Lineare Gruppenoperation/C^n/Endlich/Neues Skalarprodukt/Aufgabe/Lösung

Die Linearität in der ersten Komponente ergibt sich aus

${}{\begin{aligned}\Phi (\lambda _{1}w_{1}+\lambda _{2}w_{2},z)&={\frac {1}{\#\left(G\right)}}\sum _{\sigma \in G}\left\langle \sigma {\left(\lambda _{1}w_{1}+\lambda _{2}w_{2}\right)},\sigma z\right\rangle \\&={\frac {1}{\#\left(G\right)}}\sum _{\sigma \in G}\left\langle \lambda _{1}\sigma (w_{1})+\lambda _{2}\sigma (w_{2}),\sigma z\right\rangle \\&={\frac {1}{\#\left(G\right)}}\sum _{\sigma \in G}{\left(\lambda _{1}\left\langle \sigma (w_{1}),\sigma z\right\rangle +\lambda _{2}\left\langle \sigma (w_{2}),\sigma z\right\rangle \right)}\\&={\frac {1}{\#\left(G\right)}}\lambda _{1}\sum _{\sigma \in G}\left\langle \sigma (w_{1}),\sigma z\right\rangle +{\frac {1}{\#\left(G\right)}}\lambda _{2}\sum _{\sigma \in G}\left\langle \sigma (w_{2}),\sigma z\right\rangle \\&=\lambda _{1}\Phi (w_{1},z)+\lambda _{2}\Phi (w_{2},z).\end{aligned}}$

Ferner ist

${}{\begin{aligned}\Phi (w,z)&={\frac {1}{\#\left(G\right)}}\sum _{\sigma \in G}\left\langle \sigma w,\sigma z\right\rangle \\&={\frac {1}{\#\left(G\right)}}\sum _{\sigma \in G}{\overline {\left\langle \sigma z,\sigma w\right\rangle }}\\&={\overline {\Phi (z,w)}}.\end{aligned}}$

Daraus folgt auch

${}{\begin{aligned}\Phi (w,\lambda _{1}z_{1}+\lambda _{2}z_{2})&={\overline {\Phi (\lambda _{1}z_{1}+\lambda _{2}z_{2},w)}}\\&={\overline {\lambda _{1}\Phi (z_{1},w)+\lambda _{2}\Phi (z_{2},w)}}\\&={\overline {\lambda _{1}}}\cdot {\overline {\Phi (z_{1},w)}}+{\overline {\lambda _{2}}}\cdot {\overline {\Phi (z_{1},w)}}\\&={\overline {\lambda _{1}}}\Phi (w,z_{1})+{\overline {\lambda _{2}}}\Phi (w,z_{1}).\end{aligned}}$

Die positive Definitheit ergibt sich direkt aus

 ${}\Phi (w,w)={\frac {1}{\#\left(G\right)}}\sum _{\sigma \in G}\left\langle \sigma w,\sigma w\right\rangle$. 

Zur gelösten Aufgabe