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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.

Bobson Dugnutt
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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).

Ben Grossmann
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  • Thank you. 1) In line three you write "$M+M^T$ is a diagonal matrix with non-positive real entries on the diagonal", but isn't this matrix the zero matrix with a $-2\epsilon$ added on entry $ii$? Did you use the term "non-positive" to include the zeros on the other diagonal entries?
    1. How can one show that $A-\epsilon I$ has eigenvalues with exclusively negative real parts? I'm probably missing something obvious.
     – Bobson Dugnutt Jun 05 '17 at 17:15
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    Yes, I used the term non-positive to include the zeros on the other diagonal entries. For any matrix $A$ and eigenvector $x$ of $A$, note that $$ (A - \epsilon I)x = Ax - \epsilon x = \lambda x - \epsilon x = (\lambda - \epsilon)x $$ so, we have shifted the entire spectrum over by $\epsilon$. – Ben Grossmann Jun 05 '17 at 17:21
  • I see. $A$ must be diagonizable though, right (but there is a zero probability of hitting a singular matrix when doing simulations over cont. dist.s)? – Bobson Dugnutt Jun 05 '17 at 17:26
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    In fact, I'm not relying on diagonalizability at all. $\lambda$ is an eigenvalue if and only if there is an associated eigenvector. I suppose that what I have technically shown is that for any $\lambda$ in the spectrum of $A$, $\lambda - \epsilon$ is in the spectrum of $A - \epsilon I$. As it turns out, the converse is just as easy to show. More generally, we have the "spectral mapping theorem" (surprisingly difficult to google, but you might find it in the index of your textbook). – Ben Grossmann Jun 05 '17 at 17:31