# Case of equality in entrywise spectral radius bound

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 $$\phi\in[0,2\pi]$$?

• 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. – David Handelman Jan 29 '20 at 17:32