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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...

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  • $\begingroup$ Your title doesn't make sense. Also, what is PSD? $\endgroup$
    – Alex M.
    Commented Nov 5, 2018 at 16:26
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    $\begingroup$ @AlexM. surely "positive semidefinite". $\endgroup$
    – Nik Weaver
    Commented Nov 5, 2018 at 16:33
  • $\begingroup$ 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 $\ge \lambda\|B^tx\|^2$. $\endgroup$
    – Christian Remling
    Commented Nov 5, 2018 at 19:51
  • $\begingroup$ @AlexM. I fixed the title. PSD means positive semi definite. $\endgroup$
    – jay
    Commented Nov 5, 2018 at 20:26
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    $\begingroup$ Take $A=I$ and recall that any psd matrix can be written as $B^TB$. $\endgroup$
    – Brendan McKay
    Commented Nov 6, 2018 at 2:37

1 Answer 1

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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.

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