Product of a diagonal full rank matrix with an "almost" low rank matrix

Question

Assume $D = \operatorname{diag}(d_1,...,d_n)$ with $d_i > 0$ and $A \in \mathbb{R}^{n \times n}$.

It is known that if $\operatorname{rank}(A) = m \leq n$ then $\operatorname{rank}(DA) =m$. My question is that happens to the rank of $DA$ if $A$ is just "close" to be low rank, i.e, if the signular values of $A$ are $\sigma_1,\sigma_2,...,\sigma_m$ and the rest are $O(\varepsilon)$ for some small $\varepsilon$.

Is there a way to bound the change of the singular values of $A$ after multiplication with $D$ as a function of the entries of $D$?

Answer 1

Here's a very powerful result from Bhatia's Matrix Analysis:

Theorem III.4.5 (Gelfand Naimark) Let $A, B$ be any two matrices. Then the singular values of $A,B,$ and $AB$ satisfy the majorisation $$ \log s(AB) - \log s(B) \prec \log s(A). $$

And you probably don't need all of that. However, here's a useful little consequence (which has a nice little direct proof of its own, actually)

For $1 \leq k \leq n$, and for any $1 \leq i_1 < \cdots < i_k \leq n$, we have $$ \prod_{j=1}^k \sigma_{i_j}(A) \sigma_{n - i_j + 1}(B) \leq \prod_{j=1}^k \sigma_j(AB) \leq \prod_{j=1}^k \sigma_j(A) \sigma_j(B) $$

For convenience, take $d_1 \geq d_2 \geq \cdots \geq d_n$. Note that the singular values of $D$ are simply $\sigma_i(D) = d_i$. With that, we can get something like $$ \sigma_k(DA) = \frac{\prod_{j=1}^k \sigma_j(DA)}{\prod_{j=1}^{k-1} \sigma_j(DA)} \geq \frac{\prod_{j=1}^{k}d_{n-j}\sigma_j(A)}{\prod_{j=1}^{k-1} d_j\sigma_j(A)} = \frac{\prod_{j=1}^{k}d_{n-j}}{\prod_{j=1}^{k-1} d_j} \sigma_k(A) $$ So, we can get an inequality like $$ \frac{\prod_{j=1}^{k}d_{n-j}}{\prod_{j=1}^{k-1} d_j} \leq \frac{\sigma_k(DA)}{\sigma_k(A)} \leq \frac{\prod_{j=1}^k d_j}{\prod_{j=1}^{k-1}d_{n-j}} $$