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

2. 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 ). 1.1 Matrices and Systems of Linear Equations system of linear equations ( 線性聯立方程式 ) Ch1_2

3. Figure 1.2 No solution ( 無解 ) –2x + y = 3 –4x + 2y = 2 Lines are parallel. No point of intersection. No solutions. Solutions for system of linear equations 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. 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. Ch1_3

4. Ch1_4 Definition A linear equation in n variables x 1, x 2, x 3, …, x n has the form a 1 x 1 + a 2 x 2 + a 3 x 3 + … + a n x n = b where the coefficients ( 係數 ) a 1, a 2, a 3, …, a n 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. Ch1_5 A linear equation in three variables corresponds to a plane in three-dimensional ( 三維 ) space. Unique solution ※ Systems of three linear equations in three variables:

6. Ch1_6 No solutions Many solutions

7. Ch1_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. Ch1_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. Ch1_9 Submatrix ( 子矩陣 ) Row ( 列 ) and Column ( 行 )

10. Ch1_10 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 Location a ij 表示在 row i, column j 的元素值 也寫成 location (1,3) =  4 Size and Type

11. Ch1_11 matrix of coefficient and augmented matrix Relations between system of linear equations and matrices 係數矩陣擴大矩陣 隨堂作業： 5(f)

12. Ch1_12 給定聯立方程式後， 不會改變解的一些轉換 Elementary Transformation 1.Interchange two equations. 2.Multiply both sides of an equation by a nonzero constant. 3.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 ) 1.Interchange two rows of a matrix. ( 兩列交換 ) 2.Multiply the elements of a row by a nonzero constant. ( 某列的元素同乘一非零常數 ) 3.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 ) Elementary Row Operations of Matrices

13. Ch1_13 Example 1 Solving the following system of linear equation. Solution Equation Method Initial system: Analogous Matrix Method Augmented matrix: Eq2+(–2)Eq1 Eq3+(–1)Eq1 R2+(–2)R1 R3+(–1)R1   符號表示 row equivalent

14. Ch1_14 Eq1+(–1)Eq2 Eq3+(2)Eq2 (–1/5)Eq3 Eq1+(–2)Eq3 Eq2+Eq3 The solution is R1+(–1)R2 R3+(2)R2  (–1/5)R3  R1+(–2)R3 R2+R3  隨堂作業： 7(d)

15. Ch1_15 基本列運算符號說明： R1+(–3)R2  表示將 R1( 第一列 ) 加上 (  3)  R2 ， 所以是 R1 這列會改變， R2 不變 For example: (1) R1+2R2  (2) R2+2R1 

16. Ch1_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 不好！ 2R1+R2  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. Ch1_17 Example 2 Solving the following system of linear equation. Solution ( 請先自行練習 )

18. Ch1_18 Example 3 Solve the system Solution ( 請先自行練習 ) 隨堂作業： 10(d)(f)

19. Ch1_19 Summary AB Use row operations to [A: B] : i.e., Def. [I n : 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 I n ( A  I n ). 2. If A  I n, then the system has unique solution.

20. Ch1_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  隨堂作業： 13(b) The solutions to the three systems are

21. Ch1_21 Homework Exercise 1.1 ： 1, 2, 4, 5, 6, 7, 10, 13

22. Ch1_22 1.2 Gauss-Jordan Elimination Definition A matrix is in reduced echelon form ( 簡化梯式 ) if 1.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 ) 2.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) 3.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 ) 4.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. Ch1_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. Ch1_24 Gauss-Jordan Elimination System of linear equations  augmented matrix  reduced echelon form  solution

25. Ch1_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. Ch1_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. Ch1_27 Example 3 This example illustrates that the general solution can involve a number of parameters. Solve the system of equations Solution 變數個數 > 方程式個數  many sol.

28. Ch1_28 Example 4 This example illustrates a system that has no solution. Let us try to solve the system Solution ( 自行練習 ) The system has no solution. 0x 1 +0x 2 +0x 3 =1 隨堂作業： 5(d)

29. Ch1_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 x 1 = 0, x 2 = 0. …, x n = 0. This solution is called the trivial solution.

30. Ch1_30 Homogeneous System of linear Equations Theorem 1.2 A system of homogeneous linear equations that has more variables than equations has many solutions. Note. 除 trivial solution 外，可能還有其他解。 隨堂作業： 8(e) The system has other nontrivial solutions. Example:

31. Ch1_31 Homework Exercise 1.2: 2, 5, 8, 14

32. 1.3 The Vector Space R n Figure 1.5 Rectangular Coordinate System ( 直角座標系 ) 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) There are two ways of interpreting (5,3) - it defines the location of a point in a plane - it defines the position vector Ch1_32

33. Example 1 Sketch the position vectors. Figure 1.6 Ch1_33

34. Figure 1.7 R 2 → R 3 Ch1_34

35. Definition Let be a sequence of n real numbers. The set of all such sequences is called n-space and is denoted R n. u 1 is the first component ( 分量 ) of, u 2 is the second component and so on. Example R 4 is the sets of sequences of four real numbers. For example, (1, 2, 3, 4) and (  1, 3, 5.2, 0) are in R 4. R 5 is the set of sequences of five real numbers. For example, (  1, 2, 0, 3, 9) is in this set. Ch1_35

36. Definition Let be two elements of R n. We say that u and v are equal if u 1 = v 1, …, u n = v n. Thus two element of R n are equal if their corresponding components are equal. Definition Let be elements of R n and let c be a scalar. Addition and scalar multiplication are performed as follows Addition: Scalar multiplication : Ch1_36 Addition and Scalar Multiplication Note. (1) u, v  R n  u+v  R n (R n is closed under addition)( 加法封閉性 ) (2) u  R n, c  R  cu  R n (R n 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 R 4. 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). Ch1_37

38. Figure 1.9 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. Ch1_38

39. Example 4 Figure 1.10 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. Ch1_39

40. Figure 1.11 c > 1 0 < c < 1 –1 < c < 0 c < –1 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. Ch1_40

41. Special Vectors The vector (0, 0, …, 0), having n zero components, is called the zero vector of R n 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. Subtraction Subtraction is performed on elements of R n by subtracting corresponding components. For example, in R 3, (5, 3,  6) – (2, 1, 3) = (3, 2,  9) u uu Ch1_41

42. Theorem 1.3 Let u, v, and w be vectors in R n and let c and d be scalars. (a)u + v = v + u (b)u + (v + w) = (u + v) + w (c)u + 0 = 0 + u = u (d)u + (–u) = 0 (e) c(u + v) = cu + cv (f)(c + d)u = cu + du (g) c(du) = (cd)u (h) 1u = u Figure 1.12 Commutativity of vector addition u + v = v + u Ch1_42

43. Linear Combinations of Vectors Solution 隨堂作業： 7(e) Let u = (2, 5, –3), v = ( –4, 1, 9), w = (4, 0, 2). Determine the linear combination 2u – 3v + w. Example 5 We call au +bv + cw a linear combination ( 線性組合 ) of the vectors u, v, and w. Ch1_43

44. Column Vectors We defined addition and scalar multiplication of column vectors in R n in a componentwise manner: and 向量加法可視為矩陣運算 Row vector: Column vector: 向量的另一種表示法 Ch1_44

45. Subspaces of R n Ch1_45 Definition A subset S of R n is a subspace if it is closed under addition and under scalar multiplication. We now introduce subsets of the vector space R n that have all the algebraic properties of R n. These subsets are called subspaces. 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. Ch1_46 Example 6 Consider the subset W of R 2 of vectors of the form (a, 2a). Show that W is a subspace of R 2. 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 R 2. Observe: W is the set of vectors that can be written a(1,2). Figure 1.13 隨堂作業： 10(d)

47. Ch1_47 Example 7 Consider the homogeneous system of linear equations. It can be shown that there are many solutions x 1 =2r, x 2 =5r, x 3 =r. We can write these solutions as vectors in R 3 as (2r, 5r, r). Show that the set of solutions W is a subspace of R 3. 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 R 3. Observe: W is the set of vectors that can be written r(2,5,1). Figure 1.14 隨堂作業： 13

48. Homework Exercise 1.3: 7, 9, 10, 12, 13 Ch1_48

49. 1.4 Basis and Dimension Consider the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) in R 3. These vectors have two very important properties: (i)They are said to span R 3. 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)  R 3  (x,y,z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1) (ii)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. Ch1_49 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 R n A set of vectors that satisfies the two preceding properties is called a basis.

50. The set {(1, 0, …, 0), (0, 1, …, 0), …, (0, …, 1)} of n vectors is the standard basis for R n. The dimension ( 維度 ) of R n is n. Ch1_50 There are many bases for R 3 – sets that span R 3 and are linearly independent. For example, the set {(1, 2, 0), (0, 1,  1), (1, 1, 2)} is a basis for R 3. The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is the most important basis for R 3. It is called the standard basis of R 3. R 2 : two-dimensional space ( 二維空間 ) R 3 : three-dimensional space 觀察 : dimension 數等於 basis 中的向量個數 ( 故成為 dimension 定義 ) 隨堂作業： 1

51. Span, Linear Independence, and Basis The vectors v 1, v 2, and v 3 are said to span a space if every vector v in the space can be expressed as a linear combination of them, v = av 1 + bv 2 + cv 3. Ch1_51 The vectors v 1, v 2, and v 3 are said to be linearly independent if the identity pv 1 + qv 2 + rv m = 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. Consider the subset W of R 3 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 R 3. Example 1 Ch1_52 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 R 3.

53. Ch1_53 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. Example 1 (continue) 有加法乘法封閉性 隨堂作業： 4(a)

54. Consider the subset V of R 3 of vectors of the form (a, 2a, 3a). Show that V is a subspace of R 3 and find a basis. Example 2 Ch1_54 Sol. 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 R 3. (subspace) (basis) u = (a, 2a, 3a) = a(1, 2, 3)  {(1, 2, 3)} is a basis for V.  dim(V) = 1. 隨堂作業： 4(c)

55. Consider the following system of homogeneous linear equations. Example 3 Ch1_55 It can be shown that there are many solutions x 1 = 3r  2s, x 2 = 4r, x 3 = r, x 4 = s. Write these solution as vectors in R 4, (3r  2s, 4r, r, s). It can be shown that this set of vectors is a subspace W of R 4. Find a basis for W and give its dimension. (1) (3r  2s, 4r, r, s) = r(3, 4, 1, 0) + s(  2, 0, 0, 1)  {(3, 4, 1, 0), (  2, 0, 0, 1) } is a basis for W.  dim(W) = 2. 隨堂作業： 11 (2) If p(3, 4, 1, 0) + q(  2, 0, 0, 1) = (0, 0, 0, 0), then p=0, q=0. Sol.

56. Ch1_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 R 3. (b) The set {(1, 0, 0)} is the basis for a one-dimensional subspace of R 3. (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 R 3. (f) R 2 is a subspace of R 3.

57. Homework Exercise 1.4: 1, 4, 11, 12 Ch1_57

58. 1.5 Dot Product, Norm, Angle, and Distance Definition Let be two vectors in R n. 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. Example Find the dot product of u = (1, –2, 4) and v = (3, 0, 2) Solution Ch1_58 In this section we develop a geometry for the vector space R n. The dot product is a tool that is used to build the geometry of R n.

59. Properties of the Dot Product Let u, v, and w be vectors in R n and let c be a scalar. Then 1.u ‧ v = v ‧ u 2.(u + v) ‧ w = u ‧ w + v ‧ w 3. cu ‧ v = c(u ‧ v) = u ‧ cv 4.u ‧ u  0, and u ‧ u = 0 if and only if u = 0 Proof 1. 2. 隨堂作業： 2(b) Ch1_59

60. Norm (length) of a Vector in R n Definition The norm (length or magnitude) of a vector u = (u 1, …, u n ) in R n is denoted ||u|| and defined by Note: The norm of a vector can also be written in terms of the dot product Ch1_60 Figure 1.17

61. 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. Find the norm of the vectors u = (1, 3, 5) of R 3 and v = (3, 0, 1, 4) of R 4. Solution Example Ch1_61

62. Example 1 Solution Find the norm of the vector (2, –1, 3). Normalize this vector. 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) Ch1_62

63. Angle between Vectors (R 2 ) Figure 1.18 The law of cosines gives We get that Ch1_63 Let u=(a, b) and v=(c, d). Find the angle  between u and v. 0    

64. Definition Let u and v be two nonzero vectors in R n. 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 R 3. Solution Thus the angle between u and v is  /4 (or 45  ). 隨堂作業： 13(c) Angle between Vectors (R n ) Ch1_64

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 Proof Ch1_65 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 – 6 + 4 = 0. 可寫成 u  v

66. Properties of Standard basis of R n (1, 0), (0,1) are orthogonal unit vectors in R 2. (1, 0, 0), (0, 1, 0), (0, 0, 1) are orthogonal unit vectors in R 3. Ch1_66 We call a set of unit pairwise orthogonal vectors an orthonormal set. The standard basis for R n, {(1, 0, …, 0), (0, 1, 0, …, 0), …, (0, …, 0, 1)} is an orthonormal set. 隨堂作業： 16(d)

67. Ch1_67 Example 3 (a) Let w be a vector in R n. Let W be the set of vectors that are orthogonal to w. Show that W is a subspace of R n. (b) Find a basis for the subset W of vectors in R 3 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 = 0  u+v  w  u+v  W If c is a scalar, c(u  w) = cu  w = 0  cu  w  cu  W  W is a subspace of R n. (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. Ch1_68 Figure 1.19 Example 3 (continue) W is the plane in R 3 defined by (1, 0,  1) and (0, 1,  3). 隨堂作業： 20

69. Distance between Points Def. Let x=(x 1, x 2, …, x n ) and y=(y 1, y 2, …, y n ) be two points in R n. The distance between x and y is denoted d(x, y) and is defined by Note: We can also write this distance as follows. Example 4 Determine the distance between the points x = (1,–2, 3, 0) and y = (4, 0, –3, 5) in R 4. Solution x y xyxy Ch1_69 The distance between x=(x 1, x 2 ) and y =(y 1, y 2 ) is Generalize this expression to R n.

70. Example 5 Prove that the distance in R n has the following symmetric property: d(x, y)=d(y, x) for any x, y  R n. Solution Let Ch1_70

71. Theorem 1.5 The Cauchy-Schwartz Inequality. If u and v are vectors in R n then Here denoted the absolute value ( 絕對值 ) of the number u  v. Ch1_71

72. Theorem 1.6 Let u and v be vectors in R n. (a)Triangle Inequality ( 三角不等式 ): ||u + v||  ||u|| + ||v||. (a)Pythagorean theorem ( 畢氏定理 ): If u ‧ v = 0 then ||u + v|| 2 = ||u|| 2 + ||v|| 2. Ch1_72 Figure 1.21

73. Homework Exercise 1.5: 2, 6, 10, 13, 16, 20, 33 (1.6 節跳過 ) Ch1_73