I have the following $2n \times 2n$ block matrix
$$H = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$$
where each block is an $n \times n$ matrix and
$A$ is a strictly upper triangular matrix and $\det(A)=0$.
$B$ is an upper triangular matrix and $\det(B) \neq 0$
$C$ is an unitriangular upper matrix and $\det(C) = 1$
$D$ is an lower triangular matrix and and $\det(D) \neq 0$
I should also mention that the matrices $B$ and $D$ don't permute.
I know that using the Schur complement it is possible to write
$$\det(H) = \det \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \det(D) \det\left(A-BD^{-1}C\right)$$
I would like to know if, given these conditions, it is possible to prove that $\det(H)\neq 0$. Thanks.
Here is an counterexample: All submatrices are $2\times 2$. $$ \begin{pmatrix} 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \end{pmatrix} $$ If you replace your requirement with $D$ being upper triangular (not lower) I think you can prove your claim by row reduction.