A skew-symmetric real matrix $A$ has zeros on the diagonal and eigenvalues that are purely imaginary.
If I perturb any diagonal entry $A_{ii}=0\rightarrow -\epsilon$ where $\epsilon>0,$ simulations suggest that all the eigenvalues of the perturbed $A$ now has eigenvalues that all have negative real part. How can this be shown?
Thank you.
It is well known that if $M + M^T$ is negative semi-definite, then the eigenvalues of $M$ have non-postive real part (see this post, for instance).
In your case, we find that $M = A - \epsilon \,e_ie_i^T$ is such that $M + M^T$ is a diagonal matrix with non-positive real entries on the diagonal. We conclude that $M + M^T$ is negative semi-definite, from which we deduce that the eigenvalues have non-positive real part, as you suspected.
Notably, there are examples where we still have some imaginary eigenvalues in the resulting matrix. In particular: $$ \pmatrix{0&0&0\\0&0&-1\\0&1&0} \to \pmatrix{-\epsilon&0&0\\0&0&-1\\0&1&0} $$ has eigenvalues $-\epsilon, \pm i$. I suspect that such an "unlucky" event has probability zero if $A$ is a suitably "random" skew-symmetric matrix. An obvious perturbation to move the eigenvalues over is to consider $A - \epsilon I$ instead (that is, apply the subtraction to all diagonals).