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I have been struggling with this problem for awhile. Given that $A$ is a strictly diagonally dominant matrix with positive diagonal entries and non-positive off-diagonal entries, show that $A$ is monotone, i.e. $A^{-1} \geq 0$, meaning $a^{-1}_{ij} \geq 0$ for all $i,j$.

I have looked extensively for some help on this problem but have not come up with anything. Any help or a link to the right resource would help me out immensely!

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    $\begingroup$ What is a "strictly diagonally dominant" matrix? $\endgroup$ – Timbuc Oct 14 '14 at 3:19
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    $\begingroup$ A strictly diagonally dominant matrix is defined as: $|a_{ii}| > \sum \limits_{j \neq i} |a_{ij}|$ For all rows $i$. $\endgroup$ – Adam O'Brien Oct 14 '14 at 3:27
  • $\begingroup$ Sorry hit the enter key by accident, see the edited version now. $\endgroup$ – Adam O'Brien Oct 14 '14 at 3:30
  • $\begingroup$ Ok, but then something's odd imo: in your question, you say the off-diagonal elements of $\;A\;$ are non-positive, which means $$i\neq j\implies a_{ij}\le 0\iff a_{ij}^{-1}\le 0$$ so how do you expect to prove that $\;a_{ij}^{-1}\ge 0\;$ ? $\endgroup$ – Timbuc Oct 14 '14 at 3:32
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    $\begingroup$ The proof of $A^{-1}\ge 0$ can be found in [Berman, Plemmans - Nonnegative Matrices in the Mathematical Sciences]. The proof relies on several results scattered in the book, which makes it hard to reproduce here. $\endgroup$ – daw Oct 14 '14 at 7:08
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Take the definition:

A real n-by-n matrix $A=[a_{ij}]$ with $a_{ij}\le 0$ for all $i\neq j$ is an M-matrix if $A$ is nonsingular and $A^{-l}\le0$ (this mean that we don't have nonnegative elements)

If $A$ is strictly diagonal dominant, $|a_{jj}|>\sum_{j\neq i, i=1}^n |a_{ij}|$ and $a_{ij}\le0$ $i\neq j$. And now... why $a_{ij}^{-1}$ is nonnegative! Well if you do the Gauss-Jordan procedure to inverse you will note that, you inverse will have all elements positive.

Its it, make the inverse of matrix using elementary rows operations.

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    $\begingroup$ Well, the important part of the proof is missing. $\endgroup$ – daw Oct 14 '14 at 6:38
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$$[A|I]=\left[\begin{array}{ccc|ccc} a_{11} & ... & a_{1n} & 1 & ... & 0 \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ a_{n1} & ... & a_{nn} & 0 & ... & 1\end{array}\right]$$

Lets take the augmented system. All rows operation that will do in $A$ you do $I$ in order to transform $A$ in $I$

Perform $L_i=a_{11}L$ and $L_i=L_i-\frac{L_i}{a_{11}}$ for $i=2, \ldots,n$. This will change the matrix, since all elements below the $a_{11}$ will be zero.

Now you want to create zeros below the therm $a_{22}$ of A.

Doing the same process, $L_i=\frac{L_i}{a_{22}}$ and $L_i=L_i-\frac{L_i}{a_{22}}$ for $i=3, \ldots, n$ you will zero below the $a_{22}$. If you continue this process you will have a upper matrix in right hand of aumentad matrix and the leaf is begin to start to show your $A^{-1}$.

I so much thing to write, after this you must create zeros above the diagonal. An finally you just divide by diagonal element. Its will create $[I|A^{-1}]$.

You will see, all elements are positive. You understand? if not, i can shown you by skype. Try see this.

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Use induction. Start with $1\times 1$ matrix (can be shown trivially). Then, use the formula for the inverse of block matrix (you need to use the induction hypothesis and the diagonal dominance property for this step).

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Step 1: [A is a M-matrix (https://en.wikipedia.org/wiki/M-matrix)] Since 1) A is a Z-matrix (off diagonal is non positive) 2) A's eigenvalues have positive real part (Gershgorin’s Circle theorem) by definition, A is a M-matrix.

Step 2: [inverse of a non-singular M-matrix is nonnegative. ] See (How to prove that an M-matrix is inverse-non-negative?)

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Let $D$ be the diagonal part of $A$. We can write $$A=D(I-S)$$ where $S$ has positive elements, ($0$ on the diagonal) and the sum of elements in each row is $<1$. Let $s$ be the maximum row sum of $s$. One checks that for all $n\ge 1$ the maximum row sum of $S^n$ is $\le s^n$. Therefore we get $S^n\to 0$. That implies that the sum $I+S+\cdots +S^n$ converges to $I-S$. Since $S$ has positive entries, so do all the partial sums, and so the limit. Therefore, $(I-S)^{-1}$ has positive entries, and so does $A^{-1}$.

Obs: The proof involves an infinite process. One would like an algebraic proof. It is easy to show that all the leading minors of $A$ are $>0$. Therefore, $A$ has an $LU$ decomposition.Are the off diagonal entries of $L$, $U$ always $\le 0$ ?

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