Vtyjkyuk

How is the fundamental theorem of linear algebra stated when the inducing matrix has elements from the complex field? For example, does the usual transpose become a Hermitian transpose in a statement like this: $(nullspace(A)) = (rangespace(A^T))^perp$

Essentially, yes.

Let's say $mathbfA$ is $ntimes k$. To avoid ambiguity, $mathbfA^top$ will denote the ordinary transpose of $mathbfA$, that is, the matrix obtained by simply swapping the entries of $mathbfA$ across its main diagonal. $overlinemathbfA$ will denote the complex conjugate of $mathbfA$, that is, the matrix obtained by taking the complex conjugate of the entries of $mathbfA$. Finally, $mathbfA^*$ will denote the conjugate transpose of $mathbfA$, that is,
$$mathbfA^*=overlinemathbfA^topmbox.$$
By definition, the row space of a matrix is the span of its rows, while the column space is the span of its columns. So the row space of $mathbfA$ is always the same as the column space of $mathbfA^top$. It doesn't matter if $mathbfA$ has complex entries.

If you want to talk about the orthogonal complement of the row space of $mathbfA$, or equivalently, the orthogonal complement of the column space of $mathbfA^top$, then it does matter if $mathbfA$ has complex entries. Let's suppose it does. So the rows of $mathbfA$ lie in $mathbbC^k$, which uses the Euclidean inner product
$$left<mathbfv,mathbfwright>=mathbfv^topoverlinemathbfwmbox.$$

where $mathbfv$ and $mathbfw$ are $ktimes1$ column matrices. It doesn't matter too much if you use row vectors or column vectors. What's important is that the complex conjugate of $mathbfw$ is taken prior to the dot product. This ensures that $left<cdot,cdotright>$ satisfies all the defining properties of an inner product.

With this inner product in mind, a column vector $mathbfwinmathbbC^k$ is orthogonal to each of the rows of $mathbfA$ if and only if
$$mathbfAoverlinemathbfw=mathbf0$$
if and only if
$$overlinemathbfAmathbfw=mathbf0mbox,$$
from which we see that $mathbfw$ is in the null space of $overlinemathbfA$ (as opposed to the null space of $mathbfA$). So
$$mboxnulloverlinemathbfA=left(mboxrangemathbfA^topright)^perpmbox.$$
Replacing $mathbfA$ with $overlinemathbfA$ in the last equation shows
$$mboxnullmathbfA=left(mboxrangemathbfA^*right)^perp$$
as you suspected. Similarly, we get
$$mboxnullmathbfA^top=left(mboxrangeoverlinemathbfAright)^perpmbox.$$
Thus, if the FTLA were stated for matrices in $mathbbC^ntimes k$, then it would say something along the lines of: “The null space of $mathbfA$ is the orthogonal complement of the row space of $overlinemathbfA$, while the left null space of $mathbfA$ is the orthogonal complement of the column space of $overlinemathbfA$.”