A geometric characterization of these matrices would be the following:
An $(n\times n)$-matrix $A$ is the same as a linear transformation $A:\ {\mathbb R}^n\to{\mathbb R}^n$. Provide ${\mathbb R}^n$ with the standard scalar product.
Claim: The condition ${\rm Col}\,A={\rm Row}\,A$ is equivalent to ${\rm ker}\,A=({\rm im}\, A)^\perp$. In particular any orthogonal projection would have this property.
Proof. The condition ${\rm Col}\,A={\rm Row}\,A$ means that ${\rm im}\,A={\rm im}\,A^t$. Take an $x\in{\ker}\,A$. Then
$$0=\langle A x,y\rangle=\langle x,A^t y\rangle\qquad\forall y\ ;$$
therefore $x$ is orthogonal to ${\rm im}\,A^t={\rm im}\,A$. It follows that ${\rm ker}\,A\subset({\rm im}\,A)^\perp$, whence ${\rm ker}\,A=({\rm im}\, A)^\perp$ by counting dimensions.
Conversely, assume that ${\rm ker}\,A=({\rm im}\, A)^\perp$. Take an $x\in{\rm im}\,A^t$. Then $x=Az$ for a $z\in{\mathbb R}^n$, and we have
$$\langle u,x\rangle=\langle u,A^t z\rangle=\langle Au,z\rangle =0\qquad\forall u\in{\rm ker}\,A\ .$$
It follows that ${\rm im}\,A^t\subset({\rm ker}\,A)^\perp={\rm im}\,A$ and therefore ${\rm im}\,A^t={\rm im}\,A$, as $A$ and $A^t$ have the same rank.
You need $A \ne 0$ such that $A^2 = 0$ and $\operatorname{rank} A = n/2$.
So, for any nonsingular $S$, you can define $A = S^{-1} J S$, where $J$ is a Jordan matrix of the form
$$J = \bigoplus_{k=1}^{n/2} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$$
Best Answer
Hint Since $A^3 = 0$, every vector in the image of $A^2$ is in the null space of $A$.
(Note that the matrix size is irrelevant, and that this argument works for any nonzero nilpotent matrix by modifying the exponent appropriately.)