# [SOLVED] Generalizing the n-th power of a matrix

• May 7th 2010, 02:55 AM
bidii
[SOLVED] Generalizing the n-th power of a matrix
Question here, I'm interested in generalizing the n-th power of a matrix P, $\displaystyle P^n$. The vector in question is $\displaystyle P=\left(\begin{array}{ccc}1&0&0\\0.2&0.5&0.3\\0&0. 7&0.3\end{array}\right)$

I've found two approaches to doing this, z-transform and Diagonalization but I'm getting conflicting answers with both. What answer are you getting? also, is there a program that can calculate this for any matrix (perhaps a project I could pick up later if not)? Thanks
• May 7th 2010, 03:26 AM
HallsofIvy
Quote:

Originally Posted by bidii
Question here, I'm interested in generalizing the n-th power of a vector P, $\displaystyle P^n$. The vector in question is $\displaystyle P=\left(\begin{array}{ccc}1&0&0\\0.2&0.5&0.3\\0&0. 7&0.3\end{array}\right)$

I've found two approaches to doing this, z-transform and Diagonalization but I'm getting conflicting answers with both. What answer are you getting? also, is there a program that can calculate this for any matrix (perhaps a project I could pick up later if not)? Thanks

Your language is a bit confusing. You titled this "Generalizing the n-th power of a matrix" but then talk about a vector- except that the example you show is a matrix! And why talk about "generalizing the n-th power"? What you are really asking about is calculating the n-th power of a matrix, there is no "generalization" involved.

I don't know what you mean by using the "z-transform" to find powers but diagonalizing certainly will work. Since the matrix you give it has three distinct eigenvalues and can be diagonalized.
• May 7th 2010, 03:45 AM
bidii
Quote:

Originally Posted by HallsofIvy
Your language is a bit confusing. You titled this "Generalizing the n-th power of a matrix" but then talk about a vector- except that the example you show is a matrix! And why talk about "generalizing the n-th power"? What you are really asking about is calculating the n-th power of a matrix, there is no "generalization" involved.

I don't know what you mean by using the "z-transform" to find powers but diagonalizing certainly will work. Since the matrix you give it has three distinct eigenvalues and can be diagonalized.

Oh, I was a bit sleepy, meant "matrix" there. I also meant the use of z-transforms, not "using the z-transform". By "generalizing" i meant having a solution that can be used to calculate the nth power of the matrix, given any n.
• May 9th 2010, 08:56 PM
dwsmith
I am not sure what you mean by z transform but if you do as HallsofIvy suggested, you can diagonalize the matrix.

$\displaystyle A^n=XD^nX^{-1}$

X is the matrix of the eigenspace of the eigenvectors, D is the diagonal matrix with corresponding eigenvalues, and $\displaystyle X^{-1}$ is self-explanatory.
• May 9th 2010, 09:33 PM
bidii
Quote:

Originally Posted by dwsmith
I am not sure what you mean by z transform but if you do as HallsofIvy suggested, you can diagonalize the matrix.

$\displaystyle A^n=XD^nX^{-1}$

X is the matrix of the eigenspace of the eigenvectors, D is the diagonal matrix with corresponding eigenvalues, and $\displaystyle X^{-1}$ is self-explanatory.

I was referring to a method detailed in Kleinrock's Queueing Theory book, volume one, which used z-transforms. I chose the Diagonalization method, though.
• May 9th 2010, 09:34 PM
dwsmith
I don't have that books so I am not sure what that method is.
• May 9th 2010, 09:39 PM
bidii
No I was just mentioning my source, I wasn't expecting you to be familiar with the approach