Identify $E=\{a\in \Bbb R:\lim a^n A^n\text{exists and is different from zero}\}$ 
Consider $A=$$\left (\begin{array}{}
3 & 1\\
2 & 4
\end{array} \right)$
Consider the set $E=\{a\in \Bbb R:\lim a^n A^n\text{exists and is different from zero}\}$
What will be $E$?

The eigen values of $A$ are $5,2$. If $v$ is an eigen vector of $A$ corresponding to eigen value $\lambda$  then $A^nv=\lambda^nv\implies \lambda^n$  is an eigen value of $A^n$ corresponding to eigen vector $v$.
Now $(a^nA^n) v=a^n(A^nv)=(a^n \lambda^n)v$
Also $\lim (a^n \lambda^n)$ exists and equals non-zero only when $|a\lambda|= 1\implies |a|=\dfrac{1}{\lambda}\implies |a|=\dfrac{1}{2},\dfrac{1}{5}$
So $E=\{\pm \dfrac{1}{5},\pm\dfrac{1}{2}\}$
Will yo please check my solution?Is it okay or what are the edits required?
 A: Here is an outline of how I would solve the problem: $A$ has eigenvalues $5$ and $2$. That means that there is a basis of $\Bbb R^2$ using corresponding eigenvectors of $A$, say $v_5$ and $v_2$.
Take any vector $v\in \Bbb R^2$. It can be written as $c_5v_5+c_2v_2$ for some real numbers $c_5,c_2$. Now study $a^nA^nv$, and find for what values of $a$ both the following are satisfied:


*

*$\lim_{n\to\infty}\|a^nA^nv\|$ is not infinite for any $v$

*There is some $v$ for which $\lim_{n\to\infty}a^nA^nv$ exists and is non-zero

A: As Arthur pointed out my mistake, let me try to answer it again.
EDIT:Since eigenvalues are distinct so $A$ is diagonalizable and suppose $P$ be the corresponding non-singular modal matrix with $D=diag(2,5)$. Then
$A=P^{-1}DP\implies (aA)^n=P^{-1}(aD)^nP$ 
which exists and non-zero IFF $(aD)^n\neq O$ for which you need $a=+ \frac{1}{5}$.

$lim_{n\rightarrow\infty}(\frac{1}{5}D)^n=diag(0,1)$
$lim_{n\rightarrow\infty}(\frac{-1}{5}D)^n=diag(0,\lim_{n\rightarrow\infty}(-1)^n)$ 
doesn't exist uniquely.
$lim_{n\rightarrow\infty}(\pm\frac{1}{2}D)^n=$ doesn't exist.

