# Spectral abscissa of symmetric matrix with skew-symmetric perturbation

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.

• 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. – David Handelman Feb 21 at 13:40
• You are right. Sorry about that, I corrected the formulation. – user98563 Feb 21 at 14:16
• 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. – Jochen Glueck Feb 21 at 15:53