Checking a possible corollary of Gelfand's formula

Question

I have recently come across Gelfand's spectral formula for matrices, which Wikipedia gives as follows

For any square matrix $A$ and matrix norm, we have $\lim_{k \to \infty} \lVert A^k \rVert ^{\frac{1}{k}} = \rho(A) $, where $ \rho(A) $ is the spectral radius of $A$.

I was just thinking about this theorem and a possible conclusion, whose correctness I wanted to check. Does the theorem tell us that $ \lVert A^k \rVert $ is bounded for all natural numbers $k$ if and only if the magnitude of the largest eigenvalue in magnitude is at most $1$, this for any matrix norm?

This conclusion can help me, and I wanted to check whether or not it is correct. I thank all helpers who can check correctness.

score 2 · Accepted Answer · answered Apr 05 '22 at 02:47

2

Gelfand's formula does imply that if $\|A^k\|$ is bounded, then $\rho(A)\le1$, but the converse is not true. For instance, the norm of $\pmatrix{1&1\\ 0&1}^k=\pmatrix{1&k\\ 0&1}$ is not bounded although the spectral radius of the matrix is $1$.

A correct statement is that $\|A^k\|$ is bounded if and only if $\rho(A)\le1$ and any eigenvalue of $A$ (over $\mathbb C$) with modulus equal to $1$ is semi-simple (i.e., its algebraic and geometric multiplicities coincide). To prove the semi-simplicity in the forward implication, you may consider the Jordan form of $A$. For instance, see part (b) of my other answer.

answered Apr 05 '22 at 02:47

user1551

139,064

Thank you very much for the great answer. It is much clearer now. So can we deduce the weaker, but hopefully correct, conclusion that if $\rho(A)<1$, then $\lVert A^k \rVert$ is bounded? – Croc2Alpha Apr 05 '22 at 02:55
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@kroner Yes. In fact, $\lim_{k\to\infty}A^k=0$ when $\rho(A)<1$. There is a proof in Wikipedia. I have also devised a proof in this answer. – user1551 Apr 05 '22 at 03:22
Thank you very much. – Croc2Alpha Apr 05 '22 at 03:28

Checking a possible corollary of Gelfand's formula

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