2
$\begingroup$

Let $A,B$ denote square matrices such that $\lvert A_{ij}\rvert\le B_{ij}$ for all $i,j$, and denote the spectral radius by $\rho$. From the Gelfand spectral radius formula it is easy to see that $$\rho(A)\le\rho(\lvert A\rvert)\le\rho(B),$$ where $\lvert A\rvert$ is meant entrywise.

Question: What is known about the case of equality? Is it true that equality can only occur if $A_{ij}=e^{i\phi}B_{ij}$ for some $\phi\in[0,2\pi]$?

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ If $B$ is reducible, then all but one block can be replaced by zero blocks, with the largest eigenvalue unchanged. So you require $B$ to be irreducible, and likely primitive. In the latter case, if $0 \leq A \leq B$ entrywise, and $A \neq B$, then $\rho(A)$ is strictly less than that of $B$. Thus if $B$ is primitive and $|A_{ij}| \leq B_{ij}$, then $\rho(|A|) = \rho (B)$ implies $B = |A|$. This answers only part of the question; likely the rest ($\rho(A) = \rho(|A|)$ implies what you want) is similarly not difficult. $\endgroup$
    – David Handelman
    Commented Jan 29, 2020 at 17:32

0

Reset to default

You must log in to answer this question.