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 \((A^tw^*)B(u)=A^t(w^*)(Bu)=w^*(AB)(u)\).