My question is divided in two: $\DeclareMathOperator{\Ran}{Ran}\DeclareMathOperator{\Ker}{Ker}\DeclareMathOperator{\span}{span}$
$1$- Can $\Ran A=\Ker A^T$ for real matrices?
$2$ - What happens with complex matrices?
As for the first part, I know there is a relation that states $\Ker A=(\Ker A^*)^\perp$ as a general case (for real and complex matrices), where $A^*$ is the hermitian adjoint and $\perp$ signifies the orthogonal complement. For real matrices, this translates to $\Ker A=(\Ker A^T)^\perp$, so I'm $99\%$ sure that the answer to this question is that $\Ran A=\Ker A^T$ is not possible for real matrices. But it would be nice to have this confirmed. Also, I have no clue how I would prove this without using the aforementioned relation. Is there another way to prove this?
This leads to the obvious question of what happens for complex matrices, which is the second part of my question. I have the following example:
$$A=\begin{pmatrix}1&i\\i&-1\end{pmatrix}$$
For this matrix, we can do row operations to reduce it to echelon form:
$$A_{e}=\begin{pmatrix}1&i\\0&0\end{pmatrix}$$
So, $\Ker A=\span\left\{\begin{pmatrix}-i\\1\end{pmatrix}\right\}$ and $\Ran A=\span\left\{\begin{pmatrix}1\\i\end{pmatrix}\right\}$
If we scale the basis of $\Ker A$ by $-i$, we can clearly see that $\Ker A=\Ran A$. As this matrix is symmetric, then $\Ker A=\Ker A^T=\Ran A=\Ran A^T$. So, all the fundamental subspaces of the matrix $A$ are the same. What is the condition for this to happen? Is it coincidence that the matrix is symmetric? I'm guessing it's not, but I'd be grateful if someone could shed some light into what the the reason for this is.
EDIT: I know there have been some similar questions asked in the past, but they still don't address the topic of complex matrices very much. I'm hoping this question can do so in a bit more detail.
Suppose we have a vector $v$ with real components such that $v\in\operatorname{Ran}A$ and that $\operatorname{Ran}A\subseteq\operatorname{Ker}A^T$. Then $v=Aw$ for some $w$ and $$ 0=A^Tv=A^TAw $$ implies $w^TA^TAw=0$, that is, $v^Tv=0$ and so $v=0$. Thus $\operatorname{Ran}A=\{0\}$ and thus $\operatorname{Ran}A=\operatorname{Ker}A^T$ is not possible (unless you allow $0\times0$ matrices).
Your idea is good for complex matrices: finding a $2\times2$ matrix having rank $1$ so that its column space coincides with the null space of the transpose. Thus the matrix has to be of the form $$ A=\begin{pmatrix} a & ka \\ b & kb \end{pmatrix} $$ and we need that $$ \begin{pmatrix} a & b \\ ka & kb \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}=0 $$ that amounts to saying that $a^2+b^2=0$. Thus the choice of $a=1,b=i$ is the simplest one. Note that $k$ can be any complex number.
Since $\dim\operatorname{Ran}A=1$, the dimension of the null space of $A^T$ is $1$ as well. Hence they're equal.