# Linear Algebra - Raising a matrix to a power

by snesnerd
Tags: algebra, linear, matrix, power, raising
 P: 24 | x 1 0 | | 0 x 1 | | 0 0 x | I need to raise this matrix to the 50th power. Of course I can not solve this the extremely long way. Here is my attempt at this: Let A represent the matrix above, and let N represent the following matrix: | 0 1 0 | | 0 0 1 | | 0 0 0 | Then A = XI + N. Since XI * N = N * XI, we can apply the binomial theorem. A^50 = sum from k = 0 to 50 of (50 C k)(XI)^(50-k)N^k by the binomial theorem = (XI)^50 + 50N(XI)^49 + 1225N^(2)(XI)^48 + N^3 Now if you take N^3 you will get 0, and if you take N^2, you get the following: | 0 0 1 | | 0 0 0 | | 0 0 0 | So we get: I(X)^50 + 50N(X)^49 + 1225N^2(X)^48 which is, | X^50 50X^49 1225X^48 | | 0 X^50 50X^49 | | 0 0 X^50 | I think this is right. If my work is right could someone help me make this more thorough? I do not want to miss any details and want to show all my work. Thanks.
HW Helper
Thanks
P: 5,016
 Quote by snesnerd | x 1 0 | | 0 x 1 | | 0 0 x | I need to raise this matrix to the 50th power. Of course I can not solve this the extremely long way. Here is my attempt at this: Let A represent the matrix above, and let N represent the following matrix: | 0 1 0 | | 0 0 1 | | 0 0 0 | Then A = XI + N. Since XI * N = N * XI, we can apply the binomial theorem. A^50 = sum from k = 0 to 50 of (50 C k)(XI)^(50-k)N^k by the binomial theorem = (XI)^50 + 50N(XI)^49 + 1225N^(2)(XI)^48 + N^3 Now if you take N^3 you will get 0, and if you take N^2, you get the following: | 0 0 1 | | 0 0 0 | | 0 0 0 | So we get: I(X)^50 + 50N(X)^49 + 1225N^2(X)^48 which is, | X^50 50X^49 1225X^48 | | 0 X^50 50X^49 | | 0 0 X^50 | I think this is right. If my work is right could someone help me make this more thorough? I do not want to miss any details and want to show all my work. Thanks.
Your matrix is a so-called "Jordan Block", and raising it to large powers is a well-solved problem; see, eg., http://en.wikipedia.org/wiki/Jordan_matrix

RGV
 Sci Advisor HW Helper Thanks P: 25,228 Your solution looks just fine to me.

 Related Discussions General Math 1 Calculus & Beyond Homework 0 Calculus & Beyond Homework 7 Calculus & Beyond Homework 5 Calculus & Beyond Homework 13