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squenshl
Jun21-09, 05:28 PM
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.

slider142
Jun21-09, 06:53 PM
v is an eigenvector of M. Find the eigenvalue that it corresponds to.

HallsofIvy
Jun21-09, 07:03 PM
If Mv= \lambda v, then M^2v= M(Mv)= M(\lambda v)= \lambda Mv= \lambda^2 v. Get the point?

Pengwuino
Jun21-09, 07:04 PM
Also, remember that M^5 is just M acting on v 5 consecutive times. Eigenvectors make this quite simple :)