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Let $D$ be a diagonally dominant matrix (both row and column-wise). Assume $D$ is not necessarily symmetric.

Consider the following block representation of $D^{-1}$: \begin{equation} D^{-1}=\begin{bmatrix} B_1 & B_2 \\ B_3 & B_4 \end{bmatrix}. \end{equation} Here assume that $B_i$ are all square matrices.

My question is whether $C=(B_1^{-1} + B_1^{-T})$ is also diagonally dominant. Note that if $B_1=D^{-1}$, then we have $C=(D+D^T)$, and this trivially holds.

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Actually, I found the answer. I'm including it here in case others find it helpful as well. This paper (siam.org/meetings/la03/proceedings/LeiTGWLZ.pdf) establishes that if $D=[D_1,D_2;D_3,D_4]$ is a diagonally dominant matrix, then Schur complement of $D_4$ in $D$ given by $D_1−D_2D_4^{-1}D_3$ is also diagonally dominant. $B^{−1}_1$ is equal to this matrix (by matrix inversion lemma), hence the result follows.

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  • $\begingroup$ This should have remained a comment; closure of diagonal dominance under Schur complements has been also previously noted in some MO answers. $\endgroup$
    – Suvrit
    Commented Aug 22, 2017 at 16:59

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