Let $A$ be a diagonally dominant matrix of the form $D-L-U=D-(L_1+L_2)-(U_1+U_2)$, where $L_1,\,L_2,\,L=L_1+L_2$ are entrywise nonnegative strictly lower triangular matrices, $U_1,\,U_2,\,U=U_1+U_2$ are entrywise nonnegative strictly upper triangular matrices and $D$ is a diagonal matrix.

Denote the elements of $L_1$ and $L_2$ by $l_{1,ij}$ and $l_{2,ij}$ respectively.

We would like to show that $\|(D-L_1-U_1)^{-1} (L_2+U_2)\|_{\infty} < 1$.

To do it, we are looking to show there exists $c \in ]0,1[$ such as for all $x \in \mathbb{R}^n$, $||y||_{\infty} \leq c ||x||_{\infty}$ where $y$ is defined by $(D-L_1-U_1)y = (L_2+U_2)x$.

Even while writting $y=\dfrac{L_2+U_2}{D-L_1-U_1}x$, I still unable to prove the inequality above. Could someone help me ? Thank you in advance.

  • $\begingroup$ What would $\frac{L_2+U_2}{D-L_1-U_1}$ even mean? There's no indication that these matrices should commute with each other. $\endgroup$
    – Michael L.
    Jan 2, 2018 at 21:23
  • $\begingroup$ Yes, you are right... $\endgroup$ Jan 2, 2018 at 21:39

1 Answer 1


I suppose $A$ is strictly diagonally dominant. The problem boils down to proving this: let $A=D-F-G$ be strictly diagonally dominant, where $D$ is diagonal and $F,G$ are two entrywise nonnegative matrices with zero diagonals. Then $\|(D-F)^{-1}G\|_\infty<1$.

Suppose the contrary that $y=(D-F)^{-1}Gx$ has norm $\ge1$ for some $x$ of supremum norm $1$. Without loss of generality, let $y_1(=\|y\|_\infty\ge1)$ be the largest-sized element among all entries of $y$. Then in the first row of $Dy-Fy-Gx=0$, you may use the triangle inequality $|a-b|\ge|a|-|b|$ and the strict diagonal dominance of $A$ to obtain a contradiction: $$ 0=\left|\,d_{11}y_1-\sum_{j=2}^n(f_{1j}y_j+g_{1j}x_j)\,\right| \ge|y_1|\left(|d_{11}|-\sum_{j=2}^n(|f_{1j}+g_{1j}|)\right) >0. $$


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