If $A$ is an $m \times n$ matrix and $\operatorname{rank} (A) = m$, does that mean the kernel of $A$ is just the $0$ vector?

Question

I have been trying to figure out the answer to this question. The only approach I could think of was the rank-nullity theorem, but that only gives $m + \operatorname{nul}(A) = n$, and I'm not sure where to go from there.

Answer 1

Consider $\pmatrix{1&0}\pmatrix{0\\ 1}$.

Answer 2

No, it doesn't, unless $m=n$: indeed, $\;\operatorname{rank}A\le\min(m,n)$, so if $\;\operatorname{rank}A=m$, we know $n\ge m$, and $\;\dim(\ker A)=n-m$.