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.
Best Answer
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.