A monotone matrix is nonsingular?

Question

A monotone matrix is nonsingular.

According to Wikipedia, a proof of this is...

Let $A$ be a monotone matrix and assume there exists $x\neq0$ with $Ax=0$. Then, by monotonicity, $x\ge 0$ and $-x\ge 0$, and hence $x=0$.

1)When it says "assume that there exists $x\neq0$ with $Ax=0$", does it mean that it assumes $A$ is singular, and therefore $dim(ker(A))>0$ ?

2)I don't understand what the proof wants to say when it says "by monotonicity, $x\ge 0$ and $-x\ge 0$". From this why can you make a conclusion that it's proved?

Thanks!

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My apologies Clement, thank you very much for your help. Thanks to you now I figured this out. — Gabriel Macotti, Nov 02 '17 at 17:02

score 0 · Accepted Answer · answered Oct 08 '17 at 14:36

Yes. The proof is by contradiction: you assume $A$ is singular, pick $x\in\mathrm{ker} A\setminus \{0\}$ (which exists by your assumption $A$ is singular), and then use monotonicity of $A$ to show the contradiction: mnamely, that the $x$ you picked must be $0$, contradicting the way you chose it. (Here, a proof by contradiction is actually unnecessary: a direct proof showing that any $x\in\mathrm{ker} A$ is $0$ would work as well).
To see how monotonicity of $A$ is used: montonicity means

every vector $u$ such that $Au\geq 0$ must be such that $u \geq 0$

where $\geq$ is entry-wise.

For the vector $x$ that was picked, we have $Ax=0$ and $A(-x)=-Ax=-0=0$, so in particular $Ax\geq 0$ and $A(-x)\geq 0$. The monotonicity of $A$ then implies $x\geq 0$ and $-x\geq 0$.

But then, this implies in turn $x\geq 0$ and $x\leq -0=0$, so every entry of $x$ satisfies $0\leq x_i \leq 0$: so every entry of $x$ is $0$, so $x$ is the $0$ vector.

A monotone matrix is nonsingular?

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