If for any $x\in\mathbb{R}^n$, $\|Ax\|_\infty > 0$ then $A$ is full rank

Question

In order to show a matrix $A \in \mathbb{R}^{n \times n}$ is full rank, a paper I am reading showed that, $\|Ax\|_\infty > 0$ for any $x\in \mathbb{R}^n$. What is the reasoning behind this? I am confused why $\|Ax\|_\infty > 0$ implies full-rank-ness of $A$.

Answer 1

A square matrix $A$ has full rank if and only if its columns are linearly independent, which holds if and only if the unique solution to the equation $Ax = 0$ is $x = 0$.

In this case, we can see that whenever $x \neq 0$, we have $\|Ax\|_\infty > 0$, which means that $Ax \neq 0$. That is, $x \neq 0 \implies Ax \neq 0$ or equivalently, $Ax = 0 \implies x = 0$.

Answer 2

If $A$ is not full rank then $Ax=0$ for some nonzero $x$, and hence $||Ax||_{\infty}=0$ for some nonzero $x$.