If we know the spectrum of a Hermitian positive semidefinite $n\times n$ matrix $X$, what can be said about the minimum eigenvalue of $AXA'$, where $A$ is an arbitrary $m\times n$ complex matrix and prime denotes conjugate transpose?
If $X$ is singular, then the product will also be singular.
However, if $X$ is nonsingular, can we bound the effect of the transformation on the smallest eigenvalue? Subcases of interest are differences between tall/fat $A$, rank-deficient $A$, and nonsingular $A$.
I assume $A'$ means the Hermitian transpose, which I will write as $A^*.$ Denote the smallest singular value of $A^*$ by $\alpha,$ and the smallest singular value of $X$ by $\xi.$ For all $v$ we have $\|A^*v\|_2\geq \alpha\|v\|_2$ and hence $v^*AXA^*v\geq \xi\alpha^2\|v\|_2^2.$ This implies the minimum eigenvalue of $AXA^*$ is at least $\xi\alpha^2.$ This is optimal iff we have $\|A^*v\|_2=\alpha \|v\|_2$ and $\|Xv\|_2=\xi\|v\|_2$ simultaneously for some non-zero vector $v.$
(Since $X$ and $AXA^*$ are Hermitian positive semidefinite, their eigenvalues (=spectrum) are the same as singular values. The singular values of $A^*$ are the square roots of eigenvalues of $AA^*.$)