Linear Algebra - Raising a matrix to a power


by snesnerd
Tags: algebra, linear, matrix, power, raising
snesnerd
snesnerd is offline
#1
Nov24-12, 04:44 PM
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.
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Ray Vickson
Ray Vickson is offline
#2
Nov24-12, 04:48 PM
HW Helper
Thanks
P: 4,673
Quote Quote by snesnerd View Post
| 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
Dick
Dick is offline
#3
Nov24-12, 05:03 PM
Sci Advisor
HW Helper
Thanks
P: 25,174
Your solution looks just fine to me.


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