4
$\begingroup$

Let $A$ be a square matrix. Show that if $$\lim_{n \to \infty} A^{n} = 0$$ then $\rho(A) < 1$, where $\rho(A)$ denotes the spectral radius of $A$.

Hint: Use the Jordan canonical form.

I am self-studying and have been working through a few linear algebra exercises. I'm struggling a bit in applying the hint to this problem — I don't know where to start. Any help appreciated.

$\endgroup$
  • $\begingroup$ This follows from $A^n v = \lambda^n v$. The other direction is straighforward using the Jordan form. $\endgroup$ – copper.hat Apr 16 at 4:11
  • $\begingroup$ Any mention of the field one is working over? $\endgroup$ – Marc van Leeuwen Apr 16 at 6:59
  • $\begingroup$ The entries are allowed to be complex. $\endgroup$ – mXdX Apr 18 at 5:15
8
$\begingroup$

You don't really need Jordan canonical form. If $\rho(A) \ge 1$, $A$ has an eigenvalue $\lambda$ with $|\lambda| \ge 1$. That eigenvalue has an eigenvector $v$. Then $A^n v = \lambda^n v$, so $\|A^n v\| = |\lambda|^n \|v\| \ge \|v\|$ does not go to $0$ as $n \to \infty$, which is impossible if $A^n \to 0$.

$\endgroup$
4
$\begingroup$

Hint

$$A=PJP^{-1} \\ J=\begin{bmatrix} \lambda_1 & * & 0 & 0 & 0 & ... & 0 \\ 0& \lambda_2 & * & 0 & 0 & ... & 0 \\ ...&...&...&...&....&....&....\\ 0 & 0 & 0 & 0&0&...&\lambda_n \\ \end{bmatrix}$$ where each $*$ is either $0$ or $1$.

Prove by induction that $$J^m=\begin{bmatrix} \lambda_1^m & \star & \star & \star & \star & ... & \star \\ 0& \lambda_2^m & \star & \star & \star & ... & \star \\ ...&...&...&...&....&....&....\\ 0 & 0 & 0 & 0&0&...&\lambda_n^m \\ \end{bmatrix}$$ where the $\star$s represent numbers, that is $J^m$ is an upper triangular matrix with the $m$^th powers of the eigenvalues on the diagonal.

Note The above claim for $J^m$ is not fully using that $J$ is a Jordan cannonical form. It only uses that $J$ is upper triangular.

$\endgroup$
  • $\begingroup$ So, $A^{m} = PJ^{m}P^{-1}$. If I can show what you're asking by induction, would the limit of $J^{m} = 0$? I'm sure it is because the diagonal entries are less than one, right? $\endgroup$ – mXdX Apr 16 at 2:44
  • $\begingroup$ @mXdX Well, that is the point. First $$\lim_m J^m= \lim_m P^{-1} A^m P =0$$ Now, since $\lim J^m=0$ you can deduce that the diagonal entries converge to zero, meaning $\lambda_j^m \to 0$. This implies that $|\lambda_j |<1$ $\endgroup$ – N. S. Apr 16 at 2:49
  • $\begingroup$ I understand now. Thanks. So I would have to show, like you said, that the diagonal entries of $J^{m}$ are the $m$th powers of the eigenvalues. $\endgroup$ – mXdX Apr 16 at 2:55

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.