Finding A^5 for Symmetrical Matrix A

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Homework Help Overview

The discussion revolves around finding the fifth power of a 3x3 symmetrical matrix A. Participants are exploring methods to compute A^5 without resorting to brute force multiplication.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the potential for diagonalization of the matrix as a method to simplify the computation of A^5. There are questions about the implications of the matrix being symmetric and how that might reduce the computational effort.

Discussion Status

The conversation is ongoing, with some participants suggesting diagonalization as a viable approach while others express uncertainty about its practicality compared to brute force methods. There is no explicit consensus on the best method yet.

Contextual Notes

Participants note that the product of symmetric matrices remains symmetric, which may influence the approach to the problem. The discussion includes considerations of eigenvalues and eigenvectors in the context of diagonalization.

Geronimo85
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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?
 
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Geronimo85 said:
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.
 
Ok, how does that help?
 
Geronimo85 said:
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. :-p
 
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.
 

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