# Week 1

Opgave 1a. Let $$v \in V - \{0\}$$. Show that there exists a $$\phi \in V^{\ast}$$ such that $$\phi(v)\neq 0$$.
Bewijs.
Since $$v \neq 0$$, clearly there exists a basis element $$b_0$$ such that $$b_0^{\ast}(v) \neq 0$$. Since $$b_0^{\ast}$$ is in the dual vector space, we are done.

Opgave 1b. Show that there exists a canonical injective linear map $$i:V \to V^{**}$$.
Bewijs.
Define this map as $$i:v \mapsto V^*\to k : \phi \mapsto \phi(v)$$. Linearity follows from the linearity from elements of $$V^*$$ and from the linearity of elements of $$V^{**}$$. Suppose that $$v,w \in V$$ such that $$i(v)=i(w)$$. In other words, for all $$\phi \in V^*$$ $$\phi(v)=\phi(w)$$. As $$\phi$$ is linear, $$\phi(v-w)=0$$. But we know from 1a, that there exist a $$\phi$$ such that $$\phi(v-w) \neq 0$$ if $$v-w \neq 0$$. Therefore $$v-w$$ must be zero, and so $$v=w$$, as required.

Opgave 1c. Show that $$i$$ is an isomorphism if and only if $$V$$ is finite dimensional.
Bewijs.
($$\Leftarrow$$) Since $$V$$ is finite dimensional, $$\dim(V)=\dim(V^{**})$$. And as $$i$$ is injective, and the dimensions of domain and codomain are the same, $$i$$ must be bijective, and therefore it is an isomorphism.

($$\Rightarrow$$) ...

Opgave 2a. Show that $$\beta^* := {b^* : b \in \beta} \subset V^*$$ is a linearly independent set.
Bewijs.
Assume there are scalars such that $\sum_{b^*\in\beta^*} a_{b^*}b^* =0.$ Well, then since $$b^*$$ is in the dual vector space, for all $$c\in \beta^*$$, $a_{c^*}=\sum_{b^*\in\beta^*} a_{b^*}b^*(c) =0(c)=0.$ So all scalars must be zero, as required.