Proof $||A \underline x|| > 0 \Leftrightarrow \underline x \neq \underline 0$ If $n \geq m, A \in M(n,m)$ and $rg(A)=m$ Proof $||A \underline x|| > 0 \Leftrightarrow \underline x \neq \underline 0$ a)If $m \neq 0 \leftrightarrow A \neq 0_M$ Suppose that $\underline x = \underline 0 \Leftrightarrow ||A \cdot \underline 0||=||0||$ which is a contradiction to $||A \underline x|| > 0 $ therefore $||A \underline x|| > 0 \Leftrightarrow \underline x \neq \underline 0$ b)If $m \neq 0 \Leftrightarrow A \neq 0_M \Leftrightarrow A > 0$ per definition: $||A|| > \frac{||A \underline x||}{|| \underline x||}$ and $ \forall \underline x \neq \underline 0$ then $||A \underline x|| > 0 \Leftrightarrow \underline x \neq \underline 0$ $\Box$ Is any of these ways a correct proof? If not where are the mistakes or what can I improve?

A rigorous proof would be as follows:

a matrix $A\in M(n,m)$ has rank $m$ (i.e., maximal rank) if and only if it represents an injective homomorphism from a vector space of dimension $m$ to a vector space of dimension $n$. So, if you start with a non-zero vector $\underline x$, its image $A\underline x$ will remain different from zero, thus $||A\underline x||>0$. It is also clear that $A\underline 0=\underline 0$. \\\\\