# Powers of a Matrix and Eigenvalues proof

1. May 16, 2013

### muzziMsyed21

1. The problem statement, all variables and given/known data

Prove that if A is an nxn matrix with eigenvector v, then v is an eigenvector for Ak where kε(all positive integers)

2. Relevant equations

Av=λv

3. The attempt at a solution

Av=λv
A(Av)=A(λv)
Akv=λ(Av)

i know i may not be doing it right but this is what i can think of so far

2. May 16, 2013

### Office_Shredder

Staff Emeritus
What if I rewrote that as
A(Av) = λ(Av)

3. May 16, 2013

If you want to show that v is an eigenvector of a power of A (given that v is an eigenvector of A itself), you need to know what happens when you multiply v by some power of A. Consider the second power to begin with - the pattern you see continues for higher powers. So,

if Av = λv, what do you get when you examine

A(Av) = A(λv)

* and simplify the left side?
* and simplify the right side (remember that A(λv) = λ(Av)) when you simplify the right?

If you correctly simplify these two pieces you should be able to show that v IS an eigenvector of A-squared, AND you will know the associated eigenvalue. The latter is important, because it will give you a major hint about the link between A^k, v, and the corresponding eigenvalue. Once you see that link, you will know how to answer your question.