Inequality on spectral abscissa

1 minute read


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, 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, and it’s a self-citation! This is a good example that a theorem is eternal and may be useful to somebody in the far distant future.