## Determinant of a sum of two matrices (one dominating the other)

Let $A$ and $B$ be two $n \times n$ real matrices such that:

1. $\forall i, j: a_{ij} \geq 0, b_{ij} \geq 0$
2. let $a_\max$ be the largest entry of $A$ and $b_\min$ be the smallest nonzero entry of $B$; for some positive $K$, $b_\min \geq K \cdot a_\max$

I am interested in a statement of the type

For $A$ and $B$ as above and for $K$ large enough, there exists a $k \times k$ submatrix $C$ of $A+B$ s.t. $|\det(C)|^{1/k} \geq |\det(A)|^{1/n}$

I know that by Minkowski's determinant theorem, if $A$ and $B$ are hermitian positive semidefinite, then $\det(A + B)^{1/n} \geq \det(A)^{1/n} + \det(B)^{1/n}$.

Do you know any other related facts, possible counterexamples, helpful intuitions?

Just for information, in the special case I am interested in, $A \in \{0, 1\}^{n \times n}$ and $B \in (2\mathbb{Z})^{n \times n}$.

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We have

$$\det \left( \sum_{i=1}^N A^i \right) = \sum_{f \in F} \det A^{f}$$

where each $A^i$ is an $n\times n$ matrix for $i \in \{1, \dots, N \}$, $F = \{f: \{ 1, \dots, n \} \rightarrow {\{1, \dots, N\}} \}$ and $A^f$ is the matrix with entries $(A^f)_{ij}= A^{f(i)}_{ij}$. (You are interested in $N=2$ where there are $2^n$ terms).

You might get a start on the problem by splitting the sum over $F$ into parts depending on whether the large or small entries are present.

I get the feeling though that condition 2 is a bit weak.

(More of a comment than an answer, but I don't have the points to comment.)

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 Thanks Stephen. I was looking at the (maybe less useful) identity you get by applying Cauchy-Binet to $(A I)(I B)^T$. Just for the record, that gives $\sum_{S, T \subseteq [n]}{\det(A|_S + B|^T)}$ where $S$ and $T$ range over sets such that $|S| + |T| = n$ and $(A|_S)_{ij}$ is $A_{ij}$ if $j \in S$ and 0 otherwise, and $(B|^S)_{ij}$ is $B_{ij}$ if $i \in T$ and 0 otherwise. – Sasho Nikolov Mar 18 2012 at 19:38