Trace of composition gives a perfect pairing between opposite Hom spaces.Suggest a slogan
Let \(k\) be a field and let \(V\) and \(W\) be finite-dimensional \(k\)-vector spaces. The map \[ \begin{aligned} \Hom_k(V, W) &\to \dualof{\Hom_k(W, V)},\\ T &\mapsto \big(S\mapsto\Tr(T \circ S)\big), \end{aligned} \] is an isomorphism of \(k\)-vector spaces, where \(T\circ S\colon W\to W\).
The map is \(k\)-linear because the trace is, and both sides have dimension \(\dim_k V\cdot\dim_k W\). Choose bases, so that \(T\) is an \(m\times n\) matrix and \(S\) an \(n\times m\) matrix; then \(\Tr(TS)=\sum_{a,b}T_{ab}S_{ba}\). If \(\Tr(TS)=0\) for all \(S\), taking \(S\) to be the matrix unit \(E_{ba}\) gives \(T_{ab}=0\) for all \(a,b\), so \(T=0\). Thus the map is injective, hence an isomorphism.