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I am trying to understand how a perturbation of a skew symmetric matrix by another skew symmetric matrix affects the dominant eigenvector corresponding to $\lambda=0$.

Specifically, let $S$ be an $n \times n$ real-valued skew symmetric matrix , with $n$ being odd. Assume $S$ has eigenvectors $(e_1,...,e_n)$ and corresponding eigenvalues $(\lambda_1,...,\lambda_n)$, sorted such that $e_1=0$ (which is guaranteed because $n$ is odd). Thus, $e_1$ is the fixed point solution of $S$.

Let $M$ be another skew symmetric matrix of size $n\times n$, with eigenvalues $(b_1,...,b_n)$ and eigenvalues $(\theta_1,...,\theta_n)$.

For some small $0<\epsilon\ll1$, define:

$B=S+\epsilon M$

Then $B$ is likewise skew symmetric, and can be thought of as the perturbation of $S$ by the matrix $M$. Can we approximate or calculate the leading eigenvector of $B$, corresponding to $\lambda=0$, using the eigenvalue distribution of $B$ and $M$? In other words, can we approximate how much the fixed point of $S$ changes when perturbed by $M$?

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  • $\begingroup$ How do you know that $0$ is an eigenvalue? Are you assuming $n$ is odd? $\endgroup$ – Robert Israel Aug 17 '18 at 23:16
  • $\begingroup$ And what precisely do you mean by "the elements of $M$ are an order of magnitude smaller than the elements of $S$"? They can't be smaller than all the elements of $S$, because the diagonal elements of $S$ are $0$. $\endgroup$ – Robert Israel Aug 17 '18 at 23:22
  • $\begingroup$ @RobertIsrael Context makes it clear he means a perturbative ordering where $B = S + \epsilon M$. $\endgroup$ – eyeballfrog Aug 18 '18 at 0:30
  • $\begingroup$ Thanks, I've clarified the wording to specify n being odd and using the ϵ formulation $\endgroup$ – user3037237 Aug 20 '18 at 13:46
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Suppose $\ker(S)$ is spanned by a single vector $\bf u$ (in particular $n$ must be odd). Let $S^+$ be the Moore-Penrose pseudo-inverse of $S$.

Now it will be more convenient to write the perturbed matrix as $S + \epsilon M$ where $\epsilon$ is a parameter. Then it turns out that the following series (convergent for sufficiently small $\epsilon$) gives us a vector in $\ker(S+\epsilon M)$:

$$ \sum_{k=0}^\infty \epsilon^k (-S^+ M)^k {\bf u} $$

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  • $\begingroup$ Thanks --- I tried this computationally and unfortunately it doesn't give a good approximation. It generally end up being a near-constant vector of $\approx (1/n,....1/n)$ regardless of choose of S and M. $\endgroup$ – user3037237 Aug 20 '18 at 13:46
  • $\begingroup$ If $\epsilon$ is small it will be close to $\bf u$, which is what you want. $\endgroup$ – Robert Israel Aug 20 '18 at 15:52
  • $\begingroup$ Right, but the ultimate goal is trying to quantify how close to $\vec{u}$, and whether or not this proximity can be inferred from the structure of $S$. $\endgroup$ – user3037237 Aug 20 '18 at 20:33
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The physicist's approach to such problems is to do first-order perturbation theory (below). I am not terribly confident in the applicability of this method because of the facts that (1) the matrices are skew-symmetric and therefore must have pairs of eigenvalues of equal magnitude and (2) $S$ is not invertible by assumption. However, this answer may still be helpful.

The typical argument goes like this. For parallel notation, I'll use $\Delta S$ instead of $M$, and I'll call the eigenvector in question $v$ instead of $e_1$. Then we know $S v = 0$ and we want to find the solution to the new eigenvalue equation

$$ (S + \Delta S)(v + \Delta v) = \Delta \lambda (v + \Delta v) $$

All terms that are first-order perturbations have a $\Delta$. Now if we expand, use $Sv=0$, and retain only terms with a single $\Delta$, we get

$$ S \Delta v + \Delta S v = \Delta \lambda v $$

The next step is to multiply through by $v^T$, since we know by the fact that $S$ is skew-symmetric that $v^T S$ is also $0$. Then we get

$$ v^T \Delta S v = \Delta \lambda v^T v $$ or $$ \Delta \lambda = \frac{v^T \Delta S v}{v^T v} = 0 $$

where the last equality follows from $\Delta S$ being skew-symmetric.

At this point, the original equation has become

$$ (S + \Delta S)(v + \Delta v) = 0 $$

If we also assume that $\lambda=0$ has multiplicity 1 in $S+\Delta S$, this implies that $\Delta v$ is parallel to $v$. Otherwise, following a suggestion from the comments and from another answer, we could apply $Sv=0$ and then try to use the Roger-Penrose pseudo-inverse as an inverse for $(S+\Delta S)$. This would give

$$ \Delta v = -v - (S + \Delta S)^+ \Delta S v $$

However, I do not personally know of any theorem that would guarantee that this gives a reasonable result. I suspect that because the eigenvalue is zero, there may be problems with applying perturbation theory here in general.

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  • $\begingroup$ It should follow that △λ=0, given that e1=0. Since n is odd, there is always going to be one eigenvalue of zero. So I think you're on the right track, but the question I'm really interested in is the eigenvector. S is generally not invertible, but maybe the pseud-inverse could be used, as in the response below? $\endgroup$ – user3037237 Aug 20 '18 at 13:48
  • $\begingroup$ Sorry, I had missed the part about $e_1$ being $0$. I am not sure that this makes sense, given that part of the definition of an eigenvector is that it cannot be the zero vector. However, I realized while looking at this again that since $\Delta S$ is skew-symmetric as well, then the expression $v^T \Delta S v$ is zero for any vector, so I will edit the answer accordingly. $\endgroup$ – sasquires Aug 21 '18 at 17:14
  • $\begingroup$ I also added a comment about the pseudo-inverse, but I have no idea if it will actually work or not. I have only used pseudo-inverses in very specific circumstances that do not apply here. $\endgroup$ – sasquires Aug 21 '18 at 17:24

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