TITLE: Calculate a matrix to the power of "n" given an eigenvector
QUESTION [5 upvotes]: I have a question that I simply cannot solve. I do not want a direct answer to the question but simply an explanation as to the steps one would take to go about solving it, that way I can try it myself. To avoid getting an answer to the problem I wish to solve, I will use variables in place of actual matrices.
Calculate $A^n$x where both A and x are known , and x is an eigenvector of A. 
A is a 2-by-2 matrix.
For this question, I'm supposed to use the idea of linear combinations, which is what confuses me, I assume that means I'm supposed to solve it using the following format:
(This is my take on it, it is most likely wrong)
First find $\lambda$ -->  Ax = $\lambda$x
Then --> x = a $\lambda$x 
Solve for a and then do something like --> $A^n$ = a$\lambda^n$x 
I'm really unsure as to how to solve this, can someone point me in the right direction. Perhaps show me step by step what I need to do.

REPLY [2 votes]: If you know about diagonalizing a matrix and such, the formula $A=PDP^{-1}$ shouldn't be new to you. The cool thing about this is what happens when you take $A^2$:
\begin{align}
A&=PDP^{-1} \\
A^2&=(PDP^{-1})(PDP^{-1})
\end{align}
and because of associative law we can do $PP^{-1}$ first and that is equal to $I$. Reducing $A^2$ to $PDDP^{-1}$ which is $PD^2P^{-1}$.