Representatietheorie

**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.

**Opgave 2b.** *Show that \(\beta^*\) is a basis of \(V^*\) if and only if \(V\) is finite dimensional.*

*Bewijs.* (\(\Leftarrow\)) If \(V\) is finite dimensional, then for every \(\phi \in V^*\) we’ve got, \[\phi = \sum_{i=1}^{n} \phi(b_i)b_i^*,\] as \[\phi(b_j) = \sum_{i=1}^{n} \phi(b_i) b_i^* (b_j)\] for all \(b_j \in \beta\).

(\(\Rightarrow\)) ...

**Opgave 3a.** *Show that if \(A:V\to W\) and \(B:U\to V\) are linaer maps, that \((AB)^t=B^tA^t\).*

*Bewijs.* First observe that \(AB:U \to W\), \(A^t:W^* \to V^*\), \(B^t:V^* \to U^*\) and \((B^tA^t):W^* \to U^*\). Let \(u\in U\) and \(w^* \in W^*\) then we need to show that \[(AB)^t(w^*)(u)=(B^tA^t)(w^*)(u).\] By definition this means that we need to show that: \[w^*(AB(u))=(A^tw^*)B(u).\] This holds as \((A^tw^*)B(u)=A^t(w^*)(Bu)=w^*(AB)(u)\).

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