Condition on A for X to be positive when A X =b

Question

Given the following matrix equation: $$AX=b$$ with:

$A$ a (n,n) invertible matrix ;
$X$ and $b$ vectors of size $n$ with all elements of $b$ positive or null,

what is the condition on $A$ to have $X$ to have all his elements positive or null.

I wonder if the response is so trivial... (i also feel shameful to ask this question :) )

Square symmetric or Hermitian matrices can be positive semi-definite. I am not sure you can say that about vectors. Are you asking about cases where all the elements of $\mathbf b$ and $\mathbf X$ are positive (or non-negative)? — Henry, Oct 15 '21 at 09:41
yes, by semi-definite positive, i mean all elements of b and X are positive or null — BenG73, Oct 15 '21 at 14:21
Perhaps obviously, if $A$ is invertible then you want $A^{-1}\mathbf b$ to give non-negative results — Henry, Oct 15 '21 at 14:36
yes right. i am searching for the conditions on $A$ to get $A^{-1}b$ to be positive or null on each of its elements. — BenG73, Oct 16 '21 at 15:08
If you want this to be guaranteed for all $\mathbf b$ with non-negative elements, then you need all the elements of $A^{-1}$ to be non-negative — Henry, Oct 16 '21 at 17:31
Mmm... I may find matrix $A^{-1}$ with some negative elements that still gives a Positive x solution to the system. So you are right but this may be too restrictive... — BenG73, Oct 18 '21 at 07:04

score 1 · Answer 1 · answered Oct 22 '21 at 08:19

1

If we want $x$ to be nonnegative componentwise regardless of $b$, we need $A^{-1}$ to be nonnegative componentwise. Suppose not, then suppose $A^{-1}$ has a negative entry in column $j$, then we can let $b=e_j$ where $e_j$ is the $j$-th unit standard basis vector to construct a counter example.

You are looking for monotone matrix, matrices where if $Av \ge 0$ then $v \ge 0$ where the inequality is defined componentwise.

A subclass of monotone matrix is the M-matrix. A Z-matrix is a matrix where the off-diagonal entry is non-positive. An $M$-matrix, $A$, is a $Z$-matrix where it can be written as $A=sI-B$ where $s$ is at least as large as the maximum of the moduli of the engenvalues of $B$ where $B$ is nonnegative componentwise. Equivalence condition of $M$-matrices can be found on the wikipedia page.

answered Oct 22 '21 at 08:19

Siong Thye Goh

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Like i stated, $b$ is positive or null componentwise. So may be we don't need $A^{-1}$ to be monotone ? – BenG73 Oct 25 '21 at 08:31
If you want it to work for arbitrary $b$ that is positive or null componentwise, we need $A^{-1}$ to be positive or null componenentwise. – Siong Thye Goh Oct 25 '21 at 09:00
Beg my question: could you prove that last sentence ? – BenG73 Oct 26 '21 at 10:52
The first paragraph of my answer provides a proof. – Siong Thye Goh Oct 26 '21 at 16:02
I think your answer is right. The next question is then how to prove $A^{-1}$ is positive. At this stage, my question emerges as i have a specific linear system to solve with a need to get the solution x to be Positive. But this is beyond the question above... – BenG73 Oct 28 '21 at 11:13
here is a proof of the inverse of $M$-matrices are nonnegative. – Siong Thye Goh Oct 28 '21 at 11:26

Condition on A for X to be positive when A X =b

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