Homomorphism space/Total evaluation/Not linear/Remark

Let ${}K$ denote a field and let ${}V$ and ${}W$ denote finite-dimensional ${}K$-vector spaces. We consider the natural mapping

$\Psi \colon V\times \operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow W,(v,\varphi )\longmapsto \varphi (v),$

where we have the product space on the left. This mapping is not linear in general. We have, on one hand,

${}\varphi (v_{1}+v_{2})=\varphi (v_{1})+\varphi (v_{2})\,$

and, on the other hand,

${}{\left(\varphi _{1}+\varphi _{2}\right)}(v)=\varphi _{1}(v)+\varphi _{2}(v)\,,$

so, if we fix one component, we have additivity (and also compatibility with the scalar multiplication) in the other component. In the product space, we have

${}(v_{1},\varphi _{1})+(v_{2},\varphi _{2})=(v_{1}+v_{2},\varphi _{1}+\varphi _{2})\,,$

and, therefore,

${}{\begin{aligned}\Psi ((v_{1},\varphi _{1})+(v_{2},\varphi _{2}))&=\Psi ((v_{1}+v_{2},\varphi _{1}+\varphi _{2}))\\&={\left(\varphi _{1}+\varphi _{2}\right)}{\left(v_{1}+v_{2}\right)}\\&=\varphi _{1}{\left(v_{1}+v_{2}\right)}+\varphi _{2}{\left(v_{1}+v_{2}\right)}\\&=\varphi _{1}(v_{1})+\varphi _{1}(v_{2})+\varphi _{2}(v_{1})+\varphi _{2}(v_{2})\\&\neq \varphi _{1}(v_{1})+\varphi _{2}(v_{2})\\&=\Psi ((v_{1},\varphi _{1}))+\Psi ((v_{2},\varphi _{2}))\end{aligned}}$

(only in exceptional cases we have ${}\varphi _{1}(v_{2})+\varphi _{2}(v_{1})=0$).