$F$-linear maps $K \to K$ as a vector space over $K$(!), where $K/F$ is a finite-dimensional field extension Let $K/F$ be a finite-dimensional field extension, say $[K:F]=n$. Let $\mathcal{L}_F(K)$ denote the collection of $F$-linear transformations $K \to K$. Obviously this is an algebra over $F$. Indeed, it is just isomorphic to the $n \times n$ matrices over $F$. In particular, $\mathcal{L}_F(K)$ is an $n^2$-dimensional vector space over $F$. 
Note, however, that you can also view $\mathcal{L}_F(K)$ as a vector space over $K$. I am a bit puzzled as to what this vector space looks like though. Can somebody please clarify this? I suppose that, in a certain sense, the only thing to know about a vector space over $K$ is its dimension. So, for sure, I would like to know:

Question: What is the dimension of $\mathcal{L}_F(K)$ as a vector space over $K$?

Any other comments which shed light on the $K$-linear structure of $\mathcal{L}_F(K)$ will also be appreciated.
 A: I claim that $\dim_K(L_F(K)) = [K:F]$: To see this, choose a basis $\{e_1,e_2,\ldots, e_n\}$ of $K$ over $F$ and let $T_{i,j} \in L_F(K)$ be the operator that maps
$$
e_i \mapsto e_j \text{ and } e_k \mapsto 0 \text{ if } k\neq i
$$


*

*Since $\{T_{i,j}\}$ forms an $F$-basis for $L_F(K)$, it follows that it spans $L_F(K)$ over $K$. Furthermore, for any $i$, note that
$$
e_j^{-1}T_{i,j} - e_k^{-1}T_{i,k} = 0
$$
Hence the set
$$
S:= \{T_{1,1}, T_{2,1}, \ldots, T_{n,1}\}
$$
spans $L_F(K)$ over $K$.

*We claim that $S$ is linearly independent. To see this, suppose $\{\alpha_i : 1\leq i\leq n\}\subset K$ such that
$$
\sum \alpha_i T_{i,1} = 0
$$
Then for any $1\leq k\leq n$,
$$
\sum \alpha_i T_{i,1}(e_k) = \alpha_k e_1 = 0 \Rightarrow \alpha_k = 0
$$
Hence, $S$ is a $K$-basis for $L_F(K)$, and $\dim_K(L_F(K)) = [K:F]$
A: It is even easier:
Let $B$ be a $F$-basis of $K$. We have
$\operatorname{Hom}_F(K,K) = \operatorname{Maps}(B,K)$
The latter is well known to be a $K$-vector space with basis $B$ (since $B$ is finite!). Hence the dimension of $\operatorname{Hom}_F(K,K)$ as a $K$-vectorspace is $|B|=\dim_F K$
A: Now that the dust has settled on this question, I had might as well point out what is probably quite clear in hindsight. The fact that $\mathrm{dim}_K(\mathcal{L}_F(K)) = [K:F]$ is also an immediate consequence of the following fact.

Proposition: Let $K/F$ be a finite dimensional field extension. Let $V$ be a finite dimensional vector space over $K$. Then, the relation $\mathrm{dim}_F(V) = \mathrm{dim}_K(V) \cdot [K:F]$ is satisfied.
Proof: Say $[K:F]=m$ and $\mathrm{dim}_K(V) = n$. We just need to check that, if $\{\alpha_1,\ldots,\alpha_m\}$ is an $F$-basis for $K$ and $\{v_1,\ldots,v_n\}$ is a $K$-basis for $V$, then $\{\alpha_i v_j : 1 \leq i \leq m, 1 \leq j \leq n\}$ is a $K$-basis for $V$. This is rather straightfoward. For example, if we have $\sum_{ij} c_{ij} \alpha_i v_j = $, where $c_{ij} \in F$ then, rewriting this as $\sum_j \left(\left( \sum_i c_{ij} \alpha_i \right) e_j \right)=0$ we get that $\sum_i c_{ij} \alpha_i$ for each fixed $j$, because the $v_j$ are $K$-independent, and so it follows that each $c_{ij}=0$, because the $\alpha_i$ are $F$-independent.

So, while Prahlad Vaidyanathan's proof is very nice, reading it might leave one with the impression that the situation is more special than it really is.
