# When is a Schur complement an $M$-matrix?

Let $F=\begin{bmatrix}A & B \\\\ B^{T} & D\end{bmatrix}$ be symmetric and strictly diagonally dominant (thus an $H$-matrix). I also know that $B>0$ entrywise. What I am trying to show is that the Schur complements $F/A$ and $F/D$ are $M$-matrices. It doesn't hold in the general case but my actual matrix is quite special and I'm sure it holds for it, after checking numerous examples. Still, the raw calculations with the standard formulas are too painful. So, does anybody know of a general condition I can check?

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If I understand correctly, this is not true.

Suppose $F/A$ is an M matrix, then $-F/A$ cannot be an M matrix. If $F$ satisfies your condition, so is $\begin{bmatrix}-A & B \\ B^{T},-D\end{bmatrix}$.

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