[lin.alg]

Let V be a K-vector space, then A := \mathrm{span}\left\{ \text{linear maps}\ T : K \rightarrow V \right\} \cong V as vector spaces. (Abusing notation and treating K as a 1D vector space over K in the definition of A).

A is a vector space. Easy to check that \varphi : A \rightarrow V : T \mapsto Te where e is the single basis vector in K is a linear transformation. It is surjective: \forall v \in V,\ T_v : K \rightarrow V,\ T_v(e) = v is a linear map. It is injective: if Te = Se then Ta = Tke = kTe = kSe = Sa\ \forall a \in K, i.e. S = T.