# M^5v given that M = [3 2 4 2 0 2 4 2 3]

1. Jun 21, 2009

### squenshl

Hi,
Just wondering how to calculate M^5v given that
M = [3 2 4
2 0 2
4 2 3].
and v = (1,0,-1). v being a column vector.

2. Jun 21, 2009

### slider142

Re: E-values/E-vectors

v is an eigenvector of M. Find the eigenvalue that it corresponds to.

3. Jun 21, 2009

### HallsofIvy

Staff Emeritus
Re: E-values/E-vectors

If $Mv= \lambda v$, then $M^2v= M(Mv)= M(\lambda v)= \lambda Mv= \lambda^2 v$. Get the point?

4. Jun 21, 2009

### Pengwuino

Re: E-values/E-vectors

Also, remember that M^5 is just M acting on v 5 consecutive times. Eigenvectors make this quite simple :)