TITLE: Fundamental Theorem of Linear Algebra over Complex Field
QUESTION [6 upvotes]: 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:   $(nullspace(A)) = (rangespace(A^T))^\perp$

REPLY [5 votes]: Essentially, yes.
Let's say $\mathbf{A}$ is $n\times k$. To avoid ambiguity, $\mathbf{A}^{\top}$ will denote the ordinary transpose of $\mathbf{A}$, that is, the matrix obtained by simply swapping the entries of $\mathbf{A}$ across its main diagonal. $\overline{\mathbf{A}}$ will denote the complex conjugate of $\mathbf{A}$, that is, the matrix obtained by taking the complex conjugate of the entries of $\mathbf{A}$. Finally, $\mathbf{A}^{*}$ will denote the conjugate transpose of $\mathbf{A}$, that is,
$$\mathbf{A}^{*}=\overline{\mathbf{A}^{\top}}\mbox{.}$$
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 $\mathbf{A}$ is always the same as the column space of $\mathbf{A}^{\top}$. It doesn't matter if $\mathbf{A}$ has complex entries.
If you want to talk about the orthogonal complement of the row space of $\mathbf{A}$, or equivalently, the orthogonal complement of the column space of $\mathbf{A}^{\top}$, then it does matter if $\mathbf{A}$ has complex entries. Let's suppose it does. So the rows of $\mathbf{A}$ lie in $\mathbb{C}^{k}$, which uses the Euclidean inner product 
$$\left<\mathbf{v},\mathbf{w}\right>=\mathbf{v}^{\top}\overline{\mathbf{w}}\mbox{.}$$
where $\mathbf{v}$ and $\mathbf{w}$ are $k\times1$ 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 $\mathbf{w}$ is taken prior to the dot product. This ensures that $\left<\cdot,\cdot\right>$ satisfies all the defining properties of an inner product.
With this inner product in mind, a column vector $\mathbf{w}\in\mathbb{C}^{k}$ is orthogonal to each of the rows of $\mathbf{A}$ if and only if
$$\mathbf{A}\overline{\mathbf{w}}=\mathbf{0}$$
if and only if
$$\overline{\mathbf{A}}\mathbf{w}=\mathbf{0}\mbox{,}$$
from which we see that $\mathbf{w}$ is in the null space of $\overline{\mathbf{A}}$ (as opposed to the null space of $\mathbf{A}$). So
$$\mbox{null}\overline{\mathbf{A}}=\left(\mbox{range}\mathbf{A}^{\top}\right)^{\perp}\mbox{.}$$
Replacing $\mathbf{A}$ with $\overline{\mathbf{A}}$ in the last equation shows
$$\mbox{null}\mathbf{A}=\left(\mbox{range}\mathbf{A}^{*}\right)^{\perp}$$
as you suspected. Similarly, we get
$$\mbox{null}\mathbf{A}^{\top}=\left(\mbox{range}\overline{\mathbf{A}}\right)^{\perp}\mbox{.}$$
Thus, if the FTLA were stated for matrices in $\mathbb{C}^{n\times k}$, then it would say something along the lines of: “The null space of $\mathbf{A}$ is the orthogonal complement of the row space of $\overline{\mathbf{A}}$, while the left null space of $\mathbf{A}$ is the orthogonal complement of the column space of $\overline{\mathbf{A}}$.”
A visualization of the spaces of $\mathbf{A}$ could look like this.