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.

(\(\Rightarrow\)) ...

Opgave 2a. Show that \(\beta^* := {b^* : b \in \beta} \subset V^*\) is a linearly independent set.
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.