# Matrix product bound

Consider the following inequality \begin{align*} AB^{-1}A^\top \preceq cI \end{align*} where $A\in\mathbb{R}^{n\times m}$, $B\in\mathbb{R}^{m\times m}$, $c\in\mathbb{R}$ (given), and $I$ is the identity matrix (dim. $n$). I know that $A$ is full-rank and $B\succ 0$ (positive definite).

I am trying to place conditions on $A$ and $B$ in order to ensure that the above inequality holds. I feel like I should be able to place some bound on the eigenvalues of $B$ (or a condition on $A$) but I can't seem to rearrange the above in any useful way.

• So $A$ has full row rank and $c$ is negative? (In that case, you want to get a lower bound on the minimal eigenvalue of $AB^{-1}A^T$.) – Algebraic Pavel Aug 31 '15 at 10:13
• I've edited the question slightly to simplify things (removed the leading negative sign). Thanks for the tip. Is there any way to simplify, after the edit, the expression $\lambda_{\max}(AB^{-1}A^T)$? – jonem Aug 31 '15 at 19:41
• The simplest way how to bound the maximal eigenvalue of $AB^{-1}A^T$ is simply by its spectral norm and using the submultiplicativity and the relation to extremal eigen/singular values: $\lambda_\max(AB^{-1}A^T)=\|AB^{-1}A^T\|_2\leq\|A\|_2^2\|B^{-1}\|_2=\frac{\sigma_\max^2(A)}{\lambda_\min(B)}$. So the condition holds if $\sigma_\max^2(A)\leq c\lambda_\min(B)$. The bound is very conservative but attainable (think about scalar $A$ and $B$). – Algebraic Pavel Sep 1 '15 at 11:43