Week 1

Opgave 1a. Let \(v \in V - \{0\}\). Show that there exists a \(\phi \in V^{\ast}\) such that \(\phi(v)\neq 0\).
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 \(i:V \to V^{**}\).
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.
(\(\Leftarrow\)) 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.

(\(\Rightarrow\)) ...