TITLE: Bounding the eigenvalues of $B A B^T$ with the eigenvalues of $A$
QUESTION [6 upvotes]: 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...

REPLY [7 votes]: 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.