I asked this question earlier and a user pointed out my error with a counterexample. However, it seems like if I change my statement slightly, all the counterexamples pointed out are not an issue anymore.
Given two diagonalizable matrices $A$ and $B$ of same dimension, is it true that
$$\sigma_i(AB) \leq \sigma_i(A)\sigma_1(B),$$
where $\sigma_i(\cdot)$ are the singular values ordered in nonincreasing order? How can one show this?
Best Answer
Yes, this is true. One way to prove this is using the min-max characterization of singular values. In particular, $$ \sigma_i(PQ) = \max_{S: \dim(S) = i} \min_{x \in S, \|x\| = 1} \|PQx\|\\ \leq \max_{S: \dim(S) = i} \min_{x \in S, \|x\| = 1} \|P\| \cdot \|Qx\|\\ = \sigma_1(P) \cdot \max_{S: \dim(S) = i} \min_{x \in S, \|x\| = 1} \|Qx\| \\ = \sigma_1(P) \sigma_i(Q). $$ With that, we have $\sigma_i(PQ) \leq \sigma_1(P) \sigma_i(Q)$. It follows that $$ \sigma_i(AB) = \sigma_i([AB]^T) = \sigma_i(B^TA^T) \leq \sigma_1(B^T) \sigma_i(A^T) = \sigma_i(A) \sigma_1(B). $$