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WI08evaluesNotes

# WI08evaluesNotes - 1 Math 1b Practical Eigenvalues and...

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1 Math 1b Practical — Eigenvalues and diagonalization February 17, 2009 In certain applications, one is interested in what happens when the same square matrix is applied to a vector a number of times. Example: the rabbit epidemic. The problem in this example is taken from Linear Functions and Matrix Algebra by Bill Jacob. A disease is introduced into the rabbit population on an island somewhere. Initially there are 12 million rabbits, all ‘uninfected’. After n months, there will be U ( n ) ‘uninfected’ rabbits, S ( n ) rabbits who have become ‘sick’, and I ( n ) rabbits who have somehow become ‘immune’. So U (0) = 12 (million), S (0) = 0, I (0) = 0. The values of these numbers changes by the rule U ( n + 1) S ( n + 1) I ( n + 1) = 3 4 1 12 1 12 1 3 0 0 0 3 4 1 U ( n ) S ( n ) I ( n ) . So we have U (0) S (0) I (0) = 12 0 0 , U (1) S (1) I (1) = 9 4 0 , U (2) S (2) I (2) = 85 / 12 3 3 , . . . Here are the numerical results for the first seven months, expressed as rows for convenience: (12 , 0 , 0) = 12 . 0000 (1 . 0000 , 0 . 0000 , 0 . 0000) (9 . 0000 , 4 . 0000 , 0 . 0000) = 13 . 0000 (0 . 6923 , 0 . 3077 , 0 . 0000) (7 . 0833 , 3 . 0000 , 3 . 0000) = 13 . 0833 (0 . 5414 , 0 . 2293 , 0 . 2293) (5 . 8125 , 2 . 3611 , 5 . 2500) = 13 . 4236 (0 . 4330 , 0 . 1759 , 0 . 3911) (4 . 9936 , 1 . 9375 , 7 . 0208) = 13 . 9520 (0 . 3579 , 0 . 1389 , 0 . 5032) (4 . 4918 , 1 . 6645 , 8 . 4740) = 14 . 6303 (0 . 3070 , 0 . 1138 , 0 . 5792) (4 . 2137 , 1 . 4973 , 9 . 7224) = 15 . 4333 (0 . 2730 , 0 . 0970 , 0 . 6300) (4 . 0952 , 1 . 4046 , 10 . 845) = 16 . 3451 (0 . 2505 , 0 . 0859 , 0 . 6635) Here are the results for the 100th through 103rd month: (1405 . 82 , 439 . 147 , 4909 . 94) = 6754 . 9010 (0 . 2081 , 0 . 0650 , 0 . 7269) (1500 . 12 , 468 . 606 , 5239 . 30) = 7208 . 0226 (0 . 2081 , 0 . 0650 , 0 . 7269) (1600 . 75 , 500 . 040 , 5590 . 75) = 7691 . 5397 (0 . 2081 , 0 . 0650 , 0 . 7269) (1708 . 13 , 533 . 583 , 5965 . 78) = 8207 . 4913 (0 . 2081 , 0 . 0650 , 0 . 7269) The data show that the proportions seem to stabilize, with roughly 21% uninfected, 6% sick, and 73% immune.

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2 The numbers U ( n ), S ( n ), and I ( n ) of uninfected, sick, and immune rabits at the end of n months are the coordinates of U ( n ) S ( n ) I ( n ) = A n 12 0 0 = 3 4 1 12 1 12 1 3 0 0 0 3 4 1 n 12 0 0 . Example: Fibonacci numbers. The Fibonacci sequence is 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , 89 , 144 , 233 , . . . Each term is the sum of the previous two. Let A = 0 1 1 1 . Then (1 , 1) A = (1 , 2) = 1 (1 , 2 . 0000) , (1 , 1) A 2 = (2 , 3) = 2 (1 , 1 . 5000) , (1 , 1) A 3 = (3 , 5) = 3 (1 , 1 . 6667) , (1 , 1) A 4 = (5 , 8) = 5 (1 , 1 . 6000) , (1 , 1) A 5 = (8 , 13) = 8 (1 , 1 . 6250) , (1 , 1) A 6 = (13 , 21) = 13 (1 , 1 . 6154) , (1 , 1) A 7 = (21 , 34) = 21 (1 , 1 . 6190) , . . . There appears to be a ‘stabilizing’ of ratios of the coordinates in the above examples. But that need not happen. If, for example, A = 0 1 2 0 , and we apply A many times to the column vector of two 1’s, we get 1 2 , 2 2 , 2 4 , 4 4 , 4 8 , . . . What is going on? In many cases, everything can be explained in terms of the eigenvalues and eigenvectors of the matrix. Let A be a square matrix. A (right) eigenvector with corresponding eigenvalue λ is a nonzero (column) vector e so that A e = λ e . (1) We say λ is an eigenvalue of A when (1) holds for at least one nonzero vector e . Note that λ is an eigenvalue of A if and only if A λI is singular, since (1) is equivalent to
3 ( A λI ) e = 0 . This in turn is equivalent to det( λI A ) = 0. The polynomial det( xI A ) is caled the characteristic polynomial of A ; the zeros of the charateristic polynomial are the eigenvalues of A .

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WI08evaluesNotes - 1 Math 1b Practical Eigenvalues and...

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