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Unformatted text preview: MATH 223, Linear Algebra Winter, 2008 Solutions to Assignment 7 1. The Cayley-Hamilton Theorem says that if A is a square matrix and A ( x ) is the characteristic polynomial of A , then A ( A ) = 0. I will not be proving this in general, because we dont really need it, but the case when A is 2 2 is straightforward enough. Prove it for 2 2 matrices. Solution: If A = a b c d , then A ( x ) = x 2- ( a + d ) x + ( ad- bd ). So A ( A ) = A 2- ( a + d ) A + ( ad- bc ) I = a 2 + bc ab + bd ac + cd bc + d 2 - a 2 + ad ab + bd ac + cd ad + d 2 + ad- bc ad- bc = 0 . [This shows that the minimal polynomial of a 2 2 matrix has degree at most 2; in fact, the degree is exactly two unless the matrix is I for some scalar , and so except in this case, A = min A for 2 2 matrices. It would have been useful to know this for assignment 6.] 2. When a sequence is defined in the style of the Fibonacci sequence, an explicit formula can be found using linear algebra methods. Suppose that a and b , c and d are scalars and we define x = a , x 1 = b , and x n +2 = cx n + dx n +1 for n 0....
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