**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.