Schur stability without the Jordan canonical form

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?

Solution 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 $T_1$ with distinct eigenvalues such that $|T|\le T_1$ entrywise and $\rho(T_1)<1$. Then $\lim_{k\to\infty}T_1^k=0$ because $T_1$ is diagonalisable and $\rho(T_1)<1$. Hence $T^k$ also converges to zero, because $|T^k|\le|T|^k\le T_1^k$ entrywise. Thus $A^k$ converges to zero as well.