Proof involving the spectral radius and the Jordan canonical form

Question

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.

Answer 1

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.

Answer 2

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 $A^n \to 0$.