 )
k1
k2
 k1v1 and k2v2 span U.
隨堂作業：20
Ch04_34
Example 6
- 1 7
Determine whether the matrix 
is a linear combination
 8 - 1
of the matrices 1 0, 2 - 3, and 0 1 in the vector space
2 1 0 2 
2 0
M22 of 2  2 matrices.
Solution
Suppose
Then
1 0
2 - 3
0 1 - 1 7 



c1
c
c

2 1 2 0 2  3 2 0  8 - 1
 c1  2c2
2c  2c
2
 1
- 3c2  c3  - 1 7 


c1  2c2   8 - 1
Ch04_35
 c1  2c2  -1

- 3c2  c3  7

 2c1  2c3  8
 c1  2c2  -1
This system has the unique solution c1 = 3, c2 = -2, c3 = 1.
Therefore
- 1 7   31 0 - 22 - 3  0 1
 8 - 1
2 1 0 2  2 0
Ch04_36
Example 7
(跳過)
Show that the function h(x) =4x2+3x-7 lies in the space Span{f, g}
generated by f(x) = 2x2-5 and g(x) = x+1.
Solution
Suppose c1 f  c2 g  h.
Then
c1 (2x 2 - 5)  c2 ( x  1)  4x 2  3x - 7
2c1 x 2  c2 x - 5c1  c2  4x 2  3x - 7
2c1  4



c2  3
- 5c  c  -7
1
2

 c1  2, c2  3
 h  2 f  3g.
Therefore the function h(x) lies in Span{f, g}.
Ch04_37
Homework
Exercise 4.2:
2, 4, 6, 18, 20
Ch04_38
4.3 Linear Dependence and
Independence
The concepts of dependence and independence of vectors are
useful tools in constructing “efficient” spanning sets for vector
spaces – sets in which there are no redundant vectors.
Definition
(a) The set of vectors { v1, …, vm } in a vector space V is said to
be linearly dependent (線性相依) if there exist scalars
c1, …, cm, not all zero, such that c1v1 + … + cmvm = 0
(b) The set of vectors { v1, …, vm } is linearly independent (線性
獨立) if c1v1 + … + cmvm = 0 can only be satisfied when
c1 = 0, …, cm = 0.
Ch04_39
Example 1
Determine whether the set{(1, 2, 0), (0, 1, -1), (1, 1, 2)} is
linearly dependent in R3.
Solution
Suppose c1 (1, 2, 0)  c2 (0, 1, -1)  c3 (1, 1, 2)  0
 c3  0
 c1

  2c1  c2  c3  0 
 - c  2c  0
2
3

c1 = 0
c2 = 0 is the unique solution.
c3 = 0
Thus the set of vectors is linearly independent.
Note. 不需真的求解，只需判斷是唯一解或無限多解，故當係數矩陣是方陣時，
求算係數矩陣的行列式即可。
此題行列式0  唯一解
 線性獨立
若行列式=0  無限多解  線性相依
隨堂作業：1(c,e)
Ch04_40
Example 2
(跳過)
(a) Show that the set {x2+1, 3x–1, –4x+1} is linearly independent
in P2.
(b) Show that the set {x+1, x–1, –x+5} is linearly dependent in P1.
Solution
(a) Suppose c1(x2 + 1) + c2(3x – 1) + c3(– 4x + 1) = 0
 c1x2 +(3c2 – 4c3)x + c1 – c2 + c3 = 0
 c1 = 0, c2 = 0, c3 = 0 is the unique solution
 linearly independent
(b) Suppose c1(x+1) + c2(x –1) + c3(–x+5) = 0
 (c1+ c2 – c3)x + c1 – c2 +5c3 = 0
 many solutions
 linearly dependent
Ch04_41
Theorem 4.4
A set consisting of two or more vectors in a vector space is
linearly dependent if and only if it is possible to express one of the
vectors as a linearly combination of the other vectors.
Proof
() Let the set { v1, v2, …, vm } be linearly dependent.
Therefore, there exist scalars c1, c2, …, cm, not all zero, such that
c1v1 + c2v2 + … + cmvm = 0
Assume that c1  0. The above identity can be rewritten
 - c2 
 - cm 
v1  
v2    
vm
 c1 
 c1 
Thus, v1 is a linear combination of v2, …, vm.
Ch04_42
() Conversely, assume that v1 is a linear combination of v2, …,
vm. Therefore, there exist scalars d2, …, dm, such that
v1 = d2v2 + … + dmvm
Rewrite this equation as
1v1 + (- d2)v2 + … + (-dm)vm = 0
Thus the set {v1, v2, …, vm} is linearly dependent, completing
the proof.
Ch04_43
Linear Dependence of {v1, v2}
{v1, v2} linearly dependent;
vectors lie on a line
{v1, v2} linearly independent;
vectors do not lie on a line
Figure 4.3 Linear dependence and independence of {v1, v2} in R3.
Ch04_44
Linear Dependence of {v1, v2, v3}
{v1, v2, v3} linearly dependent;
vectors lie in a plane
{v1, v2, v3} linearly independent;
vectors do not lie in a plane
Figure 4.4 Linear dependence and independence of {v1, v2, v3} in R3.
Ch04_45
Theorem 4.5
Let V be a vector space. Any set of vectors in V that contains the
zero is linearly dependent.
Proof
Consider the set {0, v2, …, vm}, which contains the zero vectors.
Let us examine the identity
c1 0  c2 v 2    cm v m  0
We see that the identity is true for c1 = 1, c2 = 0, …, cm = 0 (not
all zero). Thus the set of vectors is linearly dependent, proving
the theorem.
Ch04_46
Theorem 4.6
Let the set {v1, …, vm} be linearly dependent in a vector space V.
Any set of vectors in V that contains these vectors will also be
linearly dependent.
Proof
Since the set {v1, …, vm} is linearly dependent, there exist scalars
c1, …, cm, not all zero, such that
c1 v1    cm v m  0
Consider the set of vectors {v1, …, vm, vm+1, …, vn}, which
contains the given vectors. There are scalars, not all zero, namely
c1, …, cm, 0, …, 0 such that
c1 v1    cm v m  0 v m 1    0 v n  0
Thus the set {v1, …, vm, vm+1, …, vn} is linearly dependent.
Ch04_47
Example 3
Let the set {v1, v2} be linearly independent.
Prove that {v1 + v2, v1 – v2} is also linearly independent.
Solution
Suppose
a( v1  v 2 )  b( v1 - v 2 )  0
(1)
We get
av1  av 2  bv1 - bv 2  0
(a  b) v1  (a - b) v 2  0
Since {v1, v2} is linearly independent
ab  0
隨堂作業：12
a -b  0
Thus system has the unique solution a = 0, b = 0. Returning to
identity (1) we get that {v1 + v2, v1 – v2} is linearly independent.
Ch04_48
Homework
Exercise 4.3:
1, 3, 8, 12
Ch04_49
4.4 Properties of Bases
Theorem 4.7
Let the vectors v1, …, vn span a vector space V. Each vector in V
can be expressed uniquely as a linear combination of these vectors
if and only if the vectors are linearly independent.
Proof
(a) () Assume that v1, …, vn are linearly independent. Let v  V.
Since v1, …, vn span V, we can express v as a linear combination of
these vectors. Suppose we can write
v  a1 v1    an v n and v  b1 v1    bn v n

a1 v1    an v n  b1 v1    bn v n

(a1 - b1 ) v1    (an - bn ) v n  0
Since v1, …, vn are linearly independent, a1 – b1 = 0, …, an – bn = 0,
implying that a1 = b1, …, an = bn.
 unique 得證
Ch04_50
(b) () Let v  V.
Assume that v can be written in only one way as a linear
combination of v1, …, vn.
Note that 0v1+ …+ 0vn= 0.
If c1v1+ …+ cnvn= 0, it implies that c1 = 0, c2 = 0, …, cn = 0.
 v1, …, vn are linearly independent.
Definition
A finite set of vectors {v1, …, vm} is called a basis for a vector
space V if the set spans V and is linearly independent.
Ch04_51
Theorem 4.8
Let {v1, …, vn } be a basis for a vector space V. If {w1, …, wm} is
a set of more than n vectors in V, then this set is linearly
dependent.
Proof
Suppose
c1w1    cm w m  0
We will show that values of c1, …, cm are not all zero.
(1)
The set {v1, …, vn} is a basis for V. Thus each of the vectors
w1, …, wm can be expressed as a linear combination of v1, …, vn.
Let
w1  a11 v1  a12 v 2    a1n v n

w m  am1 v1  am 2 v 2    amn v n
Ch04_52
Substituting for w1, …, wm into Equation (1) we get
c1 (a11v1  a12 v 2   a1n v n )   cm (am1v1  am2 v 2   amn v n )  0
Rearranging, we get
(c1a11  c2a21   cmam1 ) v1   (c1a1n  c2a2n   cmamn ) v n  0
Since v1, …, vn are linear independent,
a11c1  a21c2    am1cm  0

a1n c1  a2 n c2    amn cm  0
Since m > n, there are many solutions in this system.
Thus the set {w1, …, wm} is linearly dependent.
隨堂作業：6(b)
Ch04_53
Theorem 4.9
Any two bases for a vector space V consist of the same number
of vectors.
Proof
Let {v1, …, vn} and {w1, …, wm} be two bases for V.
By Theorem 4.8,
m  n and n  m
Thus n = m.
Definition
If a vector space V has a basis consisting of n vectors, then the
dimension of V is said to be n. We write dim(V) for the
dimension of V.
• V is finite dimensional if such a finite basis exists.
• V is infinite dimensional otherwise.
Ch04_54
Example 1
Consider the set {{1, 2, 3), (-2, 4, 1)} of vectors in R3.
These vectors generate a subspace V of R3 consisting of all
vectors of the form
v  c1 (1, 2, 3)  c2 (-2, 4, 1)
The vectors (1, 2, 3) and (-2, 4, 1) span this subspace.
Furthermore, since the second vector is not a scalar multiple of
the first vector, the vectors are linearly independent.
Therefore {{1, 2, 3), (-2, 4, 1)} is a basis for V.
Thus dim(V) = 2.
We know that V is, in fact, a plane through the origin.
Ch04_55
Theorem 4.10
(a) The origin is a subspace of R3. The dimension of this
subspace is defined to be zero.
(b) The one-dimensional subspaces of R3 are lines through the
origin.
(c) The two-dimensional subspaces of R3 are planes through the
origin.
Figure 4.5 One and two-dimensional subspaces of R3
Ch04_56
Proof
(a) Let V be the set {(0, 0, 0)}, consisting of a single elements,
the zero vector of R3. Let c be the arbitrary scalar. Since
(0, 0, 0) + (0, 0, 0) = (0, 0, 0) and c(0, 0, 0) = (0, 0, 0)
V is closed under addition and scalar multiplication. It is thus
a subspace of R3. The dimension of this subspaces is defined
to be zero.
(b) Let v be a basis for a one-dimensional subspace V of R3.
Every vector in V is thus of the form cv, for some scalar c. We
know that these vectors form a line through the origin.
(c) Let {v1, v2}be a basis for a two-dimensional subspace V of R3.
Every vector in V is of the form c1v1 + c2v2. V is thus a plane
through the origin.
隨堂作業：16(a)
Ch04_57
Theorem 4.11
Let V be a vector space of dimension n.
(a) If S = {v1, …, vn} is a set of n linearly independent vectors
in V, then S is a basis for V.
(b) If S = {v1, …, vn} is a set of n vectors V that spans V, then S
is a basis for V.
整理:
Let V be a vector space, S = {v1, …, vn} is a set of vectors in V.
若以下三點有兩點成立，則另一點也成立。
(a) dim(V) = |S|.
(b) S is a linearly independent set.
S is a basis of V.
(c) S spans V.
Ch04_58
Example 2
Prove that the set B={(1, 3, -1), (2, 1, 0), (4, 2, 1)} is a basis for R3.
Solution
Since dim(R3)=|B|=3. It suffices to show that this set is linearly
independent or it spans R3.
Let us check for linear independence. Suppose
c1 (1, 3, - 1)  c2 (2, 1, 0)  c3 (4, 2, 1)  (0, 0, 0)
This identity leads to the system of equations
c1  2c2  4c3  0
3c1  c2  2c3  0
- c1  c3  0
This system has the unique solution c1 = 0, c2 = 0, c3 = 0.
Thus the vectors are linearly independent.
The set {(1, 3, -1), (2, 1, 0), (4, 2, 1)} is therefore a basis for R3.
隨堂作業：5(a), 20(c), 21
Ch04_59
Theorem 4.12
Let V be a vector space of dimension n. Let {v1, …, vm} be a set
of m linearly independent vectors in V, where m < n. Then there
exist vectors vm+1, …, vn such that {v1, …, vm, vm+1, …, vn } is a
basis of V.
Ch04_60
Example 3
State (with a brief explanation) whether the following statements
are true or false.
(a) The vectors (1, 2), (-1, 3), (5, 2) are linearly dependent in R2.
(b) The vectors (1, 0, 0), (0, 2, 0), (1, 2, 0) span R3.
(c) {(1, 0, 2), (0, 1, -3)} is a basis for the subspace of R3 consisting
of vectors of the form (a, b, 2a -3b).
(d) Any set of two vectors can be used to generate a twodimensional subspace of R3.
Solution
(a) True: The dimension of R2 is two. Thus any three vectors are
linearly dependent.
(b) False: The three vectors are linearly dependent. Thus they
cannot span a three-dimensional space.
Ch04_61
(c) True: The vectors span the subspace since
(a, b, 2a, -3b) = a(1, 0, 2) + b(0, 1, -3)
The vectors are also linearly independent since they are not
colinear.
(d) False: The two vectors must be linearly independent.
Ch04_62
Homework
Exercise 4.4:
5, 6, 7, 16, 20, 21, 23, 25
Ch04_63
4.5 Rank
Rank enables one to relate matrices to vectors, and vice versa.
Definition
Let A be an m  n matrix. The rows of A may be viewed as row
vectors r1, …, rm, and the columns as column vectors c1, …, cn.
Each row vector will have n components, and each column vector
will have m components, The row vectors will span a subspace of
Rn called the row space of A, and the column vectors will span a
subspace of Rm called the column space of A.
Ch04_64
Example 1
Consider the matrix
 1 2 - 1 2
3 4
1 6
5 4
1 0

(1) The row vectors of A are
r1  (1, 2, - 1, 2), r2  (3, 4, 1, 6), r3  (5, 4, 1, 0)
These vectors span a subspace of R4 called the row space of A.
(2) The column vectors of A are
1 
 2
- 1
 2
c1  3 c 2  4 c3   1 c 4  6
 
 
 
 
5
4
1
 
 
 
0 
These vectors span a subspace of R3 called the column space of A.
Ch04_65
Theorem 4.13
The row space and the column space of a matrix A have the
same dimension.
Proof
Let u1, …, um be the row vectors of A. The ith vector is
u i  ( ai1 , ai 2 , ..., ain )
Let the dimension of the row space be s. Let the vectors v1, …, vs
form a basis for the row space. Let the jth vector of this set be
v j  (b j1 , b j 2 , ..., b jn )
Each of the row vectors of A is a linear combination of v1, …,
vs. Let
u1  c11 v 1  c12 v 2    c1s v s

u m  cm1 v1  cm 2 v 2    cms v s
Ch04_66
Equating the ith components of the vectors on the left and right,
we get
a1i  c11b1i  c12b2i    c1s bsi

ami  cm1b1i  cm 2b2i    cms bsi
This may be written
 a1i 
 c1s 
 c11 
 c12 
    b1i     b2i       bsi   
 
 
 
 
a
c
c
 m1 
 m2 
 mi 
cms 
This implies that each column vector of A lies in a space spanned
by a single set of s vectors. Since s is the dimension of the row
space of A, we get
dim(column space of A)  dim(row space of A)
Ch04_67
By similar reasoning, we can show that
dim(row space of A)  dim(column space of A)
Combining these two results we see that
dim(row space of A) = dim(column space of A),
proving the theorem.
Definition
The dimension of the row space and the column space of a matrix
A is called the rank (秩) of A. The rank of A is denoted rank(A).
Ch04_68
Example 2
Determine the rank of the matrix
1 2 3 
A  0 1 2
2 5 8 


Solution
The third row of A is a linear combination of the first two rows:
(2, 5, 8) = 2(1, 2, 3) + (0, 1, 2)
Hence the three rows of A are linearly dependent.
The rank of A must be less than 3. Since (1, 2, 3) is not a scalar
multiple of (0, 1, 2), these two vectors are linearly independent.
These vectors form a basis for the row space of A.
Thus rank(A) = 2.
Ch04_69
Theorem 4.14
The nonzero row vectors of a matrix A that is in reduced echelon
form are a basis for the row space of A. The rank of A is the
number of nonzero row vectors.
Proof
Let A be an m  n matrix with nonzero row vectors of r1, …, rt.
Consider the identity
k1r1  k 2r2    kt rt  0
Where k1, …, kt are scalars.
The first nonzero element of r1 is 1. r1 is the only one of the row
to have a nonzero number in this component. Thus, on adding
the vectors k1r1 , k 2r2 ,..., kt rt , we get a vector whose first
component is k1. On equating this vector to zero, we get k1 = 0.
The identity then reduced to
k 2r2    kt rt  0
Ch04_70
The first nonzero element of r2 is 1, and it is the only of these
remaining row vectors with a nonzero number in this component.
Thus k2 = 0.
Similarly, k3, …, kt are all zero.
The vector r1, …, rt are therefore linearly independent. These
vectors span the row space of A.
They thus form a basis for the row space of A. The dimension of
the row space is t. The rank of A is t, the number of nonzero row
vectors in A.
Ch04_71
Example 3
Find the rank of the matrix
1
0
A
0

0
2
0
0
0
0
1
0
0
0
0
1

0
This matrix is in reduced echelon form. There are three nonzero
row vectors, namely (1, 2, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1).
According to the previous theorem, these three vectors form a
basis for the row space of A.
Rank(A) = 3.
Ch04_72
Theorem 4.15
Let A and B be row equivalent matrices. Then A and B have the
same the row space. rank(A) = rank(B).
Theorem 4.16
Let E be a reduced echelon form of a matrix A. The nonzero row
vectors of E form a basis for the row space of A. The rank of A
is the number of nonzero row vectors in E.
Note: If A … E, 且E是reduced echelon form, then
(a) 將E的每一非零列視為向量，
這些列形成一個A的列空間的basis。
(b) rank(A) = E的非零列個數
Ch04_73
Example 4
Find a basis for the row space of the following matrix A, and
determine its rank.
1 2 3 
A   2 5 4
1 1 5 


Solution
Use elementary row operations to find a reduced echelon form of
the matrix A. We get
3  1 0
7
1 2 3   1 2
 2 5 4   0
1 - 2   0 1 - 2 
1 1 5  0 - 1
 0 0

2
0

 
 

The two vectors (1, 0, 7), (0, 1, -2) form a basis for the row
space of A. Rank(A) = 2.
隨堂作業：5(b), 12
Ch04_74
Example 5
Find a basis for the column space of the following matrix A.
A
Solution


t  



A
1
1
2 -1

3 -4
0 -2







1
1
0

2
3 -2

-1 -4 6



6
The column space of A becomes the row space of At. Let us find
a basis for the row space of At.
2 - 1  1
2 - 1  1 0
5
1
1
3 - 4  0
1 - 3  0 1 - 3
0 - 2
6 0 - 2
6 0 0
0
1   0
   
 0,  1 form a basis for the column space of A.
5  - 3
Ch04_75
Example 6
Find a basis for the subspace V of R4 spanned by the vectors
(1, 2, 3, 4), (-1, -1, -4, -2), (3, 4, 11, 8)
Solution
 1 2 3 4
Let A  - 1 - 1 - 4 - 2 .
 3 4 11 8
3
4  1
2 3
4  1 0 5 0
 1 2
- 1 - 1 - 4 - 2  0
1 -1
2  0 1 - 1 2
 3 4 11
8 0 - 2 2 - 4 0 0 0 0
 (1, 0, 5, 0) and (0, 1, -1, 2) form a basis for the subspace V.
 dim(V) = 2.
隨堂作業：7(b)
Ch04_76
Theorem 4.17
Consider a system AX=B of m equations in n variables
(a) If the augmented matrix and the matrix of coefficients have
the same rank r and r = n, the solution is unique.
(b) If the augmented matrix and the matrix of coefficients have
the same rank r and r < n, there are many solutions.
(c) If the augmented matrix and the matrix of coefficients do not
have the same rank, a solution does not exist.
隨堂作業：10(a,b,c) 僅做(i)(ii)
Ch04_77
Theorem 4.18
Let A be an n  n matrix. The following statements are equivalent.
(a) |A|  0 (A is nonsingular).
(b) A is invertible.
(c) A is row equivalent to In.
(d) The system of equations AX = B has a unique solution.
(e) rank(A) = n.
(f) The column vectors of A form a basis for Rn.
Ch04_78
Homework
Exercise 4.5:
5, 7, 10, 12
Ch04_79
4.6 Orthonormal Vectors and
Projections
Definition
A set of vectors in a vector space V is said to be an orthogonal
(正交) set if every pair of vectors in the set is orthogonal. The set
is said to be an orthonormal set if it is orthogonal and each
vector is a unit vector.
Ch04_80
Example 1
Show that the

 3 4   4 3 
(
1
,
0
,
0
),
 0, , ,  0, , - 
set 
 5 5   5 5 

Solution
(1) orthogonal:
is an orthonormal set.
 
(1,0,0)  0, 54 ,- 53   0;
0, 53 , 54  0, 54 ,- 53  0;
(1,0,0)  0, 53 , 54  0;
(2) unit vector:
(1,0,0)  12  0 2  0 2  1
   0       1
0, 54 , - 53   0  54   -53  1
0, 53 , 54
3
5
2
2
2
4
5
2
Thus the set is thus an orthonormal set.
2
2
隨堂作業：1,3(a)
Ch04_81
Theorem 4.19
An orthogonal set of nonzero vectors in a vector space is
linearly independent.
Proof Let {v1, …, vm} be an orthogonal set of nonzero vectors
in a vector space V. Let us examine the identity
c1v1 + c2v2 + … + cmvm = 0
Let vi be the ith vector of the orthogonal set. Take the dot
product of each side of the equation with vi and use the
properties of the dot product. We get
(c1v1  c2 v 2    cm v m )  v i  0  v i
c1v1  v i  c2 v 2  v i    cm v m  v i  0
Since the vectors v1, …, v2 are mutually orthogonal, vj‧vi = 0
unless j = i. Thus
ci v i  v i  0
Since vi is a nonzero, then vi‧vi  0. Thus ci = 0.
Letting i = 1, …, m, we get c1 = 0, cm = 0, proving that the
vectors are linearly independent.
Ch04_82
Definition
A basis is an orthogonal set is said to be an orthogonal basis.
A basis that is an orthonormal set is said to be an orthonormal
basis.
Standard Bases
• R2: {(1, 0), (0, 1)}
• R3: {(1, 0, 0), (0, 1, 0), (0, 0, 1)}
orthonormal bases
• Rn: {(1, …, 0), …, (0, …, 1)}
Theorem 4.20
Let {u1, …, un} be an orthonormal basis for a vector space V. Let
v be a vector in V. v can be written as a linearly combination of
these basis vectors as follows:
v  ( v  u1 )u1  ( v  u 2 )u 2    ( v  u n )u n
Ch04_83
Example 2
The following vectors u1, u2, and u3 form an orthonormal basis
for R3. Express the vector v = (7, -5, 10) as a linear combination
of these vectors.
 3 4
 4 3
u1  (1, 0, 0), u 2   0, , , u 3   0, , - 
 5 5
 5 5
Solution
v  u1  (7, - 5, 10) (1, 0, 0)  7
 3 4
v  u 2  (7, - 5, 10)  0, ,   5
 5 5
 4 3
v  u 3  (7, - 5, 10)  0, , -   -10
 5 5
Thus
v  7u1  5u2 -10u3
隨堂作業：4
Ch04_84
Orthogonal Matrices
An orthogonal matrix is an invertible matrix that has the property
A-1 = At
Theorem 4.21 (Orthogonal Matrix Theorem)
The following statements are equivalent.
(a) A is orthogonal.
(b) The column vectors of A form an orthonormal set.
(c) The row vectors of A form an orthonormal set.
Proof
(a bc)
A is orthogonal  A-1 = At  AtA= I and AAt= I
 The column vectors of A form an orthonormal set,
and the row vectors of A form an orthonormal set.
反之亦然
Ch04_85
Theorem 4.22
If A is an orthogonal matrix, then
(a) |A| = 1.
(b) A-1 is an orthogonal matrix.
Proof
(a) AAt= I  |AAt| = |A||At| = |A||A-1| = 1
 |A| = 1
(b) (A-1)t (A-1)= AAt= I
 A-1 is an orthogonal matrix
隨堂作業：9
Ch04_86
Projection of One vector onto Another Vector
Let v and u be vectors in Rn with angel a (0  a  p) between them.
Figure 4.7
OA : the projection of v onto u
OA  OB cosa  || v || cosa
v u
v u
 || v ||

|| v || || u || || u ||
v u u
v u
 OA  (
)(
)
u
|| u || || u || u  u
v u
Note : If a  p / 2 then
 0.
u u
v u
u.
So we define proj u v 
u u
Ch04_87
Definition
The projection (投影) of a vector v onto a nonzero vector u in Rn
is denoted projuv and is defined by
v u
proju v 
u
u u
O
Figure 4.8
Ch04_88
Example 3
Determine the projection of the vector v = (6, 7) onto the vector
u = (1, 4).
Solution
v  u  (6, 7)  (1, 4)  6  28  34
u  u  (1, 4)  (1, 4)  1  16  17
Thus
v u
34
proju v 
u  (1, 4)  (2, 8)
u u
17
The projection of v onto u is (2, 8).
隨堂作業：14(b)
Ch04_89
Theorem 4.23
The Gram-Schmidt Orthogonalization Process
Let {v1, …, vn} be a basis for a vector space V. The set of vectors
{u1, …, un} defined as follows is orthogonal. To obtain an
orthonormal basis for V, normalize each of the vectors u1, …, un .
u1  v1
u 2  v 2 - proju1 v 2
u 3  v 3 - proju1 v 3 - proju 2 v 3

u n  v n - proju1 v n -  - proju n -1 v n
Figure 4.9
Ch04_90
Example 4
The set {(1, 2, 0, 3), (4, 0, 5, 8), (8, 1, 5, 6)} is linearly independent
in R4. The vectors form a basis for a three-dimensional subspace V
of R4. Construct an orthonormal basis for V.
Solution
Let v1 = (1, 2, 0, 3), v2 = (4, 0, 5, 8), v3 = (8, 1, 5, 6)}.
Use the Gram-Schmidt process to construct an orthogonal set
{u1, u2, u3} from these vectors.
Let u1  v1  (1, 2, 0, 3)
Let u 2  v 2 - proju1 v 2  v 2 -
(v2  u2 )
u1  (2, - 4, 5, 2)
(u1  u1 )
Let u 3  v 3 - proju1 v 3 - proju 2 v 3
( v 3  u1 )
(v3  u2 )
 v3 u1 u 2  (4, 1, 0, - 2)
(u1  u1 )
(u 2  u 2 )
Ch04_91
The set {(1, 2, 0, 3), (2, -4, 5, 2), (4, 1, 0, -2)} is an orthogonal
basis for V.
Normalize them to get an orthonormal basis:
(1, 2, 0, 3)  12  2 2  0 2  32  14
(2, - 4, 5, 2)  2 2  (-4) 2  52  2 2  7
(4, 1, 0, - 2)  4 2  12  0 2  (-2) 2  21
 orthonormal basis for V:
2
3   2 4 5 2  4
1
2 
 1
,
, 0,
,
, 0, ,  , - , , , 


14   7 7 7 7   21 21
21 
 14 14
隨堂作業：16(a)
Ch04_92
Projection of a Vector onto a Subspace
Definition
Let W be a subspace of Rn, Let {u1, …, um} be an orthonormal
basis for W. If v is a vector in Rn, the projection of v onto W is
denoted projWv and is defined by
projW v  ( v  u1 )u1  ( v  u 2 )u 2    ( v  u m )u m
Figure 4.11
Ch04_93
Theorem 4.24
Let W be a subspace of Rn. Every vector v in Rn can be written
uniquely in the form
v = w + w
where w is in W and w is orthogonal to W. The vectors w and w
are
w = projWv and w = v – projWv
Figure 4.12
Ch04_94
Example 5
Consider the vector v = (3, 2, 6) in R3. Let W be the subspace of R3
consisting of all vectors of the form (a, b, b). Decompose v into the
sum of a vector that lies in W and a vector orthogonal to W.
Solution
We need an orthonormal basis for W. We can write an arbitrary
vector of W as follows
(a, b, b) = a(1, 0,0) + b( 0, 1, 1)
The set {(1, 0, 0), (0, 1, 1)} spans W and is linearly independent.
It forms a basis for W. The vectors are orthogonal. Normalize
each vector to get an orthonormal basis {u1, u2} for W, where
u1  (1, 0, 0),
1 1 
u 2   0,
,

2 2

Ch04_95
We get
w  projW v  ( v  u1 )u1  ( v  u 2 )u 2

 1 1   1 1 
 ((3, 2, 6)  (1, 0, 0))(1, 0, 0)   (3, 2, 6)   0,
,
,
  0,

2
2
2
2

 


 (3, 0, 0)  (0, 4, 4)  (3, 4, 4)
and
w   v - projW v  (3, 2, 6) - (3, 4, 4)  (0, - 2, 2)
Thus the desired decomposition of v is
(3, 2, 6) = (3, 4, 4) + (0, -2, 2)
In this decomposition the vector (3, 4, 4) lies in W and the vector
(0, -2, 2) is orthogonal to W.
隨堂作業：21(a)
Ch04_96
Distance of a Point from a Subspace
The distance of a point from a subspace is the distance of the
point from its projection in the subspace.
d (x, W )  x - projW x
Figure 4.13
Ch04_97
Example 6
Find the distance of the point x = (4, 1, -7) of R3 from the
subspace W consisting of all vectors of the form (a, b, b).
Solution
The previous example tells us that the set {u1, u2} where
1 1 

u1  (1, 0, 0),
u 2   0,
,

2
2


is an orthonormal basis for W. We compute projWx
projW x  (x  u1 )u1  (x  u 2 )u 2

 ((4, 1, - 7)  (1, 0, 0))(1, 0, 0)  (4, 1, - 7) 

0,
1 1
,
2 2

0,
1 1
,
2 2

 (4, 0, 0)  (0, - 3, - 3)  (4, - 3, - 3)
Thus, the distance from x to W is
x - projW x  (4, 1, - 7) - (4, - 3, - 3)
隨堂作業：26
 (0, 4, - 4)  32
Ch04_98
Homework
Exercise 4.9:
1, 3, 4, 6, 9, 14, 16, 21, 26
Ch04_99