Powers of Matrices
Date: 6/27/96 at 14:54:56 From: Jose Miguel Subject: Matrices Exponential I am in the 2nd year of the "Mathmatics teaching" course, and I'll be taking a test next Monday, and I can't get anywhere information about the chapter concerning Matrices Exponential. Would you send me some information/links about this topic? Thanks. Sofia
Date: 6/29/96 at 18:1:8 From: Doctor Anthony Subject: Re: Matrices Exponential Powers of Matrices. There are two common methods for finding powers of square matrices. These are: (1) Find the eigenvalues and eigenvectors of the matrix. If k1, k2, k3 are the eigenvalues and s1, s2, s3 the corresponding eigenvectors, then we have for the matrix A A*s1 = k1*s1 A*s2 = k2*s2 and A*s3 = k3*s3 Now make up a matrix P with s1 as the first column, s2 the second column and s3 the third column, then A*P = P*D where D is the diagonal matrix with elements k1, k2, k3 Post multiply by P^(-1) and we get A = P*D*P^(-1) A^2 = P*D*P^(-1)*P*D*P^(-1) Now P^(-1)*P = I = P*D^2*P^(-1) continuing in this style A^3 = P*D^3*P^(-1) and in general A^n = P*D^n*P^(-1) Now this a simple result to evaluate. P was simply written down knowing s1, s2 and s3, P^(-1) is the inverse matrix - not unduly difficult to find. D^n is a diagonal matrix with elements k1^n, k2^n and k3^n, results which can be read off a calculator. We note too that using this method it is as easy to find A^500 as A^3. Any standard book on algebra will show how to find the eigenvalues and eigenvectors. Briefly we solve det(A-kI) = 0 to get the eigenvalues, and having got the k values, we get the eigenvectors from A*s1 = k1*s1 etc. (2) The Cayley-Hamilton theorem and powers of matrices. If det(A-kI) = a0 + a1*k + a2*k^2 + a3*k^3 = 0 This is called the characteristic equation of the matrix, and the Cayley-Hamilton theorem states that every square matrix, A, satisfies its own characteristic equation. That is a0*I + a1*A + a2*A^2 + a3*A^3 = 0 Example; If A = |1 -1| |2 3| find A^8 The characteristic equation is k^2 - 4k + 5 = 0 So A^2 - 4A + 5I = 0 and: A^2 = 4A - 5I A^4 = 16A^2 - 40A + 25I = 16(4A-5I) - 40A + 25I = 24A - 55I A^8 = 576A^2 - 2640A + 3025I = 576(4a-5I) - 2640A + 3025I = -336A + 145I This method is clearly not as good as the previous one for finding large powers of A. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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