# Eigenvalue decomposition of a symmetric matrix minus a diagonal matrix

I have a symmetric matrix $\Sigma$ and a diagonal matrix $D$. I know that eigenvalue decomposition of $\Sigma = Q\Lambda Q^T$. Now I would like to find the eigenvalues and the eigenvectors of the following matrix: $$\Sigma_v = \Sigma-D$$ I understand that if $D=nI$, things are trivial. But in a general case, what is the best approach to understand how much the eigenvalues and vector change from that of $\Sigma$, due to the diagonal perturbation $D$?

• Hi Sai. You did well to format your question using MathJax, but please take note of the formatting changes I've made. Most of the time, you'll get a better result if you put the entire mathematical expression in $...$ – Omnomnomnom Jan 14 '17 at 16:50
• Thank you! I will keep this in mind next time. – Sai Ganesh Jan 14 '17 at 16:58

## 1 Answer

Unfortunately, there isn't really anything special about the case where $D$ is diagonal (as opposed to an arbitrary symmetric perturbation). In particular, for any $P = UDU^T$, the matrix $\Sigma - P$ is similar to $U^T \Sigma U - D$.

As far as eigenvalues go, we have Weyl's inequalities, which I think is as much as you can get without more constraints on these matrices.

For eigenvectors, the inequalities are tricky, but do exist. A good reference for that information is chapter 7 of Bhatia's Matrix Analysis.

Note that all I've really given you here are inequalities. If you want the precise eigenvalues and eigenvectors of the updated matrix, there isn't any method that's going to give that you significantly more quickly than just computing them from scratch.

• Yes, this is what I feared. But to understand the modifications, I was thinking whether it would be a good idea to see visually using different ellipsoids (for the 2 by 2 case at least). – Sai Ganesh Jan 14 '17 at 17:02
• Are you particularly interested in the $2 \times 2$ case? There's a lot you can get away with then. I assume that with "different ellipsoids", you're referring to the image of the unit circle under each transformation. Sure, that might lead you to something interesting. Since all the characterstic polynomials are degree $2$, you might actually be able to come up with an analytic expression for eigenvalues/eigenvectors as a function of the diagonal entries. – Omnomnomnom Jan 14 '17 at 17:08
• Theoretically I am interested in the general case, however as there are no analytic solutions, I thought seeing visually/ or solving for the 2 by 2 case might give some interesting insights – Sai Ganesh Jan 15 '17 at 3:12
• I mean, I'm sure that some kind of analytic solution exists up through the 4 x 4 case. It might be helpful if you gave some motivation for the problem, or explained what sort of "insights" you're looking for – Omnomnomnom Jan 15 '17 at 4:19