# Bound on product of matrices

I have two matrices, both positive definite, real symmetric $$A$$ and $$B$$. Can I impose the following lower and upper bound of the singular values of the product matrix in terms the of lower and upper bounds on singular values of those two matrices?

$$\sigma_{A_{min}} \frac{1}{\sigma_{B_{min}}} \leq || AB^{-1} || \leq \sigma_{A_{min}} \frac{1}{\sigma_{B_{min}}}$$

If yes, how can I prove this?

I assume you meant to put some $$\text{max}$$es in there since as written your lower and upper bounds are the same. For ease of notation, write the singular values as $$a_1 \ge \dots \ge a_n$$ and $$b_1 \ge \dots \ge b_n$$. Then I assume you meant to ask whether
$$\frac{a_n}{b_1} \le \| AB^{-1} \| \le \frac{a_1}{b_n}.$$
$$\| AB^{-1} \| \le \| A \| \| B^{-1} \| = \frac{a_1}{b_n}.$$
The lower bound is also straightforward: for any unit vector $$v$$ we have $$\| B^{-1} v \| \ge \frac{1}{b_1}$$ which gives $$\| AB^{-1} v \| \ge \frac{a_n}{b_1}$$. We actually don't need to assume that $$A$$ and $$B$$ are SPD, only that $$B$$ is invertible.