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.