From Weyl's theorem, i.e.:
Let $A$ and $E$ be $n\times n$ real symmetric matrices. Let $\alpha_1\geq\ldots\geq\alpha_n$ be the eigenvalues of $A$ and $\hat{\alpha}_1\geq\ldots\geq\hat{\alpha}_n$ be the eigenvalues of $\hat{A}=A+E$. Then $|\alpha_i-\hat{\alpha}_i|\leq\|E\|_2$,
we know that the perturbed eigenvalues of a symmetric matrix differ at most $\|E\|_2$ size, which means they can be calculated in a high accuracy.
But here I have the question: what if we assume $E$ be a skew-symmetric matrix, i.e.: $E^T=-E$? Do we also have $|\alpha_i-\hat{\alpha}_i|\leq c\|E\|_2$, for some constant $c$?
If it is, we may conclude that any perturbed matrix to $A$ (not only symmetric perturbation) changes little to $A$'s eigenvalues.
If not, is there any example to show that such $c$ (of $O(1)$ order) does not exists?
Best Answer
Since $A$ is diagonalizable, according to the Bauer-Fike theorem, for any eigenvalue $\lambda$ of $A$, there exists an eigenvalue $\mu$ of $A+E$ such that $$ |\lambda-\mu|\leq\kappa_2(V)\|E\|_2, $$ where $\kappa_2(V)$ is the spectral condition number of an eigenvector matrix $V$ of $A$. We can put $\kappa_2(V)=1$ because $A$ is symmetric and thus $V$ can be chosen to be orthogonal. Hence $$|\lambda-\mu|\leq\|E\|_2.$$ Note that $E$ does not need to be symmetric. If $E$ is skew-symmetric and hence normal, $\|E\|_2$ is equal to the spectral radius of $E$.