Lineare Algebra 2/Gemischte Definitionsabfrage/3/Aufgabe/Lösung

${\displaystyle {}v,w\in V}$

${\displaystyle {}d{\left(v,w\right)}:=\Vert {v-w}\Vert :={\sqrt {\left\langle v-w,v-w\right\rangle }}\,}$

den Abstand zwischen ${\displaystyle {}v}$ und ${\displaystyle {}w}$.

${\displaystyle {}v_{i}}$

${\displaystyle {}i\in I}$

${\displaystyle {}V}$

${\displaystyle \left\langle v_{i},v_{j}\right\rangle =0{\text{ für }}i\neq j}$

gilt.

${\displaystyle \psi \colon V\longrightarrow V}$

adjungiert zu ${\displaystyle {}\varphi }$, wenn

${\displaystyle {}\left\langle \varphi (v),w\right\rangle =\left\langle v,\psi (w)\right\rangle \,}$

für alle ${\displaystyle {}v,w\in V}$ gilt.

${\displaystyle {}H\subseteq G}$

${\displaystyle {}xH=Hx\,}$

für alle ${\displaystyle {}x\in G}$ ist.

${\displaystyle {}U\subseteq M}$

${\displaystyle {}x\in U}$

${\displaystyle {}\epsilon >0}$

${\displaystyle {}U{\left(x,\epsilon \right)}\subseteq U\,}$

existiert.

${\displaystyle {}K}$

${\displaystyle \varphi _{i}\colon V_{i}\longrightarrow W_{i}}$

heißt die lineare Abbildung

${\displaystyle V_{1}\otimes _{}\cdots \otimes _{}V_{n}\longrightarrow W_{1}\otimes _{}\cdots \otimes _{}W_{n},\,v_{1}\otimes _{}\cdots \otimes _{}v_{n}\longmapsto \varphi _{1}(v_{1})\otimes _{}\cdots \otimes _{}\varphi _{n}(v_{n}),}$

das Tensorprodukt der ${\displaystyle {}\varphi _{i}}$.