Invertible $2 \times 2$ block matrix.

Question

Suppose we have a $2 \times 2$ block matrix $$M =\left[ \begin{array}{cc} A & B\\ C & D \end{array}\right],$$ where the entries are positive except the diagonals where we have $0$ entries. Also we know the fact that the full matrix is symmetric, which in turn implies that $C = B^t$. Further it is known that both $A$ and $D$ are invertible.

Can we conclude from the above information that $M$ has nonzero determinant?

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My argument:

By Schur Complement we have $\det M = \det A \times \det(D - CA^{-1}B)$ since $A$ is invertible. Again since $D$ is invertible we have $\det M = \det D \times \det(A - BD^{-1}C)$. Thus $$\det A \times \det(D - CA^{-1}B) = \det D \times \det(A - BD^{-1}C)$$ $$\implies \frac{\det(D - CA^{-1}B)}{\det(A - BD^{-1}C)} = \frac{\det D}{\det A} \neq 0$$ Hence $\det(D - CA^{-1}B) \neq 0$ and $M$ is invertible.

Is my argument correct?

Answer 1

Your argument is not correct, it has a division by zero error. An equation $ax=by$ where $x,y,a,b \in \mathbb{R}$ and $a,b \neq 0$ does not allow you to deduce $x \neq 0$. Taking $x=y=0$ makes the equation true regardless of the values of $a,b$. Less directly, your argument cannot be correct because it does not use the assumption that all the nondiagonal entries are positive at any point. The claim is not true without this assumption, as the example proposed by user Chris in the comments makes clear, so your argument cannot possibly be correct if it doesn't use this assumption.

Answer 2

We cannot conclude that $M$ has a non-zero determinant. For example, consider $$ M = (J - I) \otimes J + J \otimes (J-I) \\= \left[\begin{array}{cc|cc} 0 & 1 & 1 & 2\\ 1 & 0 & 2 & 1\\ \hline 1 & 2 & 0 & 1\\ 2 & 1 & 1 & 0 \end{array}\right]. $$ Note that the sum of the first and fourth columns is equal to the sum of the second and third columns.

In the above, $J$ denotes the $2 \times 2$ matrix whose entries are all $1$ and $\otimes$ denotes the Kronecker product.