1. Chapter 4 General Vector Spaces
Linear Algebra
Chapter 4 General Vector Spaces
大葉大學 資訊工程系
黃鈴玲

2. 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！)
The sum u + v exists and is an element of V. (V is closed under addition.)
cu is an element of V. (V is closed under scalar multiplication.)

3. 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 =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

4. 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

5. A Vector Space in R3 Prove that W is a vector space. 補充： Proof
Let , for some a, bR.
Axiom 1:
u + v W. Thus W is closed under addition.
W Thus W is closed under scalar multiplication.
Axiom 2:
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

6. Vector Spaces of Matrices
Prove that M22 is a vector space.
Proof
Let
Axiom 1:
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).

7. Axiom 5:
The 2  2 zero matrix is , since
Axiom 6:
推廣：The set of m  n matrices, Mmn, is a vector space.
隨堂作業：7

8. 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.

9. 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).
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}

10. The Complex Vector Space Cn
(跳過)
It can be shown that Cn with these two operations is a complex vector space.

11. 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.
(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
Proof

12. 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。

13. (a, a, b) + (c, c, d) = (a+c, a+c, b+d)  W
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.

14. (a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)
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)

15. 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.
We get
 u + v  U.  U is closed under addition.
() Let c R. We get
cu  U.  U is closed under scalar multiplication.
U is a subspace of M22.
隨堂作業：27(a)

16. 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 + g)(x) is a polynomial of degree  n. Thus f + g  Pn. Pn is closed under addition.

17. (跳過)
() Let c R,
(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.

18. 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.

19. 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)

20. Homework
Exercise 4.1: 5, 7, 14, 18, 19, 20, 24, 27

21. 4.2 Linear Combinations Definition W={(a, a, b) | a,b R}  R3
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.

22. 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.

23. 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).
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)

24. 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

25. 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
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):

26. 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 
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)

27. 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.

28. 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  
The vectors (1, 2, 0), (0, 1, -1), and (1, 1, 2) span R3.
隨堂作業：4(a)

29. 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.

30. ()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.

31. 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.
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).

32. Figure 4.1

33. 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

34. 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
We can write
 k1v1 and k2v2 span U.
隨堂作業：20

35. Example 6 Determine whether the matrix is a linear combination
of the matrices in the vector space
M22 of 2  2 matrices.
Solution
Suppose
Then

36. This system has the unique solution c1 = 3, c2 = -2, c3 = 1.
Therefore

37. 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
Then
Therefore the function h(x) lies in Span{f, g}.

38. Homework
Exercise 4.2: 2, 4, 6, 18, 20

39. 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
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
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.

40. Example 1
Determine whether the set{(1, 2, 0), (0, 1, -1), (1, 1, 2)} is linearly dependent in R3.
Solution
Suppose
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)

41. 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

42. 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
Thus, v1 is a linear combination of v2, …, vm.

43. () Conversely, assume that v1 is a linear combination of v2, …, vm
() 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.

44. 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.

45. 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.

46. 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
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.

47. 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
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
Thus the set {v1, …, vm, vm+1, …, vn} is linearly dependent.

48. Example 3
Let the set {v1, v2} be linearly independent. Prove that {v1 + v2, v1 – v2} is also linearly independent.
Solution
Suppose
(1)
We get
Since {v1, v2} is linearly independent
Thus system has the unique solution a = 0, b = 0. Returning to identity (1) we get that {v1 + v2, v1 – v2} is linearly independent.
隨堂作業：12

49. Homework
Exercise 4.3: 1, 3, 8, 12

50. 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  
Since v1, …, vn are linearly independent, a1 – b1 = 0, …, an – bn = 0, implying that a1 = b1, …, an = bn.
 unique 得證

51. (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.

52. 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
(1)
We will show that values of c1, …, cm are not all zero.
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

53. Substituting for w1, …, wm into Equation (1) we get
Rearranging, we get
Since v1, …, vn are linear independent,
Since m > n, there are many solutions in this system.
Thus the set {w1, …, wm} is linearly dependent.
隨堂作業：6(b)

54. 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.

55. 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
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.

56. Theorem 4.10
The origin is a subspace of R3. The dimension of this subspace is defined to be zero.
The one-dimensional subspaces of R3 are lines through the origin.
The two-dimensional subspaces of R3 are planes through the origin.
Figure 4.5 One and two-dimensional subspaces of R3

57. Proof
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)

58. Theorem 4.11 Let V be a vector space of dimension n.
If S = {v1, …, vn} is a set of n linearly independent vectors in V, then S is a basis for V.
If S = {v1, …, vn} is a set of n vectors V that spans V, then S is a basis for V.
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。
dim(V) = |S|.
S is a linearly independent set.
S spans V.
整理:
S is a basis of V.

59. 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
This identity leads to the system of equations
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

60. 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.

61. 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 two-dimensional subspace of R3.
Solution
True: The dimension of R2 is two. Thus any three vectors are linearly dependent.
False: The three vectors are linearly dependent. Thus they cannot span a three-dimensional space.

62. (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.

63. Homework
Exercise 4.4: 5, 6, 7, 16, 20, 21, 23, 25

64. 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.

65. Example 1 Consider the matrix (1) The row vectors of A are
These vectors span a subspace of R4 called the row space of A.
(2) The column vectors of A are
These vectors span a subspace of R3 called the column space of A.

66. 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
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
Each of the row vectors of A is a linear combination of v1, …, vs. Let

67. dim(column space of A)  dim(row space of A)
Equating the ith components of the vectors on the left and right, we get
This may be written
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)

68. Definition 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).

69. Example 2 Determine the rank of the matrix 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.

70. 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
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 we get a vector whose first component is k1. On equating this vector to zero, we get k1 = 0. The identity then reduced to

71. 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.

72. Example 3 Find the rank of the matrix
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.

73. 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

74. Example 4
Find a basis for the row space of the following matrix A, and determine its rank.
Solution
Use elementary row operations to find a reduced echelon form of the matrix A. We get
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

75. Example 5 Find a basis for the column space of the following matrix A.
Solution
The column space of A becomes the row space of At. Let us find a basis for the row space of At.
form a basis for the column space of A. 

76. 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
Let
 (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)

77. Theorem 4.17 Consider a system AX=B of m equations in n variables
If the augmented matrix and the matrix of coefficients have the same rank r and r = n, the solution is unique.
If the augmented matrix and the matrix of coefficients have the same rank r and r < n, there are many solutions.
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)

78. Theorem 4.18
Let A be an n  n matrix. The following statements are equivalent.
|A|  0 (A is nonsingular).
A is invertible.
A is row equivalent to In.
The system of equations AX = B has a unique solution.
rank(A) = n.
The column vectors of A form a basis for Rn.

79. Homework
Exercise 4.5: 5, 7, 10, 12

80. 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.

81. Example 1 Show that the set is an orthonormal set. Solution
(1) orthogonal:
(2) unit vector:
Thus the set is thus an orthonormal set.
隨堂作業：1,3(a)

82. 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
Since the vectors v1, …, v2 are mutually orthogonal, vj‧vi = 0 unless j = i. Thus
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.

83. 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:

84. 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.
Solution
Thus
隨堂作業：4

85. 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

86. 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| =  |A| = 1
(b) (A-1)t (A-1)= AAt= I  A-1 is an orthogonal matrix
隨堂作業：9

87. 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
So we define

88. Definition
The projection (投影) of a vector v onto a nonzero vector u in Rn is denoted projuv and is defined by O
Figure 4.8

89. Example 3
Determine the projection of the vector v = (6, 7) onto the vector u = (1, 4).
Solution
Thus
The projection of v onto u is (2, 8).
隨堂作業：14(b)

90. 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 .
Figure 4.9

91. 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.

92. 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:
 orthonormal basis for V:
隨堂作業：16(a)

93. 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
Figure 4.11

94. w = projWv and w = v – projWv
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

95. 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

96. We get
and
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)

97. 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.
Figure 4.13

98. 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
is an orthonormal basis for W. We compute projWx
Thus, the distance from x to W is
隨堂作業：26

99. Homework
Exercise 4.9: 1, 3, 4, 6, 9, 14, 16, 21, 26