Suppose that we have a symmetric matrix ${\bf S}$ with eigenvalue decomposition ${\bf S} = {\bf Q}{\bf \Lambda}{\bf Q}^T$. Assume that we have two diagonal matrices ${\bf D}_1$ and ${\bf D}_2$ that are multiplying ${\bf S}$ from the left and right, i.e. ${\bf A} = {\bf D}_1{\bf S}{\bf D}_2$. Can we relate the eigenvalues of ${\bf S}$ to the ones of ${\bf A}$? How about the case where ${\bf S}$ is not symmetric?
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ Did you mean ${\bf A} = {\bf D}_1{\bf S}{\bf D}_2$? If not, what is ${\bf V}$? $\endgroup$– Joseph O'RourkeSep 21, 2011 at 1:37
-
$\begingroup$ Yes, let me correct it, thank you :) $\endgroup$– AnadimSep 21, 2011 at 1:53
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
In general there is no relation: for example consider the simplest case $S$ itself is diagonal and invertible. Letting $D_1=S^{-1}$ then $A$ can be any diagonal matrix $D_2$. The only considerations you can do are related to the presence of the zero eigenvalues using Binet formula for determinants. Notice also that in general $A$ itself can be nonsymmetric, and its eigenvalues can be complex. However small perturbations, i.e. small $D_1$ and $D_2$, result in a small perturbation of the eigenvalues of $S$ in the complex plane.
-
3$\begingroup$ Another case in which you can tell at least something is: when $D_1=D_2$ has positive entries, then the inertia of $S$ is preserved. $\endgroup$ Sep 23, 2011 at 11:06
-
$\begingroup$ Thank you. I agree that if ${\bf S}$ is a diagonal then the eigenvalues can be arbitrary, based on the ${\bf D}_i$s. But then again, you were able to exactly describe the eigenvalues of ${\bf A}$ as a function of the eigenvalues of ${\bf S}$ and ${\bf D}$. Is there a way to do that in general? i.e., can I pick the diagonals ${\bf D}_i$ so that I can spectrally shape ${\bf S}$ (nondiagonal) in any way that I like? $\endgroup$– AnadimSep 24, 2011 at 0:38