Is it possible to have Ran A = Ker $A^{T}$

Question

Is it possible to have $\text{rank }A$ = $\ker A^{T}$ when $A$ is a real matrix? I think the answer is no, but I'm not sure about how to prove it. What about when $A$ is a complex matrix? Here, I think the answer is yes, but I cannot come up with an example.

Can anyone please help me?

Answer 1

The range of $A^t$ is $(\operatorname{ker}A)^{\perp}$, in general.

Thus there are no (nontrivial) examples. That is, this implies we're in the zero vector space.

Answer 2

$Ran(A)=Ker A^T$ implies that $AA^T=0$, this implies that $\langle AA^T(x),x\rangle=\langle A^T(x),A^T(x)\rangle=0$ implies $A^T=A=0$.