# Determining all real numbers $a$ for which the limit $\lim\limits_{n \to \infty} a^nA^n$ exists and is non-zero for a $3 \times 3$ matrix $A$.

Q. Let $A$ be a $3 \times 3$ matrix $$A= \left(\begin{matrix} 1 & -1 & 0 \\ -1 & 2 &-1 \\ 0 & -1 & 1 \\ \end{matrix}\right)$$

Determine all real numbers $a$ for which the limit $\lim\limits_{n \to \infty} a^nA^n$ exists and is non-zero.[For a sequence of $3 \times 3$ matrices $\{B_n\}$ and a $3 \times 3$ matrix $B$, $\lim\limits_{n \to \infty} B_n = B$ means that, for all vectors $x \in \Bbb R^3$, we have $\lim\limits_{n \to \infty} B_nx=Bx$ in $\Bbb R^3$.]

My approach :

I found eigenvalues of $A$ to be $0,1,3$ with eigenvectors $(1,1,1)$,$(1,0,-1)$ and $(1,-2,1)$ respectively.

Let $x \in \Bbb R^3$, then $x=b(1,1,1)+c(1,0,-1)+d(1,-2,1)$ for some unique scalars $b,c,d$ since the eigenvectors form a basis of $\Bbb R^3$.

Thus, $\lim\limits_{n \to \infty} a^n A^n x=\lim\limits_{n \to \infty} a^n A^n[b(1,1,1)+c(1,0,-1)+d(1,-2,1)]=\lim\limits_{n \to \infty} [a^n\;b\; A^n(1,1,1)+a^n\;c\;A^n(1,0,-1)+a^n\;d\;A^n(1,-2,1)]=\lim\limits_{n \to \infty} [a^n\;b\; 0^n(1,1,1)+a^n\;c\;1^n(1,0,-1)+a^n\;d\;3^n(1,-2,1)]=(0,0,0)+c \lim\limits_{n \to \infty} a^n(1,0,-1)+d\lim\limits_{n \to \infty} (3a)^n(1,-2,1)$ which exists.

Since $\lim\limits_{n \to \infty} a^n A^n$ exists, so from second term of the last expression, $|a| \le 1$ and from third term of the last expression $|3a| \ \le 1 \Rightarrow |a| \le \frac 13$. This implies that $|a| \le \frac 13$.

But if $|a| \lt \frac 13$, then $\lim\limits_{n \to \infty} a^n A^nx=(0,0,0) \; \forall \; x \in \Bbb R^3.$

$\therefore a=\frac 13$.

Is this approach correct? Particularly I am concerned with the basis I used in the beginning. Since everything in the question is in the standard basis, did I do right by writing $x$ in the eigenbasis?

Another approach could be finding modal matrix $P$ and using $A^n=PD^nP^{-1}$ where $D$ is the diagonal matrix $\left(\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\ \end{matrix}\right)$ which also yields the same answer $a=\frac 13$.

• It looks fine to me. – José Carlos Santos Nov 25 '17 at 20:14
• Your solution is good. You should post it as an answer so that this question can be resolved. – Ken Duna Nov 25 '17 at 20:22

As per suggested in comments by @KenDuna, I am posting the same solution in OP as an answer so that it is clear to any future reader that the solution is indeed correct.

We find eigenvalues of $A$. Which come out to be $0,1,3$ with eigenvectors $(1,1,1)$,$(1,0,-1)$ and $(1,-2,1)$ respectively.

Let $x \in \Bbb R^3$, then $x=b(1,1,1)+c(1,0,-1)+d(1,-2,1)$ for some unique scalars $b,c,d$ since the eigenvectors form a basis of $\Bbb R^3$.

Thus, $\lim\limits_{n \to \infty} a^n A^n x=\lim\limits_{n \to \infty} a^n A^n[b(1,1,1)+c(1,0,-1)+d(1,-2,1)]=\lim\limits_{n \to \infty} [a^n\;b\; A^n(1,1,1)+a^n\;c\;A^n(1,0,-1)+a^n\;d\;A^n(1,-2,1)]=\lim\limits_{n \to \infty} [a^n\;b\; 0^n(1,1,1)+a^n\;c\;1^n(1,0,-1)+a^n\;d\;3^n(1,-2,1)]=(0,0,0)+c \lim\limits_{n \to \infty} a^n(1,0,-1)+d\lim\limits_{n \to \infty} (3a)^n(1,-2,1)$ which we want to exist.

Since $\lim\limits_{n \to \infty} a^n A^n$ exists, so from second term of the last expression, $|a| \le 1$ and from third term of the last expression $|3a| \ \le 1 \Rightarrow |a| \le \frac 13$. This implies that $|a| \le \frac 13$.

But if $|a| \lt \frac 13$, then $\lim\limits_{n \to \infty} a^n A^nx=(0,0,0) \; \forall \; x \in \Bbb R^3.$

$\therefore a=\frac 13$.

Another approach could be finding modal matrix $P$ and using $A^n=PD^nP^{-1}$ where $D$ is the diagonal matrix $\left(\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\ \end{matrix}\right)$ which also yields the same answer $a=\frac 13$.