What is the characterization of matrices $B$(not necessarily squared) such that $BA$ has the same largest singular value as $A$? How about when $BA$ preserves the same $k$ largest singular values of $A$?
What if $A$ is a positive definite matrix?
I know that $B$ where the columns of $B$ are orthonormal satisfies this, but is that all?
I would really appreciate it if someone can guide me in the right direction.
I think it depends on the matching between left and right singular vectors. We can look at Singular Value Decomposition of both A and B: \begin{align} A &= U\Sigma V^T\\ B &= Y\Psi Z^T \\ BA &= Y\Psi Z^T U\Sigma V^T \end{align} Since we look for singular values, we can get rid of orthogonal matrices at the margins: $$\sigma(BA) \equiv \sigma( \Psi Z^T U\Sigma) $$ We do not know anything about the singular values (or eigenvalues in square case) of the matrix $D_1 C D_2$ where $D_1$ and $D_2$ are diagonal matrices and C is any matrix with known singular values. If $C$ is also diagonal matrix which is $Z^TU=I$ in your case, then of course, you can find the singular values. As a result of this analysis, positive definiteness of A is irrelevant. If B is an orthogonal matrix then we have: $$\sigma(BA) \equiv \sigma( B U\Sigma) $$ so, as a special case, if $B^TB = I$ i.e. column of $B$ constitutes an orthogonal set of vectors (no need B to be orthonormal) then: $$\sigma(BA) \equiv \sigma( B U\Sigma) =\sigma(U\Sigma) = \sigma(\Sigma)$$ Moreover, if all singular values of $BU$ is 1, then again $$\sigma(BA) = \sigma(\Sigma).$$ These are the cases that we can say something about singular values of $D_1 C D_2$.