Suppose a symmetric positive definite matrix $A \in \mathbb R^{n\times n}$ and a full-rank matrix $B \in \mathbb R^{n \times m}$ are given. In this question, the matter of computing the inverse $(B^\top A B)^{-1}$ is considered extensively.
Is there a way to bound from above the quantity
$$\|(B^\top AB)^{-1}\|\leq \textbf{?}$$
where $\|\cdot\| \in \{\|\cdot\|_2,\|\cdot\|_F\}$ is either the spectral or the Frobenious norm? I suspect an upper bound would involve the singular values of $A$, and the norm of $B$.
Best Answer
Taking $\|\cdot\|$ to be the spectral norm, we have the following: $$ \|(B^TAB)^{-1}\| = \frac{1}{\sigma_\min(B^TAB)}. $$ Now, using the fact that $$ \sigma_\min(M) = \min_{\|x\| = 1} \|Mx\|, $$ we find that $\sigma_{\min}(PQ) \geq \sigma_{\min}(P)\sigma_{\min}(Q)$. Thus, $$ \sigma_\min(B^TAB) \geq \sigma_{\min}(B^T)\sigma_{\min}(AB) \\ \geq \sigma_{\min}(B^T)\sigma_{\min}(A)\sigma_{\min}(B) \\ = \sigma_{\min}(A) \sigma_{\min}(B)^2. $$ Thus, we arrive at the bound from md5's comment: $$ \|(B^TAB)^{-1}\| = \frac{1}{\sigma_\min(B^TAB)} \leq \frac 1{\sigma_{\min}(A) \sigma_{\min}(B)^2}. $$