# Inequality on spectral abscissa

Let $A, B$ be square (complex) matrices such that $|B| \le A$. Then it is well known that $\rho(B) \le \rho(|B|) \le \rho(A)$, where $\rho$ denotes the spectral radius (largest absolute value of all eigenvalues). See, for example, Theorem 8.4.5 of Horn and Johnson (2013). In my recent paper with Brendan Beare and Won-Ki Seo, we needed to use the spectral abscissa (largest real part of all eigenvalues) instead of the spectral radius. By analogy, we can make the following conjecture: if $A, B$ are square complex matrices such that $\mathrm{Re} b_{nn} \le a_{nn}$ for all $n$ and $|b_{nn’}| \le a_{nn’}$ for all $n \neq n’$, then is it true that $\zeta(B) \le \zeta(A)$, where $\zeta$ denotes the spectral abscissa?
It turns out that the answer is yes. The main part of the proof uses the fact that $\zeta(B) \le \zeta(C)$, where $C$ is the matrix with diagonal entries $c_{nn} = \mathrm{Re} b_{nn}$ and off-diagonal entries $c_{nn’} = |b_{nn’}|$. This result is Corollary 1 of Deutsch (1975). The reason I am writing this post is because this result is not well known: in fact, Deutsch (1975) has been cited only once before we cited it, and it’s a self-citation! This is a good example that a theorem is eternal and may be useful to somebody else in the far distant future.