TITLE: If $a$ is an integer, $A=\begin{bmatrix}a+1&2\\-1&a-2\end{bmatrix},\quad P=\begin{bmatrix}1&2\\-1&-1\end{bmatrix},\quad Q=PAP$, Find $P^2$ and $Q$
QUESTION [2 upvotes]: If $a$ is an integer,
$$A=\begin{bmatrix}a+1&2\\-1&a-2\end{bmatrix},\quad P=\begin{bmatrix}1&2\\-1&-1\end{bmatrix},\quad Q=PAP$$
1.) Find $P^2$ and $Q$ 
2.) If $n$ is an integer, find $Q^n$ AND $A^n$ 
3.) $\lim \limits_ {n\to \infty}\ {A^n}=O$, where $O$ is the null matrix
1.) $Q=PAP=\begin{bmatrix}-a+1&0\\0&-a\end{bmatrix}\quad$
$P^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}\quad$
2.) $Q^n =$$A=\begin{bmatrix}(-a+1)^n&0\\0&(-a)^n\end{bmatrix}\quad$ 
Because it is diagonal matrix , is my assumption right ? But I don't know how to find $A^n$? Should I use $A=PDP^{-1}$? I see that $Q$ and $A$ are similar matrices, because determinant is same, is it $P^{-1}QP^{-1}=A$ and related to eigenvector or eigenvalues?

REPLY [1 votes]: Note that $P^2=-I$, hence $P^{-1}=-P$.
$$Q=PAP$$
$$(-Q)=(-P)AP$$
Hence $-Q$ and $A$ are similar.
$$(-Q)^n = (-P)A^nP$$
$$(-1)^n Q^n = -PA^nP$$
$$A^n = (-1)^nPQ^n(-P)=(-1)^{n+1}PQ^nP$$
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