# Bounding a determinant ratio

Let $A=[A_{0}\ E;E^{T} \ B]$ be a real positive definite matrix and let $B$ be a principal submatrix. I am interested in tightly bounding $\frac{|B|}{|A|}$ from below in some "explicit" way that will involve the entries of $A$. ($A$ is actually a strictly diagonally dominant $M$-matrix, if that helps).

An obvious first approach is to try the Fischer-Hadamard inequality which gives:

$$\frac{|B|}{|A|} \geq \frac{1}{|A_{0}|}.$$

This is not bad at all, but insufficient for my purposes (it gets the order of magnitude right in my examples but I need more than that).

Another approach is to let $C=A/B$ be the Schur complement and to observe that $|A|=|B||C|$ and so $\frac{|B|}{|A|} \geq \frac{1}{|C|}$; however, I can't find appropriate bounds on $|C|$ to use as the next step.

Do you know of such bounds? Or another way to tackle this problem?

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Of course, the first approach I described is a special case of the second. – Felix Goldberg Dec 6 '12 at 21:38

From the standard block-matrix determinant formula, we get $$\det(A) = \det(B) \det(A_0 - E B^{-1} E^T)$$ so $$\frac{\det(B)}{\det(A)} = \frac{1}{\det(A_0 - E B^{-1} E^T)} = \frac{1}{\det(A_0) \det(I - A_0^{-1/2} E B^{-1} E^T A_0^{-1/2})}$$
Since $$u^T (A_0 - E B^{-1} E^T) u = (u^T, \ (B^{-1} E^T u)^T) A \pmatrix{u\cr B^{-1} E^T u\cr} > 0 \ \text{for u \ne 0}$$ we get that $A_0 - E B^{-1} E^T$ is positive definite; together with the fact that $A_0^{-1/2}E B^{-1} E^T A_0^{1/2}$ is positive semidefinite this says $I \ge I - A_0^{-1/2} E B^{-1} E^T A_0^{-1/2} > 0$ and gives your lower bound $1/\det(A_0)$. Of course that lower bound is exact if $E = 0$. You could improve the lower bound if you could find some nonzero lower bounds on eigenvalues of $A_0^{-1/2} E B^{-1} E^T A_0^{-1/2}$.