Homomorphism space/Linear subspaces/Fact/Proof

Proof

(1). It is clear that we have a linear subspace. In order to prove the statement about the dimension, let ${}U'$ be an direct complement of ${}U$ in ${}V$, so that

${}V=U\oplus U'\,.$

Let ${}u_{1},\ldots ,u_{r}$ be a basis of ${}U'$. Every linear mapping from ${}S$ maps ${}U$ to ${}0$, and on ${}U'$ (or on a basis thereof) we have free choice. Therefore

${}S\cong \operatorname {Hom} _{K}{\left(U',W\right)}\,$

and the statement follows from fact.

(2). It is clear that we have a linear subspace. The natural mapping

$\operatorname {Hom} _{K}{\left(V,U\right)}\longrightarrow \operatorname {Hom} _{K}{\left(V,W\right)}$

of fact  (2) is injective in this case. Therefore,

${}T\cong \operatorname {Hom} _{K}{\left(V,U\right)}\,.$

(3). It is clear that we have a linear subspace. In the finite-dimensional case, let

${}V=V_{1}\oplus V_{2}\,$

be a direct sum decomposition. Due to fact, we have

${}\operatorname {Hom} _{K}{\left(V,W\right)}=\operatorname {Hom} _{K}{\left(V_{1},W\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,$

and

${}H=\operatorname {Hom} _{K}{\left(V_{1},W_{1}\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,.$

Therefore, the dimension equals

$\dim _{K}{\left(V_{1}\right)}\cdot \dim _{K}{\left(W_{1}\right)}+\dim _{K}{\left(V_{2}\right)}\cdot \dim _{K}{\left(W\right)}=\dim _{K}{\left(V_{1}\right)}\dim _{K}{\left(W_{1}\right)}+{\left(\dim _{K}{\left(V\right)}-\dim _{K}{\left(V_{1}\right)}\right)}\dim _{K}{\left(W\right)}\,.$

(4). Setting

${}L_{i}={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{i})\subseteq W_{i}\right\}}\,,$

we have ${}L=\bigcap _{i=1}^{k}L_{i}$. Hence, (4) follows from (3).

To fact