Timeline for When is a matrix power nonnegative

5 events

when toggle format	what		by	license	comment
Mar 5, 2018 at 7:49	comment	added	Robert Israel		Since $A$ is real, if $\lambda$ is an eigenvalue of $A$, so is $\overline{\lambda}$. Since $\left\|\overline{\lambda}^q\right\| = \left\|\lambda^q\right\| = \mu$, we must have $\overline{\lambda}^q = \mu$. But if $\overline{\lambda} \ne \lambda$, $A^q$ has at least two linearly independent eigenvectors for eigenvalue $\mu$, namely eigenvectors of $A$ for $\lambda$ and $\overline{\lambda}$.
Mar 4, 2018 at 23:02	comment	added	Hans		Could you please explicate the claim "since $A$ is real and $\mu$ is a simple eigenvalue, $\lambda$ must be real"?
Oct 18, 2011 at 19:33	comment	added	Richard Stanley		I believe that the following is still open: given an $n\times n$ integer matrix $A$ and $1\leq i,j\leq n$, is it decidable whether there is some $q\geq 1$ for which $(A^q)_{ij}=0$? See citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.155.2606. This suggests that the question posed here might also not be known to be decidable, but I could be wrong about this.
Oct 18, 2011 at 17:44	history	edited	Robert Israel	CC BY-SA 3.0	added 751 characters in body
Oct 10, 2011 at 19:05	history	answered	Robert Israel	CC BY-SA 3.0