Fundamental Theorem of Linear Algebra over Complex Field

Question

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: $(nullspace(A)) = (rangespace(A^T))^\perp$

Answer 1

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.