# Linear Transformation over Subfield

Letting $F\subseteq K$ be fields, and $V$ a vector space over $K$. Certainly, $V$ is also a vector space over $F$. And if $\{e_1,...,e_n\}$ is a basis for $K$ over $F$ and $\mathcal{B}=\{v_1,...,v_m\}$ a basis for $V$ over $K$, we can show that the set of $e_iv_j$ is a basis for $V$ over $F$.

Question: If I have a transformation $T:V\rightarrow V$ over $K$ whose matrix with respect to $\mathcal B$ has entries $a_{ij}$, how can I find a nice matrix representation over $F$ with respect to the basis defined above? Certainly, there should be a nice way to order this induced basis to invoke a prettier block matrix type solution, but I'm having some trouble working out the details...

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Each entry of $T$ is a linear transformation $K \to K$ which is in particular $F$-linear, so can be expressed as a linear transformation over $F$ with respect to the basis $\{ e_1, ... e_n \}$. These are the blocks you want.

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Ah, cool. But how do i express the $a_{ij}$ as transformations over $F$? Certainly each is a linear combination of the $e_i$, but I don't know where to go from there... – AsinglePANCAKE Oct 9 '12 at 18:28
@AsinglePANCAKE: the same way as any other linear transformation. Look at what it does to a basis of $K$. – Qiaochu Yuan Oct 9 '12 at 18:30
My ignorance shows through more and more. So if I have $a_{11}= b_{11}e_1+...+b_{n1}e_n$ then applying this to $e_1$ gives me the vector $(b_{11}e_1,...,b_{n1}e_n)^T$ with respect to the basis for $K$. This gives me an n by n block in place of each $a{ij}$, but something feels pretty fishy still...(apologies, I'm on a bus on my phone) – AsinglePANCAKE Oct 9 '12 at 18:49
No, applying this to $e_1$ gives $b_{11} e_1 e_1 + ... + b_{n1} e_n e_1$. Then you need to expand the products $e_i e_j$ in terms of the $e_i$. – Qiaochu Yuan Oct 9 '12 at 20:04