Bounding the determinant of a principal submatrix

Question

Suppose that $A$ is a positive definite $n\times n$ matrix and $B$ is the $k\times k$ top left corner of $A$. Here $k<n$. Suppose that I know $|\det(A)|>\delta$ for some $\delta>0$. Does a similar inequality hold for $B$? That is, is there $\epsilon>0$ depending on $n,k$ and $\delta$ only such that $|\det(B)|>\epsilon$? I only know that $B$ is also positive definite.

Answer 1

Suppose that $C$ is bottom right corner $n-k\times n-k$ sub-matrix of $C$.Then, the following inequality holds: $$ \det(A)\leq \det(B)\det(C). $$ which means $\det(B)\det(C)\geq \delta$. But that's all one can say. Because if $A$ is a block diagonal matrix of $B$ and $C$ then $\det(B)$ can be arbitrarily small if $\det(C)$ is made enough large.