1. Chapter 1 Linear Equations and Vectors
Linear Algebra
Chapter 1 Linear Equations and Vectors
大葉大學 資訊工程系
黃鈴玲

2. 1.1 Matrices and Systems of Linear Equations
Definition
An equation (方程式) in the variables (變數) x and y that can be written in the form ax+by=c, where a, b, and c are real constants (實數常數) (a and b not both zero), is called a linear equation (線性方程式).
The graph of this equation is a straight line in the x-y plane.
A pair of values of x and y that satisfy the equation is called a solution (解).
system of linear equations (線性聯立方程式)

3. Solutions for system of linear equations
Figure 1.2
No solution (無解)
–2x + y = 3
–4x + 2y = 2
Lines are parallel. No point of intersection. No solutions.
Figure 1.3
Many solution (無限多解)
4x – 2y = 6
6x – 3y = 9
Both equations have the same graph. Any point on the graph is a solution. Many solutions.
Figure 1.1
Unique solution (唯一解)
x + y = 5
2x - y = 4
Lines intersect at (3, 2)
Unique solution: x = 3, y = 2.

4. Definition
A linear equation in n variables x1, x2, x3, …, xn has the form
a1 x1 + a2 x2 + a3 x3 + … + an xn = b
where the coefficients (係數) a1, a2, a3, …, an and b are constants.
常見數系的英文名稱：
natural number (自然數), integer (整數), rational number (有理數),
real number (實數), complex number (複數)
positive (正), negative (負)

5. A linear equation in three variables corresponds to a plane in three-dimensional (三維) space.
※ Systems of three linear equations in three variables:
Unique solution

6. No solutions
Many solutions

7. How to solve a system of linear equations?
Gauss-Jordan elimination. (高斯-喬登消去法)
1.2節會介紹

8. Definition A matrix (矩陣) is a rectangular array of numbers.
The numbers in the array are called the elements (元素) of the matrix.
Matrices
注意矩陣左右兩邊是中括號不是直線，直線表示的是行列式。

9. Row (列) and Column (行)
Submatrix (子矩陣)

10. Identity Matrices (單位矩陣)
Size and Type
Location
aij表示在row i, column j 的元素值
也寫成 location (1,3) = -4
Identity Matrices (單位矩陣)
diagonal (對角線) 上都是1，其餘都是0，I 的下標表示size

11. Relations between system of linear equations and matrices
matrix of coefficient and augmented matrix
係數矩陣
擴大矩陣
隨堂作業：5(f)

12. Elementary Row Operations of Matrices
給定聯立方程式後，
不會改變解的一些轉換
Elementary Transformation
Interchange two equations.
Multiply both sides of an equation by a nonzero constant.
Add a multiple of one equation to another equation.
將左邊的轉換對應到矩陣上
Elementary Row Operation (基本列運算)
Interchange two rows of a matrix. (兩列交換)
Multiply the elements of a row by a nonzero constant. (某列的元素同乘一非零常數)
Add a multiple of the elements of one row to the corresponding elements of another row. (將一個列的倍數加進另一列裡)

13. Example 1 Solving the following system of linear equation. Solution
 符號表示row equivalent
Solution
Equation Method
Initial system:
Analogous Matrix Method
Augmented matrix:
Eq2+(–2)Eq1
Eq3+(–1)Eq1
R2+(–2)R1
R3+(–1)R1 

14. Eq1+(–1)Eq2
Eq3+(2)Eq2
R1+(–1)R2
R3+(2)R2 
(–1/5)Eq3
Eq1+(–2)Eq3
(–1/5)R3 
Eq2+Eq3
R1+(–2)R3
R2+R3 
The solution is
The solution is
隨堂作業：7(d)

15. 基本列運算符號說明：  表示將R1(第一列) 加上 (-3) R2， 所以是R1這列會改變，R2不變 R1+(–3)R2
For example:
(1)
R1+2R2 
(2)
R2+2R1 

16. 基本列運算符號說明： 避免將要修改的列乘以某個倍數，這樣容易出錯 2R1+R2  不好！ 2R1  R1+R2 較好！
基本列運算的步驟： (盡量使矩陣一步步變成以下形式)
…
…
…

17. Example 2 Solving the following system of linear equation.
Solution (請先自行練習)

18. Example 3
Solve the system
Solution (請先自行練習)
隨堂作業：10(d)(f)

19. Summary A B Use row operations to [A: B] : i.e.,
Def. [In : X] is called the reduced echelon form (簡化梯式) of [A : B].
Note. 1. If A is the matrix of coefficients of a system of n equations in n variables that has a unique solution, then A is row equivalent to In (A  In).
2. If A  In, then the system has unique solution.

20. Example 4 Many Systems
Solving the following three systems of linear equation, all of which have the same matrix of coefficients.
Solution
R2+(–2)R1
R3+R1 
The solutions to the three systems are
隨堂作業：13(b)

21. Homework
Exercise 1.1： 1, 2, 4, 5, 6, 7, 10, 13

22. 1.2 Gauss-Jordan Elimination
Definition
A matrix is in reduced echelon form (簡化梯式) if
Any rows consisting entirely of zeros are grouped at the bottom of the matrix. (全為零的列都放在矩陣的下層)
The first nonzero element of each other row is 1. This element is called a leading 1. (每列第一個非零元素是1，稱做leading 1)
The leading 1 of each row after the first is positioned to the right of the leading 1 of the previous row. (每列的leading 1出現在前一列leading 1的右邊，也就是所有的leading 1會呈現由左上到右下的排列)
All other elements in a column that contains a leading 1 are zero. (包含leading 1的行裡所有其他元素都是0)

23. Examples for reduced echelon form
()
()
()
()
利用一連串的 elementary row operations，可讓任何矩陣變成 reduced echelon form
The reduced echelon form of a matrix is unique.
隨堂作業：2(b)(d)(h)

24. Gauss-Jordan Elimination
System of linear equations  augmented matrix  reduced echelon form  solution

25. Example 1
Use the method of Gauss-Jordan elimination to find reduced echelon form of the following matrix.
Solution
pivot (樞軸，未來的 leading 1)
pivot
The matrix is the reduced echelon form of the given matrix.

26. Example 2 Solve, if possible, the system of equations
Solution (請先自行練習)
The general solution (通解) to the system is
隨堂作業：5(c)

27. Example 3
This example illustrates that the general solution can involve a number
of parameters. Solve the system of equations
變數個數 > 方程式個數  many sol.
Solution

28. Example 4
This example illustrates a system that has no solution. Let us try to solve the system
Solution(自行練習)
0x1+0x2+0x3=1
The system has no solution.
隨堂作業：5(d)

29. Homogeneous System of linear Equations
Definition
A system of linear equations is said to be homogeneous (齊次) if all the constant terms (等號右邊的常數項) are zeros.
Example:
Observe that is a solution.
Theorem 1.1
A system of homogeneous linear equations in n variables always has the solution x1 = 0, x2 = 0. …, xn = 0. This solution is called the trivial solution.

30. Homogeneous System of linear Equations
Note. 除 trivial solution 外，可能還有其他解。
The system has other nontrivial solutions.
Example:
Theorem 1.2
A system of homogeneous linear equations that has more variables than equations has many solutions.
隨堂作業：8(e)

31. Homework
Exercise 1.2: 2, 5, 8, 14

32. 1.3 The Vector Space Rn Rectangular Coordinate System (直角座標系)
There are two ways of interpreting (5,3) - it defines the location of a point in a plane - it defines the position vector
the origin (原點)：(0, 0)
the position vector：
the initial point of : O
the terminal point of : A(5, 3)
ordered pair (序對): (a, b)
Figure 1.5

33. Example 1
Sketch the position vectors
Figure 1.6

34. R2 → R3
Figure 1.7

35. Definition
Let be a sequence of n real numbers. The set of all such sequences is called n-space and is denoted Rn.
u1 is the first component (分量) of , u2 is the second component and so on.
Example
R4 is the sets of sequences of four real numbers. For example, (1, 2, 3, 4) and (-1, 3, 5.2, 0) are in R4.
R5 is the set of sequences of five real numbers. For example, (-1, 2, 0, 3, 9) is in this set.

36. Addition and Scalar Multiplication
Definition
Let be two elements of Rn.
We say that u and v are equal if u1 = v1, …, un = vn. Thus two element of Rn are equal if their corresponding components are equal.
Definition
Let be elements of Rn and let c be a scalar. Addition and scalar multiplication are performed as follows
Addition:
Scalar multiplication :
Note.
(1) u, v Rn  u+v  Rn (Rn is closed under addition)(加法封閉性)
(2) u Rn, c R  cu  Rn (Rn is closed under scalar multiplication)(純量乘法封閉性)

37. Example 2
Let u = ( –1, 4, 3, 7) and v = ( –2, –3, 1, 0) be vector in R4.
Find u + v and 3u.
Solution
Example 3
Figure 1.8
Consider the vector (4, 1) and (2, 3), we get
(4, 1) + (2, 3) = (6, 4).

38. In general, if u and v are vectors in the same vector space, then u + v is the diagonal (對角線) of the parallelogram (平行四邊形) defined by u and v.
Figure 1.9

39. Example 4
Consider the scalar multiple of the vector (3, 2) by 2, we get
2(3, 2) = (6, 4)
Observe in Figure 4.6 that (6, 4) is a vector in the same direction as (3, 2), and 2 times it in length.
Figure 1.10

40. c > 1 0 < c < 1 –1 < c < 0 c < –1
Figure 1.11
In general, the direction of cu will be the same as the direction of u if c > 0, and the opposite direction to u if c < 0.
The length of cu is |c| times the length of u.

41. Special Vectors
The vector (0, 0, …, 0), having n zero components, is called the zero vector of Rn and is denoted 0.
Negative Vector
The vector (–1)u is written –u and is called the negative of u. It is a vector having the same magnitude (量) as u, but lies in the opposite direction to u. u -u
Subtraction
Subtraction is performed on elements of Rn by subtracting corresponding components. For example, in R3, (5, 3, -6) – (2, 1, 3) = (3, 2, -9)

42. Theorem 1.3
Let u, v, and w be vectors in Rn and let c and d be scalars.
u + v = v + u
u + (v + w) = (u + v) + w
u + 0 = 0 + u = u
u + (–u) = 0
c(u + v) = cu + cv
(c + d)u = cu + du
c(du) = (cd)u
1u = u
Figure 1.12 Commutativity of vector addition u + v = v + u

43. Linear Combinations of Vectors
We call au +bv + cw a linear combination (線性組合) of the vectors u, v, and w.
Let u = (2, 5, –3), v = ( –4, 1, 9), w = (4, 0, 2). Determine the linear combination 2u – 3v + w.
Example 5
Solution
隨堂作業：7(e)

44. Column Vectors Row vector: Column vector: 向量的另一種表示法
We defined addition and scalar multiplication of column vectors in Rn in a componentwise manner:
and
向量加法可視為矩陣運算

45. Subspaces of Rn Definition
We now introduce subsets of the vector space Rn that have all the algebraic properties of Rn. These subsets are called subspaces.
Definition
A subset S of Rn is a subspace if it is closed under addition and under scalar multiplication.
Recall:
(1) u, v S  u+v  S (S is closed under addition)(加法封閉性)
(2) u S, c R  cu  S (S is closed under scalar multiplication) (純量乘法封閉性)

46. Example 6
Consider the subset W of R2 of vectors of the form (a, 2a). Show that W is a subspace of R2.
隨堂作業：10(d)
Proof
Let u = (a, 2a), v = (b, 2b)  W, and k R.
u + v = (a, 2a) + (b, 2b) = (a+ b, 2a + 2b) = (a + b, 2(a + b))  W
and
ku = k(a, 2a) = (ka, 2ka)  W
Thus u + v  W and ku  W.
W is closed under addition and scalar multiplication.
W is a subspace of R2.
Observe: W is the set of vectors that can be written a(1,2).
Figure 1.13

47. Example 7
Consider the homogeneous system of linear equations. It can be shown that there are many solutions x1=2r, x2=5r, x3=r.
We can write these solutions as vectors in R3 as (2r, 5r, r). Show that the set of solutions W is a subspace of R3.
隨堂作業：13
Proof
Let u = (2r, 5r, r), v = (2s, 5s, s) W, and k R.
u + v = (2(r+s), 5(r+s), r+s)  W
and ku = (2kr, 5kr, kr)  W
Thus u + v  W and ku  W.
W is a subspace of R3.
Observe: W is the set of vectors that can be written r(2,5,1).
Figure 1.14

48. Homework
Exercise 1.3: 7, 9, 10, 12, 13

49. 1.4 Basis and Dimension Standard Basis of Rn
向量空間的某些子集合(subset)本身也是向量空間，稱為子空間(subspace)
Basis: a set of vectors which are used to describe a vector space.
Standard Basis of Rn
Consider the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) in R3.
These vectors have two very important properties:
They are said to span R3. That is, we can write an arbitrary vector (x, y, z) as a linear combination (線性組合) of the three vectors: For any (x,y,z) R3  (x,y,z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)
They are said to be linearly independent (線性獨立). If p(1, 0, 0) + q(0, 1, 0) +r(0, 0, 1) = (0, 0, 0)  p = 0, q = 0, r = 0 is the unique solution.
A set of vectors that satisfies the two preceding properties is called a basis.

50. There are many bases for R3 – sets that span R3 and are linearly independent.
For example, the set {(1, 2, 0), (0, 1, -1), (1, 1, 2)} is a basis for R3.
The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is the most important basis for R3. It is called the standard basis of R3.
R2 : two-dimensional space (二維空間)
R3 : three-dimensional space
觀察: dimension數等於basis中的向量個數 (故成為dimension定義)
The set {(1, 0, …, 0), (0, 1, …, 0), …, (0, …, 1)} of n vectors is the standard basis for Rn. The dimension (維度) of Rn is n.
隨堂作業：1

51. Span, Linear Independence, and Basis
The vectors v1, v2, and v3 are said to span a space if every vector v in the space can be expressed as a linear combination of them, v = av1 + bv2 + cv3.
The vectors v1, v2, and v3 are said to be linearly independent if the identity pv1 + qv2 + rvm = 0 is only true for p = 0, q = 0, r = 0.
A basis for a space is a set that spans the space and is linearly independent. The number of vectors in a basis is called the dimension of the space.

52. Example 1
Consider the subset W of R3 consisting of vectors of the form (a, b, a+b). The vector (2, 5, 7)W, whereas (2, 5, 9)W. Show that W is a subspace of R3.
Proof.
Let u=(a, b, a+b) and v=(c, d, c+d) be vectors in W and k be a scalar.
(1) u+v = (a, b, a+b) + (c, d, c+d) = (a+c, b+d, (a+c)+(b+d))
u+v W
(2) ku = k(a, b, a+b) = (ka, kb, ka+kb)
ku W
W is closed under addition and scalar multiplication.
W is a subspace of R3.

53. Example 1 (continue)
Separate the variables in the above arbitrary vector u.
u = (a, b, a+b) = (a, 0, a) + (0, b, b) = a(1, 0, 1) + b(0, 1, 1)
The vectors (1, 0, 1) and (0, 1, 1) thus span W.
Furthermore, p(1, 0, 1) + q(0, 1, 1) = (0, 0, 0) leads to p=0 and q=0.
The two vectors (1, 0, 1) and (0, 1, 1) are thus linearly independent.
The set {(1, 0, 1), (0, 1, 1)} is therefore a basis for W.
The dimension of W, dim(W)= 2.
有加法乘法封閉性
隨堂作業：4(a)

54. Example 2
Consider the subset V of R3 of vectors of the form (a, 2a, 3a). Show that V is a subspace of R3 and find a basis.
Sol.
(subspace)
Let u=(a, 2a, 3a) and v=(b, 2b, 3b) be vectors in V and k be a scalar.
(1) u+v = (a+b, 2(a+b), 3(a+b))
u+v V
(2) ku = k(a, 2a, 3a) = (ka, 2ka, 3ka)
ku V
V is closed under addition and scalar multiplication.
V is a subspace of R3.
(basis)
u = (a, 2a, 3a) = a(1, 2, 3)
 {(1, 2, 3)} is a basis for V.
隨堂作業：4(c)
 dim(V) = 1.

55. Example 3
Consider the following system of homogeneous linear equations.
It can be shown that there are many solutions x1 = 3r - 2s, x2 = 4r, x3 = r, x4 = s.
Write these solution as vectors in R4, (3r - 2s, 4r, r, s). It can be shown that this set of vectors is a subspace W of R4.
Find a basis for W and give its dimension.
Sol.
(1) (3r - 2s, 4r, r, s) = r(3, 4, 1, 0) + s( -2, 0, 0, 1)
(2) If p(3, 4, 1, 0) + q( -2, 0, 0, 1) = (0, 0, 0, 0), then p=0, q=0.
 {(3, 4, 1, 0), ( -2, 0, 0, 1) } is a basis for W.
 dim(W) = 2.
隨堂作業：11

56. Exercise 7
State with a brief explanation whether the following statements are true or false.
(a) The set {(1, 0, 0), (0, 1, 0)} is the basis for a two-dimensional subspace of R3.
(b) The set {(1, 0, 0)} is the basis for a one-dimensional subspace of R3.
(c) The vector (a, 2a, b) is an arbitrary vector in the plane spanned by the vectors (1, 2, 0) and (0, 0, 1).
(d) The vector (a, b, 2a-b) is an arbitrary vector in the plane spanned by the vectors (1, 0, 2) and (0, 1, -1).
(e) The set {(1, 0, 0), (0, 1, 0), (1, 1, 0)} is a basis for a subspace of R3.
(f) R2 is a subspace of R3.

57. Homework
Exercise 1.4: 1, 4, 11, 12

58. 1.5 Dot Product, Norm, Angle, and Distance
In this section we develop a geometry for the vector space Rn.
Definition
Let be two vectors in Rn. The dot product (內積) of u and v is denoted u‧v and is defined by
The dot product assigns a real number to each pair of vectors.
The dot product is a tool that is used to build the geometry of Rn.
Example
Find the dot product of u = (1, –2, 4) and v = (3, 0, 2)
Solution

59. Properties of the Dot Product
Let u, v, and w be vectors in Rn and let c be a scalar. Then
u‧v = v‧u
(u + v)‧w = u‧w + v‧w
cu‧v = c(u‧v) = u‧cv
u‧u  0, and u‧u = 0 if and only if u = 0
Proof 1. 2.
隨堂作業：2(b)

60. Norm (length) of a Vector in Rn
Figure 1.17
Definition
The norm (length or magnitude) of a vector u = (u1, …, un) in Rn is denoted ||u|| and defined by
Note: The norm of a vector can also be written in terms of the dot product

61. Example
Find the norm of the vectors u = (1, 3, 5) of R3 and v = (3, 0, 1, 4) of R4.
Solution
Definition
A unit vector is a vector whose norm is 1. If v is a nonzero vector, then the vector
is a unit vector in the direction of v.
This procedure of constructing a unit vector in the same direction as a given vector is called normalizing the vector.

62. Example 1
Find the norm of the vector (2, –1, 3). Normalize this vector.
Solution
The norm of (2, –1, 3) is
The normalized vector is
The vector may also be written
This vector is a unit vector in the direction of (2, –1, 3).
隨堂作業：6(b), 10(d)

63. Angle between Vectors (R2)
Let u=(a, b) and v=(c, d). Find the angle q between u and v.
The law of cosines gives
We get that
Figure 1.18
0    

64. Angle between Vectors (R n)
Definition
Let u and v be two nonzero vectors in Rn. The cosine of the angle  between these vectors is
Example 2
Determine the angle between the vectors u = (1, 0, 0) and v = (1, 0, 1) in R3.
Solution
Thus the angle between u and v is p/4 (or 45).
隨堂作業：13(c)

65. Definition
Two nonzero vectors are orthogonal (正交) if the angle between them is a right angle (直角).
Two nonzero vectors u and v are orthogonal if and only if u‧v = 0.
Theorem 1.4
可寫成uv
Proof
Example
The vectors (2, –3, 1) and (1, 2, 4) are orthogonal since
(2, –3, 1)‧(1, 2, 4) = (2  1) + (–3  0) + (1  4) = 2 – = 0.

66. Properties of Standard basis of Rn
(1, 0), (0,1) are orthogonal unit vectors in R2.
(1, 0, 0), (0, 1, 0), (0, 0, 1) are orthogonal unit vectors in R3.
We call a set of unit pairwise orthogonal vectors an orthonormal set.
The standard basis for Rn, {(1, 0, …, 0), (0, 1, 0, …, 0), …, (0, …, 0, 1)} is an orthonormal set.
隨堂作業：16(d)

67. Example 3
(a) Let w be a vector in Rn. Let W be the set of vectors that are orthogonal to w. Show that W is a subspace of Rn.
(b) Find a basis for the subset W of vectors in R3 that are orthogonal to w=(1, 3, 1). Give the dimension and a geometrical description of W.
Solution
(a) Let u, vW. Since uw and vw, we have uw=0 and vw=0.
(u+v)w = uw + vw =  u+v w  u+v W
If c is a scalar, c(uw) = cuw =  cu w  cu W
W is a subspace of Rn.
(b) Let (a, b, c)W and (a, b, c)w, then (a, b, c)(1, 3, 1)=0
 a+3b+c=0
 W is the set {(a, b, -a - 3b) | a, b R}
Since (a, b, -a - 3b) = a(1, 0, -1) + b(0, 1, -3). It is clear that {(1, 0, -1), (0, 1, -3)} is a basis for W  dim(W)=2

68. Example 3 (continue) W is the plane in R3 defined by
(1, 0, -1) and (0, 1, -3).
Figure 1.19
隨堂作業：20

69. Distance between Points
The distance between x=(x1, x2) and y =(y1, y2) is
Generalize this expression to Rn.
Def. Let x=(x1, x2, …, xn) and y=(y1, y2, …, yn) be two points in Rn. The distance between x and y is denoted d(x, y) and is defined by
Note: We can also write this distance as follows. x y
x-y
Example 4
Determine the distance between the points x = (1,–2 , 3, 0) and y = (4, 0, –3, 5) in R4.
Solution

70. Example 5
Prove that the distance in Rn has the following symmetric property: d(x, y)=d(y, x) for any x, y  Rn.
Solution
Let

71. Theorem 1.5 The Cauchy-Schwartz Inequality.
If u and v are vectors in Rn then
Here denoted the absolute value (絕對值) of the number uv.

72. Theorem 1.6 Let u and v be vectors in Rn. Triangle Inequality (三角不等式):
||u + v||  ||u|| + ||v||.
Pythagorean theorem (畢氏定理): If u‧v = 0 then ||u + v||2 = ||u||2 + ||v||2.
Figure 1.21

73. Homework
Exercise 1.5: 2, 6, 10, 13, 16, 20, 33
(1.6節跳過)