Schur stability without the Jordan canonical form

Question

Theorem If all eigenvalues of $A\in\mathbb{R}^{n\times n}$ have modulus strictly smaller than 1, then $$\lim_{k\to\infty} A^k = 0.$$

The proofs I know of this result all require the Jordan canonical form. Is there a simpler one that doesn't?

Answer 1

Suppose $\rho(A)<1$. Let $A$ be unitarily similar to an upper triangular matrix $T$ over $\mathbb C$. Let $|T|$ be the entrywise absolute value of $T$. Pick an entrywise nonnegative triangular matrix $R$ with distinct eigenvalues such that $|T|\le R$ entrywise and $\rho(R)<1$. Then $\lim_{k\to\infty}R^k=0$ because $R$ is diagonalisable and $\rho(R)<1$. Hence $T^k$ also converges to zero, because $|T^k|\le|T|^k\le R^k$ entrywise. Thus $A^k$ converges to zero as well.

Remark. The proof above also gives rise to a simple but loose upper bound to $\|A^m\|$. Let $r=\rho(R)$ and $PDP^{-1}$ be a diagonalisation of $R$. Then $$ \|A^m\|_F =\|T^m\|_F \le\big\||T|^m\big\|_F \le\|R^m\|_F =\|PD^mP^{-1}\|_F \le\|P\|_F\|P^{-1}\|_F\sqrt{n}r^m. $$ It follows that for every matrix norm $\|\cdot\|$, there exists a constant $C>0$ such that $$ \|A^m\|\le Cr^m $$ for all nonnegative integers $m$. The constant $C$ may depend on $A$ and $n$ but not $m$, while $r\ge\rho(A)$ can be chosen to be arbitrarily close to $\rho(A)$. (If all dominant eigenvalues of $A$ are semi-simple, we may actually pick $r=\rho(A)$, but the proof will require some modifications.)