# Proof involving the spectral radius and the Jordan canonical form

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.

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

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 $$A^n \to 0$$.

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.

• 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? – mXdX Apr 16 at 2:44
• @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 $|\lambda_j |<1$ – N. S. Apr 16 at 2:49
• 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. – mXdX Apr 16 at 2:55