Find real and complex $A_{m \times n}$ such that $\operatorname{Ran} A = \operatorname{Ker} A^T$ where $\operatorname{Ran} A$ is column space of $A$. $\newcommand{\Ran}{\operatorname{Ran}} \newcommand{\Ker}{\operatorname{Ker}}\newcommand{\b}{\mathbf}$
If $\b y \in \Ran A$ and $\b y \in \Ker A^T$, then $\b y = A \b x$ and $A^T \b y = \b 0 $ or $A^TA \b x = \b 0$.
Therefore either $\b x = \b 0$ or $A^TA=0$. I need a $A$ such that $A^TA = 0_{m \times m}$. 
For complex matrix $A$, I easily found $$\begin{bmatrix}1 & -i\\i &1\end{bmatrix}$$
but I can't find a real matrix $A$.
If I expand the product, $A^TA = 0$, I get $$0= (A^TA)_{ij} = \sum^m_{k=1} \left(A^T\right)_{ik}\left(A\right)_{kj} = \sum^m_{k=1} \left(A\right)_{ki}\left(A\right)_{kj}$$
for $i = j$, $\sum^m_{k=1} \left(A\right)_{ki}\left(A\right)_{ki} = 0$ which implies $A_{ki} = 0$ for all $1\le k \le m$. Since $1\le i \le m$, therefore $A_{ki} = 0$ for all $1 \le k,i\le m$.
The matrix $$B = \begin{bmatrix} 0 & 0 &0 \\ 0 &0 & 2 \end{bmatrix}$$ satisfies the condition $B_{ij} = 0$ for all $1 \le i,j\le m = 2$ but $B^TB \ne 0$.

My question :
Is there a real matrix (non-trivial) for which $\Ran A = \Ker A^T$ ? 

Here is an argument for a real Euclidean space. Note that if $A \neq 0$ then $\langle A v, A v \rangle > 0$ for some $v$. Then since $\langle A^{\mathrm{t}} A v, v \rangle = \langle A v, A v \rangle > 0$ it follows that $A^{\mathrm{t}} A v \neq 0$. So $A v$ is in the range of $A$ but not in the kernel of $A^{\mathrm{t}}$.
