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
$$\frac{a_n}{b_1} \le \| AB^{-1} \| \le \frac{a_1}{b_n}.$$
The upper bound is easy: we have
$$\| AB^{-1} \| \le \| A \| \| B^{-1} \| = \frac{a_1}{b_n}.$$
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.
