3
$\begingroup$

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 $a=\frac 13$.

$\endgroup$
  • 2
    $\begingroup$ It looks fine to me. $\endgroup$ – José Carlos Santos Nov 25 '17 at 20:14
  • 1
    $\begingroup$ Your solution is good. You should post it as an answer so that this question can be resolved. $\endgroup$ – Ken Duna Nov 25 '17 at 20:22
1
$\begingroup$

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 $a=\frac 13$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.