Suppose $S$ is a tall-and-skinny $m \times n$ matrix with iid Gaussian entries and $D$ is a $m \times m$ deterministic diagonal matrix. What can be said about the bounds on the largest and smallest singular values of a product $DS$? This post mentions a relevant "classical result", but I can't find a proof of that statement.
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See The smallest singular value of deformed random rectangular matrices.
If the diagonal elements of $D$ are of order unity then the smallest singular value of $DS$ is of order $\sqrt{m}-\sqrt{n}$ with high probability. The largest singular value is of order $\sqrt{m}+\sqrt{n}$.
If there is no constraint on $D$ the upper bound is the product of the largest singular value of $D$ and the largest singular value of $S$ (the latter being of order $\sqrt{m}+\sqrt{n}$).
answered Jul 22, 2022 at 19:48
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$\begingroup$ What about a case when there is no constrain on the values of elements of D except that they are placed in a decreasing order? $\endgroup$– MaxCommented Jul 22, 2022 at 20:20
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$\begingroup$ since the distribution of $D$ is unitarily invariant, the order of the elements of $D$ is irrelevant; the cited paper considers the more general case that the diagonal elements of $D$ lie in an interval $(c,1/c)$ for some $c\in(0,1)$. $\endgroup$ Commented Jul 22, 2022 at 20:34