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 $B$.

  • $\begingroup$ It is a given real matrix of compatible size (n x m) $\endgroup$
    – G. Gare
    Jun 29 at 10:18
  • $\begingroup$ I write "quadratic form" because it reminds me of $v^\top A v$ with $v$ a $n$-dimensional vector. Indeed that is the case if $m = 1$. Moreover, I would like to obtain an upper bound involving e.g. the norm of $B$ and the real eigenvalues of $A$. $\endgroup$
    – G. Gare
    Jun 29 at 10:51
  • $\begingroup$ Ok I'll give it a thought. I was thinking that since any matrix norm is equivalent in the finite-dimensional case, there should be no problem in not choosing a specific norm. Concerning your previous comment: You can think of a quadratic form in the matrix $B$ (call it $B$, call it $X$, it is still an unknown matrix, and the bound will depend on the norm of $B$ as much as it would have depended on the norm of $X$) $\endgroup$
    – G. Gare
    Jun 29 at 11:23
  • 1
    $\begingroup$ What about an upper bound like $1/\sigma_{\text{min}}(B)^2 \sigma_{\text{min}}(A)$? $\endgroup$
    – md5
    Jun 29 at 14:21
  • $\begingroup$ @md5 it would be fantastic. Does it work? Could you write down the details in an answer? $\endgroup$
    – G. Gare
    Jun 29 at 14:22

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: $$ \|(B^TAB)^{-1}\| = \frac{1}{\sigma_\min(B^TAB)} \leq \frac 1{\sigma_{\min}(A) \sigma_{\min}(B)^2}. $$


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