Q: Gelfand's corrolaries counterexample? Gelfand's corrolaries (https://en.wikipedia.org/wiki/Spectral_radius#Gelfand_corollaries) state that, for any $2$ matrices $\mathbf{A}_1$, $\mathbf{A}_2$, the following relation is true:
$ \rho(\mathbf{A}_1\mathbf{A}_2) \leq \rho(\mathbf{A}_1)\rho(\mathbf{A}_2) $
However, assume the following matrices:
$\mathbf{A}_1 = \begin{bmatrix} 0     &0     &0     &0\\
     0     &1     &0     &0\\
     0     &0     &1     &0\\
     0     &0     &0     &1 \end{bmatrix} $,       $\mathbf{A}_2 = \begin{bmatrix} 0.2974   &-0.8912   &-0.5780    &0.5958\\
   -0.0314   &-0.8287   &-0.0872    &0.1178\\
   -0.0796    &0.1481   &-0.4975   &-0.2049\\
   -0.9233    &0.7149    &0.5110   &-1.1726 \end{bmatrix} $.
If you check the spectral radius of those matrices, as well as the spectral radius of their product you will find that the inequality doesn't hold.
I can not understand what's wrong, are there any constraints for the correlaries to be true? I couldn't find any.

A: Gelfand's corollaries do not in fact state something as general as you have here. The Wikipedia statement also says "assuming that they all commute we obtain" before the inequality. Indeed, the two matrices $\mathbf{A}_1$ and $\mathbf{A}_2$ do not commute. A simpler example of matrices that don't commute and whose product have higher spectral radius is given here.
