1. CHAPTER 8.9 ~ 8.16
Matrices

2. Contents 8.9 Power of Matrices 8.10 Orthogonal Matrices
8.11 Approximation of Eigenvalues
8.12 Diagonalization
8.13 Cryptography
8.14 An Error-Correcting Code
8.15 Method of Least Squares
8.16 Discrete Compartmental Models

3. 8.9 Power of Matrices
Introduction It is sometimes important to be able to quickly compute a power Am, m a positive integer, of an n × n matrix A: A2 = AA, A3 = AAA = A2A A4 = AAAA = A3A = A2A2 and so on.

4. A matrix A satisfies its own characteristic equation.
THEOREM 8.26
A matrix A satisfies its own characteristic equation.
Cayley-Hamilton Theorem
If the characteristic equation is then (1)

5. Matrix of Order 2
Suppose then 2 −  – 2 = 0. From Theorem 8.26, A2 − A – 2I = 0 or A2 = A + 2I (2) and also A3 = A2 + 2A = 2I + 3A A4 = A3 + 2A2 = 6I + 5A A5 = 10I + 11A A6 = 22I + 21A (3)

6. From the above discussions, we can write
From the above discussions, we can write Am = c0I + c1A and m = c0 + c1  (5) Using 1 = −1 , 2 = −2, we have then (6)

7. Matrix of Order n
Similar to the previous discussions, we have Am = c0I + c1A + c2A2 +…+ cn–1An–1 where ck, k = 0, 1,…, n–1, depend on m.

8. Example 1
Compute Am for
Solution The characteristic equation is 3 + 2 2 +  – 2 = 0, then 1 = –1, 2 = 1, 3 = 2. Thus Am = c0I + c1A +c2A2, m = c0 + c1 + c2 2 (7) In turn letting 1 = –1, 2 = 1, 3 = 2, we obtain (–1)m = c0 – c1 + c = c0 + c1 + c2 　 (8) m = c0 +2c1 + 4c2

9. Solving (8),
Since Am = c0I + c1A +c2A2, we have
eg. m = 10

10. Finding the Inverse
then A2 – A – 2I = 0, I = (1/2)A2 – (1/2)A, Multiplying both sides by A–1, then A–1 = (1/2)A – (1/2)I Thus

11. 8.10 Orthogonal Matrices
DEFINITION 8.14
An n  n matrix A is symmetric if A = AT, where AT is
The transpose of A.
Symmetric Matrix

12. Let A be a symmetric matrix with real entries. Then the
THEOREM 8.27
Let A be a symmetric matrix with real entries. Then the
eigenvalues of A are real.
Rear Eigenvalues
Proof Let AK = K, then (1) Since A is real, (2)

13. Take the transpose of (2), use the fact that A is symmetric and multiply on the right by K (3) Now AK = K, we have (4) Using (4) – (3) gives (5)

14. Since we have

15. Inner Product
x  y = x1 y1 + x2 y2 + … + xn yn (6) Similarly X  Y = XTY = x1 y1 + x2 y2 + … + xn yn (7)

16. Let A be a n × n symmetric matrix. The eigenvectors
THEOREM 8.28
Let A be a n × n symmetric matrix. The eigenvectors
corresponding to distinct (different) eigenvalues are
orthogonal.
Orthogonal Eigenvectors
Proof Let 1,, 2 be two distinct eigenvalues corresponding to eigenvectors K1 and K2. Since AK1 = 1K1 , AK2 = 2K2 (8) (AK1)T = K1TAT = K1TA = 1K1T

17. THEOREM 8.28
K1TAK2 = 1K1TK2 (9)
Since AK2 = 2K2, K1TAK2 = 2K1TK2 (10) (10) – (9) gives 0 = 1K1TK2 − 2K1TK2 or 0 = (1 − 2) K1TK2 Since 1  2, then K1TK2 = 0.

18. Example 1
The matrix has  = 0, 1, −2 and

19. Example 1 (2)
We find

20. An n × n nonsingular matrix A is orthogonal if A-1 = AT
DEFINITION 8.15
An n × n nonsingular matrix A is orthogonal if A-1 = AT
Orthogonal Matrix
A is orthogonal if ATA = I.

21. Example 2 (a) I is an orthogonal matrix, since ITI = II = I
(b) So, A is orthogonal.

22. An n × n matrix A is orthogonal if and only if its column
THEOREM 8.29
An n × n matrix A is orthogonal if and only if its column
X1, X2, …, Xn form an orthonormal set.
Criterion for an Orthogonal Matrix
Partial Proof We have A = (X1, X2, …, Xn), and A is orthogonal then

23. THEOREM 8.29
It follows that XiTXj = 0, i  j , i, j =1, 2, …, n XiTXi = 1, i =1, 2, …, n Thus all Xi form an orthonormal set.

24. Consider the matrix in example 2

26. And are unit vectors:

27. Example 3
In example 1, we have Since

28. Example 3 (2)
Thus, an orthonormal set is

29. Example 3 (3)
We have the orthogonal matrix Please verify that PT = P-1.

30. Example 4
For the symmetric matrix We can find  = −9, −9, 9. As in Sec 8.8, we have

31. Example 4 (2)
From the last matrix we see
Now for

32. Example 4 (3)
We find K3  K1 = K3  K2 = 0, K1  K2 = – 4  0 Using Gram-Schmidt process, V1 = K Now we have an orthogonal set and we can also make them an orthonormal set as

33. Example 4 (4)
Then is orthogonal.

34. 8.11 Approximation of Eigenvalues
DEFINITION 8.16
Let denote the eigenvalues of an
n × n matrix A. The eigenvalues is said to be the
dominant eigenvalues of A if An eigenvector corresponding to is called the
dominant eigenvector of A.
Dominant Eigenvalue

35. Example 1
(a) The matrix has eigenvalues Since , it follows that there is dominant eigenvalue.
(b) The eigenvalues of the matrix Again, the matrix has no dominant eigenvalue.

36. Power Method
Look at the sequence (1) where X0 is a nonzero n1 vector that is an initial guess or approximation and A has a dominant eigenvalue.
Therefore, (2)

37. Let us make some further assumptions:
Let us make some further assumptions: |1| > |2|  …  |n| and the corresponding eigenvectors K1, K2, …, Kn are linearly independent and can be a basis for Rn. Thus (3) here we also assume that c1  0.
Since AKi = iKi , then AX0 = c1AK1 + c2AK2 + … + cnAKn becomes (4)

38. Multiplying (4) by A,
(5) (6) Since |1| > |i|, i = 2, 3, …, n, as m  , we have (7)

39. However, the constant multiple of an eigenvector is also an eigenvector, then
Xm = Am X0 is an approximation to a dominant eigenvector. Since AK = K, AK  K= K  K then (8) which is called the Rayleigh quotient.

40. Example 2
For the initial guess

41. Example 2 (2) i 3 4 5 6 7 Xi
We have It appears then that the vectors are approaching scalar multiples of

42. Example 2 (3)

43. The remainder of this section is neglected since it is of less importance.

44. Scaling

45. Example 3
Repeat the iterations of Examples 2 using scaled-down vectors.
Solution From

46. Example 3 (2)
We defined We continuous in this manner to construct the following table: In contrast to the table in Example 3, it is apparent from this table that the vectors are approaching i 3 4 5 6 7 Xi

47. Method of Deflation
The Procedure we shall consider next is a modification of the power method and is called the method of deflation. We will limit the discussion to the case where A is a symmetric matrix.
Suppose 1 and K1 are the dominant eigenvalue and a corresponding normalized eigenvector of a symmetric matrix A. Furthermore, suppose the eigenvalues of A are such that It can be proved that the matrix

48. 8.12 Diagonalization
Diagonalizable Matrices If there exist a matrix P, such that P-1AP = D is diagonal, then A is said to be diagonalizable.
If an n × n matrix A has n linearly independent
Eigenvectors K1, K2, …, Kn, then A is diagonalizable.
THEOREM 8.30
Sufficient Condition for Diagonalizability

49. THEOREM 8.30
Proof Since P = (K1, K2, K3) is nonsingular, then P-1 exists, and Thus, P-1AP = D

50. If an n  n matrix A is a diagonalizable of and only if
THEOREM 8.31
If an n  n matrix A is a diagonalizable of and only if
A has n linearly independent eigenvalues.
Criterion for Diagonalizability
If an n  n matrix A has n distinct eigenvalues, it is
aiagonalizable.
THEOREM 8.32
Sufficient Condition for Diagonalizability

51. Example 1
Diagonalize
Solution  = 1, 4. Using the same process, we have then

52. Example 2
Consider We have

53. Example 2 (2)
Now

54. Example 2 (3)
Thus, P-1AP = D.

55. Example 3
Consider We have  = 5, 5. Since we can only find a single eigenvector this matrix can not be diagonalizable.

56. Example 4 Consider We have  = −1, 1, 1. For  = −1,
For  = 1, Use Gauss-Jordan method

57. Example 4 (2)
We can have Since we have three linearly independent eigenvectors, A is diagonalizable.
Let then P-1AP = D.

58. Example 4 (3)
Since we have three linearly independent eigenvectors, A is diagonalizable. Let then P-1AP = D.

59. Orthogonally Diagonalizable
There exists an orthogonal matrix P, which can diagonalize A. Then A is said to be orthogonally diagonalizable.
An n  n matrix A can be orthogonally diagonalizable
If and only if A is symmetric.
THEOREM 8.33
Criterion for Orthogonal Diagonalizability

60. THEOREM 8.33
Partial Proof Assume an nn matrix A can be orthogonally diagonalizable, then there exits an orthogonal matrix P such that P-1AP = D. A = PDP-1. Since P is orthogonal, P-1 = PT, then A = PDPT. However, A = (PDPT)T = PDTPT = PDPT = AT Thus A is symmetric.

61. Example 5
Consider From Example 4 of Sec 8.8, we find However, they are not mutually orthogonal.

62. Example 5 (2)
Now redo for  = We have k1 + k2 + k3 = 0, choosing k2 = 1, k3 = 0, we get K2; choosing k2 = 0, k3 = 1, we get K3. If we choose them by another way: k2 = 1, k3 = 1 and k2 = 1, k3 = – 1.

63. Example 5 (3)
We obtain two entirely different but orthogonal Thus an orthogonal set is

64. Example 5 (4)
Since we obtain an orthonormal set.

65. Example 5 (5)
Then and D = P-1AP

66. Example 5 (6)
This is verified form

67. Quadratic Forms
An algebraic expression of the form ax2 + bxy + cy2 (4) is called a quadratic form. If we let then (4) can be written as (5)
Note: is symmetric.

68. Example 6
Identify the conic section whose equation is 2x2 + 4xy − y2 = 1
Solution From (5) we have or XTAX = 1 (6) where

69. Example 6 (2)
For A, we have and K1, K2 are orthogonal. Moreover, an orthonormal set is

70. Example 6 (3)
Hence we have the orthogonal matrix If we let X = PX’ where , then (7)

71. Example 6 (4)
Using (7), (6) becomes or – 2X2 + 3Y2 = See Fig 8.11

72. Fig 8.11

73. 8.13 Cryptography Introduction Secret writing means code.
A simple code Let the letters a, b, c, …., z be represented by the numbers 1, 2, 3, …, 26. A sequence of letters cab then be a sequence of numbers. Arrange these numbers into an m  n matrix M. Then we select a nonsingular m  m matrix A. The new sent message becomes Y = AM, then M = A-1Y.

74. 8.14 An Error Correcting Code
Parity Encoding Add an extra bit to make the number of one is even

75. Example 2 (a) W = (1 0 0 0 1 1) (b) W = (1 1 1 0 0 1) Solution
(a) The extra bit will be 1 to make the number of one is 4 (even). The code word is then C = ( ).
(b) The extra bit will be 0 to make the number of one is 4 (even). So the edcoded word is C = ( ).

76. Fig 8.12

77. Example 3
Decoding the following (a) R = ( ) (b) R = ( )
Solution
(a) The number of one is 4 (even), we just drop the last bit to get ( ).
(b) The number of one is 3 (odd). It is a parity error.

78. Hamming Code
where c1, c2, and c3 denote the parity check bits.

79. Encoding

80. Example 4
Encode the word W = ( ).
Solution

81. Decoding

82. Example 5
Compute the syndrome of (a) R = ( ) and (b) R = ( )
Solution (a) we conclude that R is a code word. By the check bits in ( ), we get the decoded message ( ).

83. Example 5 (2)
(b) Since S  0, the received message R is not a code word.

86. Example 6
Changing zero to one gives the code word C = ( ). Hence the first, second, and fourth bits from C we arrive at the decoded message ( ).

87. 8.15 Method of Least Squares
Example 2 If we have the data (1, 1), (2, 3), (3, 4), (4, 6), (5,5), we want to fit the function f(x) =ax + b. Then a + b = 1 2a + b = 3 3a + b = a + b = a + b = 5

88. Example 2 (2)
Let we have

89. Example 2 (3)

90. Example 2 (4)
We have AX = Y. Then the best solution of X will be X = (ATA)-1ATY = (1.1, 0.5)T. For this line the sum of the square error is The fit function is y = 1.1x + 0.5

91. Fig 8.15

92. 8.16 Discrete Compartmental Models
The General Two-Compartment Model

93. Fig 8.16

94. Discrete Compartmental Model

96. Fig 8.17

97. Example 1
See Fig The initial amount is 100, 250, 80 for these three compartment. For Compartment 1 (C1): % to C2 0% to C3 then 80% to C1 For C2: 5% to C % to C3 then 65% to C2 For C3: 25% to C1 0% to C3 then 75% to C3

98. Fig 8.18

99. Example 1 (2)
That is, New C1 = 0.8C C C3 New C2 = 0.2C C2 + 0C3 New C3 = 0C C C3 We get the transfer matrix as

100. Example 1 (3)
Then one day later,

101. Note: m days later, Y = TmX0

102. Example 2

103. Example 2 (2)

104. Example 2 (3)

105. Thank You !