Is the row space the annihilator of the null space?

Is the row space of a matrix $A$ the annihilator of the null space of the matrix? I guess I'm assuming that solutions to the vector equation $AX = 0$ are vectors in some vector space $V$ and that the rows of matrix $A$ may represent linear functionals so that the span of the rows of $A$ represents a subspace of $V^*$.

Yes, if we interpret linear functionals in the usual way for matrices (dot product). This is one of the so-called "fundamental theorems" for linear algebra; see this article. In particular, if $A$ is a matrix (viewed as a linear transformation between Euclidean vector spaces), then

$$ R(A^T) = N(A)^{\perp}, $$ where $R(A^T)$ is the range of the transpose of $A$, which is the same as the row space, $N(A)$ is the null space, and the symbol $\perp$ indicates the orthogonal complement, which is the same as the annihilator when using the dot product as the realization of the dual space.

To prove this: If $x \in R(A^T)$, then $x = A^T z$ for some vector $z$. Now take any $y \in N(A)$. Then

$$ \langle x,y \rangle = x^T y = z^T A y = z^T (0) = 0, $$ where we used the fact that $Ay = 0$ since $y \in N(A)$. This proves that each element of $R(A^T)$ is orthogonal to every element of $N(A)$; that is, $R(A^T) \subseteq N(A)^{\perp}$.

To get equality, it suffices to prove that the two subspaces have equal dimension.