I am interested in bounds on the minimal distance between the spectral abscissa $\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ of a matrix $A$ and the eigenvalues of its perturbated version $A+S$. In my case, the matrix $A$ is symmetric and the perturbation $S$ is skew-symmetric. The matrix has eigenvalues with algebraic multiplicity higher than 1 (this is an unnice property that excludes some results). The matrix is a $N$-block matrix with $n\times n$ blocks and the perturbation is block diagonal with $n\times n$ blocks.

A known result in the literature states:

If $A$ and $B$ are normal with $\|A-B\|=\varepsilon$ in a unitarily invariant norm $\|\cdot\|$ (e.g. the Frobenius norm), then for any eigenvalue $\lambda$ of $A$ there is an eigenvalue $\mu$ of $B$ such that $|\lambda-\mu|\leq\varepsilon$.

Let $A$ be symmetric and $\nu=\arg\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ (eigenvalue with spectral abscissa of $A$). Let $B=A+S$ where $S$ is skew-symmetric. By the above result, there exists an eigenvalue $\mu$ of $A+S$ such that

$$ |\mathrm{Re}\nu-\mathrm{Re}\mu|\leq|\nu-\mu|\leq\|S\|. $$

I wonder if there is some results that utilises that one matrix is symmetric and the other skew-symmetric and not just normality? I've found many results about symmetric (Hermitian) matrices, but not the symmetric and skew-symmetric combination.

My intuition says that the skew-symmetric matrix should contribute to the imaginary part of the eigenvalue at first. Therefore, the minimum distance between the spectral abscissa of $A$ and the spectrum of $A+S$ is reduced compared to the case when only normality holds.

Edit: Another result states that for a matrix $C(z)$, continuously differentiable around $z=0$, with eigenvalue $\xi$ and left, right eigenvectors $u,v$, it holds for $C(z)v(z)=\xi(z)v(z)$ that $$ \xi^\prime(0)=\frac{u^*C^\prime(0)v}{u^*v} $$ If we let $C(z)=A+Sz$, then $C^\prime(0)=S$, $u=v$ whereby $\xi^\prime(0)=0$. This supports the idea that the change in the eigenvalues of $C$ is smaller for a symmetric matrix with skew-symmetric perturbation than it is in the general case.

  • $\begingroup$ The result for normal matrices is misstated---take diagonal matrices. The best I can see is that for any eigenvalue $\lambda$ of $A$, there exists an eigenvalue $\mu$ of $B$ such that $|\lambda - \mu| \leq \epsilon$. If $A$ and $B$ are symmetric, then the eigenvalues are real and can be ordered, and a stronger result is available. $\endgroup$ – David Handelman Feb 21 at 13:40
  • $\begingroup$ You are right. Sorry about that, I corrected the formulation. $\endgroup$ – user98563 Feb 21 at 14:16
  • $\begingroup$ Welcome to MathOverflow! I don't quite see what kind of result you are looking for. By choosing $A=0$ and by choosing a skew-symmetric matrix $S$ that has only $i\|S\|$ and $-i\|S\|$ as eigenvalues, you can see that the result that you mentioned (below the paragraph that is written italic) is optimal. $\endgroup$ – Jochen Glueck Feb 21 at 15:53

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