Powers of Matrices, Putzer's MethodDate: 03/09/97 at 17:30:34 From: Dina Pesenson Subject: Matrices and exponents Hi, I'm a junior in college. In my elementary differential equations course, we received an interesting extra credit problem. I was wondering if you could tell me where to look for help. Here's the problem: 3 0 0 -1 -2 1 3 -1 0 -3 2 0 2 -2 -3 1 -1 1 0 0 0 0 0 0 2 e^ I'm not supposed to use the Taylor expansion to approximate. The only other way I found so far is by diagonalizing the matrix. The problem is that when you try to find the eigenvalues, you get 2,2,2,2,2 (I used Mathematica to find them). This means that I cannot find 5 linearly independent eigenvectors, which in turn means I cannot find an inverse to the matrix consisting of eigenvectors. Therefore, I cannot diagonalize the original matrix. I also tried to do the entire problem on Mathematica, but it wouldn't solve it - the most it offers is the Taylor approximation. In our class, we learned how to deal with eigenvalues of multiplicity 2 only. Where could I find more information on how to work with eigenvalues of higher multiplicity in a case like this? What would you do about this? Date: 03/10/97 at 14:54:16 From: Doctor Pete Subject: Re: Matrices and exponents Look for the following textbook in your university library: Apostol, Tom M. _Calculus_, Vol II, 2nd Ed. pgs. 205-211 In case you cannot find this book (it is excellent), I will briefly describe what the above pages discuss, which is Putzer's method for calculating e^(tA), where t is a constant and A is an n x n matrix. In particular, a special case of Putzer's method applies, for, as you have noted, the matrix in question has equal eigenvalues. First, the Cayley-Hamilton theorem states: Let A be an n x n matrix and let f(L) = det(LI-A) = L^n + c[n-1]L^(n-1) + c[n-2]L^(n-2) + ... + c[1]L + c[0] be its characteristic polynomial. Then f(A) = 0. I hope you are familiar with this statement, and better yet, its proof - though we won't go into it. The basic idea we want to get out of this is that the (n)th power of any n x n matrix A is expressible as a linear combination of lower powers I, A, A^2, ... , A^(n-1). Then it immediately follows that A^(n+1), A^(n+2), and in general, all higher powers of A are also expressible as linear combinations of these lower powers. Since e^(tA) has a convergent Taylor series expansion as an infinite sum of powers of A, it is reasonable to expect that we can reexpress this infinite sum as a finite sum over powers of 0 to (n-1). That is, we may have: e^(tA) = Sum[q[k,t]A^k,{k,0,n-1}] (I am using a "pseudo-Mathematica" notation here) where q[k,t] are scalar coefficients which depend on t. Putzer's method, then, is really a theorem which demonstrates that q[k,t] exists, and finds it. In its general form, Putzer's method is outlined as follows: let L[1], L[2], ... L[n] be the eigenvalues of the n x n matrix A. Define a sequence of polynomials in A: P[0,A] = I, P[k,A] = Product[A-L[m]I,{m,1,k}], for k = 1, 2, ..., n. Then: e^(tA) = Sum[r[k+1,t]P[k,A],{k,0,n-1}] where the scalar coefficients r[1,t], ..., r[n,t] are determined recursively from the system of linear differential equations: r'[1,t] = L[1]r[1,t], r[1,0] = 1, r'[k+1,t] = L[k+1]r[k+1,t]+r[k,t], r[k+1,0] = 0, k = 1, 2, ..., n-1. I will not prove this; try to prove it yourself (if you can, you'd probably get more than extra credit!), or refer to Apostol. Now, this is not very convenient to work with; however, there is a special case of the above, when L[1] = L[2] = ... = L[n] = L; i.e., all the eigenvalues of A are equal. If we call this common eigenvalue L, then we have: e^(tA) = e^(Lt) Sum[t^k/k! (A-LI)^k, {k,0,n-1}]. The proof of this is not too difficult--observe that the matrices LtI and t(A-LI) commute, and apply the Cayley-Hamilton theorem. Of course, in your example, t=1, L=2. Since I've essentially told you the answer, I would strongly suggest you find out more about Putzer's method, and take a whack at proving it. It isn't extraordinarily difficult, but it's a bit computational. I also highly recommend Apostol's text, both volumes - it is calculus done right. Best wishes, -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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