# 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.