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](https://math.stackexchange.com/questions/178877/eigenvalues-singular-values-and-the-angles-between-eigenvectors) on math.SE about a week ago. 



Eigenvalues, singular values, and the angles between eigenvectors
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 .)


