I am interested in bounds on the difference in spectral abscissa $\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ between a matrix $A$ and its perturbated version $A+S$ in the case when 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 any pair of eigenvalues $\lambda,\mu$ of $A$ and $B$ satisfy $|\lambda-\mu|\leq\varepsilon$.

For $A$ symmetric and $B=A+S$, where $S$ is skew-symmetric, then $\mathrm{Re}\lambda(A)=\lambda(A)$ and

$$ |\mathrm{Re}\lambda(A)-\mathrm{Re}\mu(A+S)|\leq|\lambda(A)-\mu(A+S)|\leq\|S\|. $$

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, and therefore the change in the spectral abscissa is reduced. Does this seem resonable?