chapter 2 basic linear algebra ( 基本線性代數 ) 2 2.1 2.1 matrices ( 矩陣 ) & vectors (...

Chapter 2

Basic Linear Algebra( 基本線性代數 )

2

2.12.1 Matrices( 矩陣 ) & Vectors( 向量 )

A matrix( 矩陣 )is any rectangular array of numbers

If a matrix If a matrix AA has has mm rows and rows and nn columns it is columns it is referred to as an referred to as an mm x x nn matrix. matrix.

mm x x nn is the is the order( 階 ) of the matrix. It is typically of the matrix. It is typically written aswritten as

1

3

2

4

1

4

2

5

3

6

1

2

2 1( )

A

a 11

a 21

....

a m1

a 12

a 22

....

a m2

....

....

....

....

a 1n

a 2n

....

a mn

p.11

3

The number in the The number in the iith row and th row and jjth column of A is th column of A is called the called the ijth element of of AA and is written and is written aaij..

Two matrices A = [aij] and B = [bij] are equal if and only if A and B are the same order and for all i and j, aij = bij.

A = B if and only if x = 1, y = 2, w = 3, and z = 4A = B if and only if x = 1, y = 2, w = 3, and z = 4

If A1

3

2

4

and Bx

w

y

z

4

Any matrix with only one column is a column vector( 行向量 ) or column matrix ( 行矩陣 ) .The number of rows in a column vector is the dimension of the column vector.

C=

Rm will denote the set all m-dimensional column vectors

Any matrix with only one row (a 1 x n matrix) is a row vector ( 列向量 ) or row matrix ( 列矩陣 ).

The dimension of a row vector is the number of columns. R=(1 2 3)

1

2

5

Any m-dimensional vector (either row or column) in which all the elements equal zero is called a zero vector ( 零向量 )or zero matrix ( 零矩陣 ) (written 0).

Any m-dimensional vector corresponds to a directed line segment in the m-dimensional plane. For example, the two-dimensional vector uu

corresponds to the line segment joining the point (0,0) to the point (1,2)

0 0 0( )0

0

6

Other Forms

Diagonal matrix ( 對角線矩陣 )

Identity matrix ( 單位矩陣 )

Upper triangular matrix( 上三角矩陣 )

Lower triangular matrix ( 下三角矩陣 )

)nj,i1 ,ji ( 0a ,]a[A ijnnij ≤≤≠∀

nj,i1 ; ji ,0

ji ,1a ,]a[A ijnnij

)nj,i1 ,ji ( 0a ,]a[A ijnnij ≤≤∀

)nj,i1 ,ji ( 0a ,]a[A ijnnij ≤≤∀

7

Example :

8

Transpose Matrix ( 轉置矩陣 )

,A nnaij

.atrixranspose mcalled a t is ]a[=A

then )nj1 m,i1(a=a ,]a[=A If

m×nTij

T

ijTijn×mij ≤≤≤≤

p.15

9

Example :

34

10

17

41

A =A T⇒

3114

4071

10

轉置矩陣之性質

786

921

543

A

708060

902010

504030

B

11

Square Matrix of Order n( 方陣之乘冪 )

令 A=[aij]nxn, 則矩陣 A 的乘命冪 ( 非負 ) 定義為 A0=In,

A1=A,

且對 K≥2, AK=(AK-1)(A)

若 A 為一方陣 , 且若 r 與 s 均為非負整數 , 則

(1) Ar|s=(Ar)(As)

(2) (Ar)s=Ars=(As)r

12

31

42A

51

23B

1955

460160

238

167

55

200BA

but

2890

432100

170

1610)AB(

22

2

2

Example : (AB)2≠A2B2 (AB = BA)?

13

Symmetric Matrix ( 對稱矩陣 ) & Skew-symmetric Matrix ( 斜對稱矩陣 )

14

Example :

029

205

950

653

542

321

B

Ais a symmetric matrix

is a skew-symmetric matrix

15

對稱矩陣與斜對稱矩陣之性質

16

Matrix Operations ( 矩陣之基本運算 )

The scalar product( 純量積 ) is the result of multiplying two vectors where one vector is a column vector and the other is a row vector. For the scalar product to be defined, the

dimensions of both vectors must be the same. The scalar product of u and v is written:

u v u 1 v1 u 2 v2 .... u n vn

u 1 2 3( ) v212

u v 1 2( ) 2 1( ) 3 2( ) 10

p.13

17

The Scalar Multiple of a MatrixGiven any matrix A and any number c, the matrix

cA is obtained from the matrix A by multiplying each element of A by c.

Addition of Two MatricesLet A = [aij] and B =[bij] be two matrixes with the

same order. Then the matrix C = A + B is defined to be the m x n matrix whose ijth element is aij + bij.

Thus, to obtain the sum of two matrixes A and B, we add the corresponding elements of A and B.

A1

1

2

0

3 A3

3

6

0

A1

0

2

1

3

1

B1

2

2

1

3

1

C1 1

0 2

2 2

1 1

3 3

1 1

0

2

0

0

0

0

p.14

18

矩陣相加

6753

4812

7543

A

3276

8593

4257

B ?C

19

This rule for matrix addition may be used to add vectors of the same dimension.

Vectors may be added using the parallelogram law or by using matrix addition.

v 2 1( )

u 1 2( )

u v 3 3( )

X1

X2

u

v

u+v

1 2 3

1

2

3

(1,2)

(2,1)

(3,3)

20

Line segments can be defined using scalar multiplication and the addition of matrices. If u=(1,2) and v=(2,1), the line segment

joining u and v (called uv) is the set of all points in the m-dimensional plane corresponding to the vectorscu +(1-c)v, where 0 ≤ c ≤ 1.

X1

X2

u

v

1 2

1

2c=1

c=1/2

c=0

21

矩陣加法之性質

22

常數乘以矩陣之性質

23

Matrix Multiplication ( 矩陣相乘 ) Given to matrices A and B, the matrix

product of A and B (written AB) is defined if and only if the number of columns in A = the number of rows in B.

The matrix product C = AB of A and B is the m x n matrix C whose ijth element is determined as follows: ijth element of C = scalar product of row i of

A x column j of B

p.16

24

矩陣相乘

26

矩陣乘法之性質

27

Example 1: Matrix Multiplication

Computer C = AB for

Solution Because A is a 2x3 matrix and B is a 3x2 matrix,

AB is defined, and C will be a 2x2 matrix.

A1

2

1

1

2

3

B

1

2

1

1

3

2

C5

7

8

11

C11

1 1 2( )

1

2

1

5 C12

1 1 2( )

1

3

2

8

C21

2 1 3( )

1

2

1

7 C22

2 1 3( )

1

3

2

11

p.16

28

LU Decomposition (LU 分解法 )

30

例題：

32

Consider a system of linear equations.

The variables, or unknowns, are referred to as x1, x2, …, xn while the aij’s and bj’s are constants.

A set of such equations is called a linear system of m equations in n variables.

A solution to a linear set of m equations in n unknowns is a set of values for the unknowns that satisfies each of the system’s m equations.

a11 x1 a12 x2 .... a1n xn b1

a21 x1 a22 x2 .... a2n xn b1

.... .... .... ....

am1 x1 am2 x2 .... amn xn bm

2.2 Matrices and Systems of Linear Equations

p.20

33

Example 5: Solution to Linear System

Show that

a solution to the linear system

and that

is not a solution to the linear system.

x1

2

1

3x

p.21

0=x+x2

5=x2+x

21

21

34

Example 5 ： Solution

To show that

is a solution, x1=1 and x2=2 must be substituted in both equations. The equations must be satisfied.

The vector

is not a solution, because x1=3 and x2=1 fail to satisfy 2x1-x2=0

x1

2

1

3x

35

Matrices can simplify and compactly represent a system of linear equations.

This system of linear equations may be written as This system of linear equations may be written as AAxx==bb and is called it’s and is called it’s matrix representation..

AA : : coefficent matrix ( 係數矩陣 ) X X : : variable matrix ( 變數矩陣 ) bb : : constant matrix ( 常數矩陣 ) A|b : : augmented matrix ( 擴增矩陣 )

A

a 11

a 21

....

a m1

a12

a 22

....

a m2

....

....

....

....

a 1n

a 2n

....

a mn

x

x1

x2

....

xn

b

b 1

b 2

....

b m

a11 x1 a12 x2 .... a1n xn b1

a21 x1 a22 x2 .... a2n xn b1

.... .... .... ....

am1 x1 am2 x2 .... amn xn bm

p.21

36

Row equivalent( 列同義 ) & Elementary matrix( 基本矩陣 )

37

交換

乘常數

相加

38

Elementary row operations( 基本列運算 )

Elementary row operations (ERO) transforms agiven matrix A into a new matrix A’ via one of thefollowing operations: Type 1 ERO –A’ is obtained by multiplying any row

of A by a nonzero scalar. ( 乘上一非零定數 )Type 2 ERO – Multiply any row of A (say, row i) by a

nonzero scalar c. For some j ≠ i, let row j of A’ = c*(row i of A) + row j of A and the other rows of A’ be the same as the rows of A. ( 乘上一非零定數 + 另一列 )

Type 3 ERO – Interchange any two rows of A. ( 任二列交換 )

p.22

39

Reduced Row Echelon Form( 簡約列梯陣 )

40

可求一方陣的逆方陣 (A-1)

可解線性方程組。。基本列運算將矩陣變形成另一矩陣，，使所得矩陣適合

某一特殊形式 ( 同義 ) 。。原始矩陣與所得矩陣並無相等關係

基本列運算 (EROs) 之功能

41

Example :Example:

43

2.3 The Gauss-Jordan Method

Also Gauss-Jordan Elimination( 高斯 - 約旦消去法 ).

Using the Gauss-Jordan method, it can be shown that any system of linear equations must satisfy one of the following three cases:Case 1. no solution.Case 2. a unique solution.Case 3. an infinite number of solutions.

The Gauss-Jordan method is important because many of the manipulations used in this method are used when solving linear programming problems by the simplex algorithm.

The Gauss-Jordan method solves a linear equation system by utilizing EROs in a systematic fashion.

p.22

44

Case1 ： a unique solution

P.24 The steps to using the Gauss-Jordan method

The augmented matrix representation is

2

2

1

2

1

1

1

2

2

9

6

5

A|b =A|b =

5=2x+ x x

6=2x+ x 2x

9=x + 2x + 2x

321

321

321

p.24

45

Step 1 Multiply row 1 by ½. This Type 1 ERO yields

Step 2 Replace row 2 of A1|b1 by -2(row 1 A1|b1) + row 2 of A1|b1. The result of this Type 2 ERO is

5211

6212

11

bA29

21

11 |

5211

3130

11

bA29

21

22 |

46

Step 3 Replace row 3 of A2|b2 by -1(row 1 of A2|b2) + row 3 of A2|b2. The result of this Type 2 ERO is

The first column has now been transformed into

1

0

0

21

23

29

21

33

20

3130

11

bA |

47

Step 4 Multiply row 2 of A3|b3 by -1/3. The result of this Type 1 ERO is

Step 5 Replace row 1 of A4|b4 by -1(row 2 of A4|b4) + row 1 of A4|b4. The result of this Type 2 ERO is

21

23

31

29

21

44

20

110

11

bA |

21

23

31

27

65

55

20

110

01

bA |

48

Step 6 Place row 3 of A5|b5 by 2(row 2 of A5|b5) + row 3 of A5|b5. The result of this Type 2 ERO is

Column 2 has now been transformed into

1

0

0

25

65

31

27

65

66

00

110

01

bA |

49

Step 7 Multiply row 3 of A6|b6 by 6/5. The result of this Type 1 ERO is

Step 8 Replace row 1 of A7|b7 by -5/6 (row 3 of A7|b7)+A7|b7. The result of this Type 2 ERO is

3300

13

100

2

7

6

501

3300

110

01

bA 31

27

65

77 |

3300

110

1001

bA 31

88 |

51

Case2 ： no solution

Example 6 (p.28)

p.28

x1+ 2x2 =3

2x1+ 4x2 =4

52

Case3 ： an infinite number of solutions

Example 7 (p.28)

p.28

x1+ x2 =1

x2+ x3 =3

x1+2x2+ x3 =4

53

Exercise : Solving linear system by Gauss-Jordan Elimination

55

原擴增矩陣經有限次之基本列運算後，同義於一上三角矩陣，故由後代法 (backward-substitution) 解得 x1,x2,x3 。

56

若繼續矩陣之基本列運算，使其係數矩陣同義於一單位矩陣，則可直接求得 x1,x2,x3 ，而不必使用後代法的運算步驟，繼續矩陣之基本列運算。

58

After the Gauss Jordan method has been applied to any linear system, a variable that appears with a coefficient of 1 in a single equation and a coefficient of 0 in all other equations is called a basic variable ( 基本變數 , BV).

Any variable that is not a basic variable is called a nonbasic variable ( 非基本變數 ,NBV).

Basic Variablesp.30

59

kcx

cx

kx

cx

kx

bA

3

22

1

000000

110100

202010

311001

'|'

1

2

3

4

5

20000

00000

13100

12010

11001

'|' bA

3

2

1

3100

2010

1001

'|'

3

2

1

x

x

x

bA

No solution

Unique solution

BV={x1,x2,x3}, NBV is empty.

Infinite number of solutions

BV={x1,x2,x3}

NBV={x4,x5}

p.30

60

Summary of Gauss-Jordan Method

Does A' | b ' have a row [ 0, 0 , ..., 0 | c ] (c < > 0 ) ?

Ax = b has nosolution.

Find BV and NBV.

Is NBV empty?

Ax = b has a uniquesolution.

Ax = b has an infinitenumber of solutions.

Yes No

Yes

No

p.32

Figrue 6

61

Exercise ：

方程組，求所有 a 之值。

(1)no solution

(2)a unique solution

(3) infinite number of solutions

a10)xa(xx

43x xx

3x xx

32

21

321

321

(3)a=3

(2) 除了 a=3,-3 以外

(1) a=-3

62

Exercise ： p.32 #1 ~ #8

1. X1+ X2+ X4 =3 X2+ X3 =4 X1+2X2+ X3+ X4 =8

2. X1+ X2+ X3 =4 X1+2X2 =6

3. X1+ X2 =12X1+ X2 =33X1+2X2 =4

4. 2X1- X2+ X3+ X4 =6 X1+ X2+ X3 =4

5. X1+ X4 =5 X2+ 2X4 =5 X3+ 0.5X4 =1 2X3+ X4 =3

6. 2X2+ 2X3 =4 X1+2X2+ X3 =4 X2- X3 =0

7. X1+ X2 =2 -X2+ 2X3 =3 X2+ X3 =3

8. X1+ X2+ X3 =1 X2+ 2X3+ X4 =2 X4 =3

63

2.4 Linear Independence and Linear Dependence

A linear combination( 線性組合 ) of the vectors in V is any vector of the form c1v1 + c2v2 + … + ckvk where c1, c2, …, ck are arbitrary scalars.

A set of V of m-dimensional vectors is linearly independent( 線性獨立 ) if the only linear combination of vectors in V that equals 0 is the trivial linear combination.

A set of V of m-dimensional vectors is linearly dependent ( 線性相依 ) if there is a nontrivial linear combination of vectors in V that adds up to 0.

P.32

64

Example 10: LD Set of Vectors

Show that V = {[ 1 , 2 ] , [ 2 , 4 ]} is a linearly dependent set of vectors.

Solution Since 2([ 1 , 2 ])–1([ 2 , 4 ])=(0 0), there is a

nontrivial linear combination with c1=2 and c2= -1 that yields 0. Thus V is a linear dependent set of vectors.

65

What does it mean for a set of vectors to linearly dependent?

A set of vectors is linearly dependent only if some vector in V can be written as a nontrivial linear combination of other vectors in V.

If a set of vectors in V are linearly dependent, the vectors in V are, in some way, NOT all “different” vectors. By “different we mean that the direction specified by any vector in V cannot be expressed by adding together multiples of other vectors in V. For example, in two dimensions, two linearly dependent vectors lie on the same line.

Linear dependent

66

Let A be any m x n matrix, and denote the rows of A by r1, r2, …, rm. Define R = {r1, r2, …, rm}.

The rank( 秩 ) of A is the number of vectors in the largest linearly independent subset of R.

To find the rank of matrix A, apply the Gauss-Jordan method to matrix A. (Example 14, p.35)

Let A’ be the final result. It can be shown that the rank of A’ = rank of A. The rank of A’ = the number of nonzero rows in A’. Therefore, rank A = rank A’ = number of nonzero rows in A’.

The Rank of a MatrixP.34

67

A method of determining whether a set of vectors V = {v1, v2, …, vm} is linearly dependent is to form a matrix A whose ith row is vi. (p.35)

If the rank A = m, then V is a linearly independent set of vectors.

If the rank A < m, then V is a linearly dependent set of vectors.

Whether a Set of Vectors Is Linear Independent

68

2.52.5 The Inverse of a Matrix ( 反矩陣 )

A square matrix( 方陣 ) is any matrix that has an equal number of rows and columns.

The diagonal elements ( 對角元素 ) of a square matrix are those elements aij such that i=j.

A square matrix for which all diagonal elements are equal to 1 and all non-diagonal elements are equal to 0 is called an identity matrix( 單位矩陣 ). An identity matrix is written as Im.

P.36

69

For any m x m matrix A, the m x m matrix B is the inverse of A if BA=AB=Im.

Some square matrices do not have inverses. If there does exist an m x m matrix B that satisfies BA=AB=Im, then we write B=A-1. (p.39)

The Gauss-Jordan Method for inverting an m x m Matrix A isStep 1. Write down the m x 2m matrix A|ImStep 2. Use EROs to transform A|Im intoIm|B.

This will be possible only if rank A=m. If rank A<m, then A has no inverse. (p.40)

70

Inverse Matrix ( 反矩陣 )

71

反矩陣之性質

72

Example :

例題：

74

AX=b 為一線性方程組 , 若 b1=b2=…=bm=0 ，

則稱為齊次方程組 (homogeneous) , 以 AX=0 表之。

1.若 x1=x2=…=xn=0 為其中一組解，則稱為必然解 (trivial solution)

2.若 x1 、 x2 、… xn 不全為 0 ，

則稱為非必然解 (nontrivial solution)

Homogeneous System Equations

76

Example: ? solution trival a has

101

011

326

A ,0AX

solution. trival a has system the

00A]x,x,[xX

matrix,singular non is A 1.1-T

321

solution. trival a has system the ,0]xxx[X

0100

0010

0001

~

0100

0010

0101

~

0100

0110

0101

~

0010

0110

0101

~

0320

0110

0101

~

0320

0011

0101

~

0326

0011

0101

~

0326

0011

0101

~

0101

0011

0326

.2

T321

77

Example: ? solution trival a has

222

012

111

A ,0AX

solution. nontrival a has system theRt

tx

t32

x

3t

x

3

2

1

78

Example : Solving linear system by inverse matrix

82

Exercise : Solving linear system by inverse matrix

1x x 3x

62xx 5 2x

7x x 2 x

321

321

321

19x 8,x 4,x 321 Answer:

83

Exercise : p.41 #5~#6

1. Use A-1 to solve the following linear system:

2. Use A-1 to solve the following linear system:

7=x5+x2

4=x3+x

21

21

2=x-x+x3

0=x2-x+x4

4=x+x

321

321

31

84

2.6 Determinants

Associated with any square matrix A is a number called the determinant ( 行列式 ) of A (often abbreviated as det A or |A|).

If A is an m x m matrix, then for any values of i and j, the ijth minor of A (written Aij) is the (m - 1) x (m - 1) submatrix of A obtained by deleting row i and column j of A.

Determinants can be used to invert square matrices and to solve linear equation systems.

p.42

86

餘因子展開式

87

行列式之性質 (1)

88


89


90

Exercise : p.43 #1~#4

1. 2. 3. A matrix is said to be upper triangular if i>j.aij=0.

Show that the determinant of any upper triangular 3x3 matrix is equal the product of the matrix’s diagonal elements.(This result is true for any upper triangular matrix.)

4. a. Show that for any 1x1 and 3x3 matrix. det(-A)=-detA .b. Show that for any 2x2 and 4x4 matrix. det(-A)=detA .c. Generalize the results of part (a) and (b).

91

Exercise :

若 det(λI3-A)=0, λ= ？

212

030

313

A

λ =-5 或 λ=0 或 λ=3

93

Cramer’s Rule

94

Cramer’s Rule 注意事項若 det(A) ≠0 ，則 n 元一次方程組為相容方程組，

其唯一解為

若 det(A) =det(A1) =det(A2)= …=det(An) =0 ，則 n 元一次方程組為相依方程組，其有無限多組解

若 det(A) =0 ，而 det(A1) ≠0 或 det(A2) ≠0 …，，或det(An) ≠0 ，則 n 元一次方程組為矛盾方程組，其為無解。

)Adet()Adet(

=x ..., ,)Adet()Adet(

=x ,)Adet()Adet(

=x nn

22

11

95

Example : Solving

635

212

211

A

6x6x3x5

1x2x x2

4x2x x

321

321

321

Sol:

0

635

112

411

)Adet(

,0

665

212

241

)Adet( ,0

636

211

214

)Adet( ,0

635

212

211

)Adet(

3

21

The system has an infinite number of solutions

96

Exercise : Solving linear system by Cramer’s rule

1x x 3x

62xx 5 2x

7x x 2 x

321

321

321

19=x 8,=x 4,=x 321Answer:

97

LU-Decompositions

Factoring the coefficient matrix into a product of a lower and upper triangular matrices.

This method is well suited for computers and is the basis for many practical computer programs.

98

Solving linear systems by factoring Let Ax = b ， and coefficient matrix A can be

factored into a product of n × n matrices as A = LU ， where L is lower triangular and U is upper triangular, then the system Ax = b can be solved by as follows ：

Step 1. Rewrite Ax = b as LUx = b (1)

Step 2. Define a n × 1 new matrix y by Ux = y (2)

Step 3. Use (2) to rewrite (1) as Ly = b and solve this system for y.

Step 4. Substitude y in(2) and solve for x

99

Example : Solving linear system by

LU decomposition

3

2

2

x

x

x

100

310

131

734

013

002

100

310

131

734

013

002

294

083

262

3

2

1

3

2

2

294

083

262

3

2

1

x

x

x

100

Step 2. Ux=y

Step 3. Ly=b

Solving y by forward-substitution. y1=1, y2=5, y3=2

37y3y 4y

2 y 3y

2 2y

3

2

2

734

013

002

321

21

1

3

2

1

y

y

y

3

2

1

3

2

1

100

310

131

y

y

y

x

x

x

101

Step 4. y1=1, y2=5, y3=2

Solving y by backward-substitution. x1=2, x2=-1, x3=2

2x

53xx

1x 3xx

2

5

1

x

x

x

100

310

131

3

32

321

3

2

1

3

2

1

3

2

1

100

310

131

y

y

y

x

x

x

102

Assume that A has a Doolittle factorization A = LU.L ： unit lower-△ ， U ： upper-△

The solution X to the linear system AX = b , is found in three steps:

1. Construct the matrices L and U, if possible.

2. Solve LY = b for Y using forward substitution.

3. Solve UX = Y for X using back substitution.

Doolittle method

=

103

=

Assume that A has a Doolittle factorization A = LU. L ： lower-△ ， U ： unit upper-△

The solution X to the linear system AX = b , is found in three steps:

1. Construct the matrices L and U, if possible.

2. Solve LY = b for Y using forward substitution.

3. Solve UX = Y for X using back substitution.

Crout method

104

Solving linear System By

Gauss-Jordan Elimination Inverse Matrix Cramer’s Rule LU-Decompositions

Doolittle Method Crout Method

chapter 2 basic linear algebra ( 基本線性代數 ) 2 2.1 2.1 matrices ( 矩陣 ) & vectors (...

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