Finding A^5 for Symmetrical Matrix A

[Geronimo85]

I have a 3x3 symmetrical matrix A:


1, 8^.5, 0
8^.5, 1, 8^.5
0, 8^.5, 1

that I need to find A^5 for. Is there a method aside from brute force multiplication to do so?

 

[radou]

I have a 3x3 symmetrical matrix A:


1, 8^.5, 0
8^.5, 1, 8^.5
0, 8^.5, 1

that I need to find A^5 for. Is there a method aside from brute force multiplication to do so?

A fact which should be useful is that the product of symmetric matrices is symmetric itself.

 

[Geronimo85]

Ok, how does that help?

 

[radou]

Ok, how does that help?

It reduces the number of multiplications you have to do. But it's still a semi-brute multiplication method. Hope someone passes by with a better suggestion perhaps.

 

[HallsofIvy]

Since that is a symmetric matrix, it can be "diagonalized". That is there exist an invertible matrix C and a diagonal matrix D such that D= CAC^-1. Then A= C^-1DC. Then A⁵= C^-1D⁵C. D is the matrix having the eigenvalues of A (which are easy to find) on its main diagonal, 0s off the diagonal. C is a matrix having the corresponding eigenvectors of A as columns.