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Suppose the $n \times n$ matrix $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ and singular values $\sigma_1, \ldots, \sigma_n$. It seems plausible that by comparing the singular values and eigenvalues we gets some sort of information about eigenvectors. Consider:

a. The singular values are equal to the absolute values of eigenvalues if and only if the matrix is normal, i.e., the eigenvectors are orthogonal (see http://en.wikipedia.org/wiki/Normal_matrix , item 11 of the "Equivalent definitions" section ).

b. Suppose we have two distinct eigenvalues $\lambda_1, \lambda_2$ with eigenvectors $v_1, v_2$. Suppose, hypothetically, we let $v_1$ approach $v_2$, while keeping all the other eigenvalues and eigenvectors the same. Then the largest singular value approaches infinity. This follows since $\sigma_{\rm max} = ||A||_2$ and $A$ maps the vector $v_1 - v_2$, which approaches $0$, to $\lambda_1 v_1 - \lambda_2 v_2$, which does not approach $0$.

It seems reasonable to guess that the ``more equal'' $|\lambda_1|, \ldots, |\lambda_n|$ and $\sigma_1, \ldots, \sigma_n$ are, the more the eigenvectors look like an orthogonal collection. So naturally my question is whether there is a formal statement to this effect.

Note: I asked this question on math.SE about a week ago.

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  • $\begingroup$ I do not understand what does letting one vector approach another mean. The matrix $A$ would deform severely in the process, would it not? $\endgroup$
    – Felix Goldberg
    Aug 28, 2012 at 22:12

1 Answer 1

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Suprprisingly, no, or at least not using any obvious measure of eigenvalue equality. Let $M$ be a matrix that is very non-normal, and consider the matrix:

$\left(\begin{array}{cc} I+\epsilon M & 0 \\ 0 & 2I + \epsilon M\end{array}\right)$

This matrix has a bunch of eigenvalues very close to $1$ and a bunch of eigenvalues very close to $2$, and similarly for the singular values. Thus, the difference between the eigenvalues and singular values, properly normalized to account for the total variance of eigenvalues, is very very small.

But the eigenvectors of this matrix are the same as the eigenvectors of $M$, repeated twice, and thus are very non-orthogonal. (Except, I guess, that half the eigenvectors are orthogonal to the other half .)

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  • $\begingroup$ Nice answer. Huh: so if you know that $\sigma_i = |\lambda_i|$ you can conclude that the matrix is normal, but if you only know that $|\sigma_i - |\lambda_i|| \leq \epsilon$, then you can't conclude much, no matter how small $\epsilon$ is. Perhaps I shouldn't be surprised by this since eigenvectors obviously are not continuous functions of the matrix elements, but I can't help but find this surprising. $\endgroup$
    – user21162
    Aug 10, 2012 at 23:22
  • $\begingroup$ I think it might help to consider some measure of orthogonality of a pair of eigenvectors weighted by the difference between the two corresponding eigenvalues, or the squared difference. This makes sense, because if the eigenvalues are so close that they are the same, the question of whether the eigenvectors are orthogonal no longer makes sense at all, so the weight should be zero. But at that point I think you can just say that the difference of the singular values and the eigenvalues is a measure of orthagonality. $\endgroup$
    – Will Sawin
    Aug 10, 2012 at 23:31
  • $\begingroup$ Alternatively, I suppose another variation of the question might be to consider only the case when the eigenvalues are distinct and the minimum distance between them is $\delta>0$. $\endgroup$
    – alex o.
    Aug 11, 2012 at 0:00
  • $\begingroup$ I like this answer, but can you be more precise about what it means that $ M $ is very non-normal? Do you have a measure of non-normality in mind? $\endgroup$
    – Andrew T. Barker
    Aug 11, 2012 at 14:03
  • $\begingroup$ Say, all pairs of eigenvectors of $M$ have an angle below $.01$ radians. $\endgroup$
    – Will Sawin
    Aug 11, 2012 at 14:20

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