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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?

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  • $\begingroup$ Please recalculate $P^2$, it is incorrect. The actual answer makes a great difference to your problem. $\endgroup$ – астон вілла олоф мэллбэрг Feb 22 '18 at 4:59
  • $\begingroup$ $P^2 = -I$ 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 $A^n = (-1)^{n}PQP$ $\endgroup$ – Doug M Feb 22 '18 at 5:01
  • $\begingroup$ thankyou so much! can I write it $A^n = (-1)^{n-1}PQP$ ? $\endgroup$ – fiksx Feb 22 '18 at 11:08
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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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  • $\begingroup$ 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 $a<1$ ? $\endgroup$ – fiksx Feb 22 '18 at 11:01
  • $\begingroup$ $(-Q)=P^{-1}AP=(-P)AP$ since $P^{-1}-=-P$. It is not the same as $A^n=(-1)^{n-1}PQP$ but it is the same as $A^n=(-1)^{n-1}PQ^{\color{red}{n}}P$ $\endgroup$ – Siong Thye Goh Feb 22 '18 at 16:39
  • $\begingroup$ We need $|a-1|< 1$ and $|a|<1$, I don't think it is possible to converge to $0$ if $a$ is an integer. $\endgroup$ – Siong Thye Goh Feb 22 '18 at 17:31
  • $\begingroup$ okay thankyou so much for the help!!!! :D $\endgroup$ – fiksx Feb 23 '18 at 13:16

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