Homomorphism space/Total evaluation/Not linear/Remark

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

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

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

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

and, on the other hand,

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

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

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

and, therefore,

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

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