CHAPTER 8.9 ~ 8.16 Matrices

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

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.

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)

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)

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)

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.

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 + c2 1 = c0 + c1 + c2 　 (8) 2m = c0 +2c1 + 4c2

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

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

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)

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)

Since we have

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

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

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.

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

Example 1 (2) We find

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.

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

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

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.

Consider the matrix in example 2

And are unit vectors:

Example 3 In example 1, we have Since

Example 3 (2) Thus, an orthonormal set is

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

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

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

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

Example 4 (4) Then is orthogonal.

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

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.

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)

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)

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

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.

Example 2 For the initial guess

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

Example 2 (3)

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

Scaling

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

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

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

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

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

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

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

Example 2 Consider We have

Example 2 (2) Now

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

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

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

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

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

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

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.

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

Example 5 (2) Now redo for  = 8 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.

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

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

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

Example 5 (6) This is verified form

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.

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

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

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

Example 6 (4) Using (7), (6) becomes or – 2X2 + 3Y2 = 1. See Fig 8.11

Fig 8.11

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.

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

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 = (1 0 0 0 1 1 1). (b) The extra bit will be 0 to make the number of one is 4 (even). So the edcoded word is C = (1 1 1 0 0 1 0).

Fig 8.12

Example 3 Decoding the following (a) R = (1 1 0 0 1 0 1) (b) R = (1 0 1 1 0 0 0) Solution (a) The number of one is 4 (even), we just drop the last bit to get (1 1 0 0 1 0). (b) The number of one is 3 (odd). It is a parity error.

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

Encoding

Example 4 Encode the word W = (1 0 1 1). Solution

Decoding

Example 5 Compute the syndrome of (a) R = (1 1 0 1 0 0 1) and (b) R = (1 0 0 1 0 1 0) Solution (a) we conclude that R is a code word. By the check bits in (1 1 0 1 0 0 1), we get the decoded message (0 0 0 1).

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

Example 6 Changing zero to one gives the code word C = (1 0 1 1 0 1 0). Hence the first, second, and fourth bits from C we arrive at the decoded message (1 0 1 0).

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 = 4 4a + b = 6 5a + b = 5

Example 2 (2) Let we have

Example 2 (3)

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

Fig 8.15

8.16 Discrete Compartmental Models The General Two-Compartment Model

Fig 8.16

Discrete Compartmental Model

Fig 8.17

Example 1 See Fig 8.18. The initial amount is 100, 250, 80 for these three compartment. For Compartment 1 (C1): 20% to C2 0% to C3 then 80% to C1 For C2: 5% to C1 30% to C3 then 65% to C2 For C3: 25% to C1 0% to C3 then 75% to C3

Fig 8.18

Example 1 (2) That is, New C1 = 0.8C1 + 0.05C2 + 0.25C3 New C2 = 0.2C1 + 0.65C2 + 0C3 New C3 = 0C1 + 0.3C2 + 0.75C3 We get the transfer matrix as

Example 1 (3) Then one day later,

Note: m days later, Y = TmX0

Example 2

Example 2 (2)

Example 2 (3)

Thank You !