So I have the following 2N $\times$ 2N block matrix

$H=\begin{bmatrix} A & B \\ C & D \end{bmatrix}$

where each block in an N$\times$N matrix.

Each block have the following characteristics :

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 D and B don't permute.

I Know that using the Schur complement 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 is possible to prove that $det(H) \neq 0$


  • $\begingroup$ I doubt that you can prove it without assuming something more! If some pair of submatrices commute, you can prove using hostemostel.com/pdf/MTH/0.pdf If you replace $D$ being lower tri with $D$ upper tri, you can prove using row reductions. But with your conditions you can make an counterexample. $\endgroup$ – kjetil b halvorsen Jan 13 '14 at 11:12

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.


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