# 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 $Q$

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 =$$A=\begin{bmatrix}(-a+1)^n&0\\0&(-a)^n\end{bmatrix}\quad Because it is diagonal matrix , is my assumption right ? But I don't know how to find A^n? Should I use A=PDP^{-1}? I see that Q and A are similar matrices, because determinant is same, is it P^{-1}QP^{-1}=A and related to eigenvector or eigenvalues? • Please recalculate P^2, it is incorrect. The actual answer makes a great difference to your problem. – астон вілла олоф мэллбэрг Feb 22 '18 at 4:59 • P^2 = -I or P^{-1} = -P Your logic is correct regarding Q^n is correct. Regarding A^2 note that A^2 = P^{-1}QP^{-1}P^{-1}QP^{-1} and since P^{-1} = -P, A^2 = -PQ^2P = - and A^n = (-1)^{n}PQP – Doug M Feb 22 '18 at 5:01 • thankyou so much! can I write it A^n = (-1)^{n-1}PQP ? – fiksx Feb 22 '18 at 11:08 ## 1 Answer Note that P^2=-I, hence P^{-1}=-P.$$Q=PAP(-Q)=(-P)AP$$Hence -Q and A are similar.$$(-Q)^n = (-P)A^nP(-1)^n Q^n = -PA^nPA^n = (-1)^nPQ^n(-P)=(-1)^{n+1}PQ^nP$$• thankyou so much, but do you mean similar by$(-Q)= P^{-1}AP$, because there is matrix P that diagonalize A? is it the same to write$A^n = (-1)^{n-1}PQP$? and I tried to compute range of a in order$\lim \limits_ {n\to \infty}\ {A^n}=O$.is it okay to wrote$A^n$like this ?$P \lim \limits_ {n\to \infty}\ {(-Q)^n}P$=$\lim \limits_ {n\to \infty}\ {A^n}$=$P \lim \limits_ {n\to \infty}\ {\begin{bmatrix}(a-1)^n&0\\0&(a)^n\end{bmatrix}\quad}P=O$, is this right? then take limit for$Q^n$, so the range is$a<1$? – fiksx Feb 22 '18 at 11:01 •$(-Q)=P^{-1}AP=(-P)AP$since$P^{-1}-=-P$. It is not the same as$A^n=(-1)^{n-1}PQP$but it is the same as$A^n=(-1)^{n-1}PQ^{\color{red}{n}}P$– Siong Thye Goh Feb 22 '18 at 16:39 • We need$|a-1|< 1$and$|a|<1$, I don't think it is possible to converge to$0$if$a\$ is an integer. – Siong Thye Goh Feb 22 '18 at 17:31
• okay thankyou so much for the help!!!! :D – fiksx Feb 23 '18 at 13:16