Let $V$ be a vector space over a field $k$, and $F$ a subfield of $k$. An $F$-submodule $V_0$ of $V$ is called an $F$-structure if the natural $k$-linear map $V_0 \otimes_F k \rightarrow V$ is an isomorphism. Surjectivity means that $V_0$ contains a spanning set of $V$. Injectivity means that things in $V_0$ which are linearly independent over $F$ remain so over $k$.
If $P$ is a $k$-vector subspace of $V$, then $P$ is said to be defined over $F$ (with respect to the $F$-structure $V_0$) if $V_0 \cap P$ is an $F$-structure for $P$, or equivalently if the span of $V_0 \cap P$ over $k$ is $P$.
Let $f: V \rightarrow W$ be a $k$-linear transformation of vector spaces with $F$-structures, $V_0, W_0$. We say that $f$ is defined over $F$ (with respect to the given $F$-structures) if $f(V_0) \subseteq W_0$.
If $f$ is defined over $F$, is the kernel $\textrm{Ker } f$ also defined over $F$? A textbook I'm reading indicates that this is the case, but I haven't been able to prove it.
In order to avoid posting a long question, I will write my attempt at a proof in an answer below.
Let $f_0$ be the $F$-linear map induced by $f$ between $V_0$ and $W_0$ and $\varphi_V$ be the natural isomorphism $V_0\otimes_F k \to V$ (and $\varphi_W$ is defined similarly).
Then, since $\varphi_V$ and $\varphi_W$ are induced by the canonical injections $V_0 \hookrightarrow V$ and $W_0 \hookrightarrow W$, we can see that $$f=\varphi_W\circ(f_0\otimes \mathrm{Id})\circ \varphi_V^{-1}$$ and $$\mathrm{Ker} f = \varphi_V(\mathrm{Ker} (f_0\otimes \mathrm{Id})) = \varphi_V((\mathrm{Ker} f_0)\otimes_F k).$$
Since $\mathrm{Ker}f_0 = V_0\cap \mathrm{Ker}f$, we obtain the desired result.