# Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$

Given a Hermitian positive semi-definite $$n \times n$$ matrix $$A$$ and a rectangular $$m \times n$$ matrix $$B$$, is there anything that can be said about the eigenvalues of the matrix $$B A B^T$$?

It seems to me like one can regroup the product with a test vector $$x$$ to show that $$(x^T B)A(B^T x)$$ is at least the smallest eigenvalue of $$A$$ and at most the largest eigenvalue of $$A$$. However, this seems like it’s too easy of a solution...

• Your title doesn't make sense. Also, what is PSD? Nov 5 '18 at 16:26
• @AlexM. surely "positive semidefinite". Nov 5 '18 at 16:33
• In your second paragraph, you forgot that $B^tx$ need not have the same norm as $x$. But of course you can make trivial observations along these lines, for example if $B^t x$, $\|x\|=1$, is an eigenvector of the min ev $\lambda$ of $A$, then the smallest ev of $BAB^t$ is $\ge \lambda\|B^tx\|^2$. Nov 5 '18 at 19:51
• @AlexM. I fixed the title. PSD means positive semi definite. Nov 5 '18 at 20:26
• Take $A=I$ and recall that any psd matrix can be written as $B^TB$. Nov 6 '18 at 2:37

## 1 Answer

The following paper studies relations between $$\lambda(BAB^T)$$ and $$\lambda(A)$$:

Li, Mathias (1999). The Lidskii-Mirsky-Wielandt theorem – additive and multiplicative versions. Numerische Mathematik. January 1999, Volume 81, Issue 3, pp 377–413.