91Ó°ÊÓ

The proof is carried out in four steps. Step 1: The linear transformation \(\psi: V \to V^{**}\), is given by \(\psi(v)(f) = f(v)\). Step 2: Given an ordered basis \(\beta = \{v_1,v_2,\dots,v_n\}\) of a finite-dimensional vector space \(V\), the dual basis \(\beta^* = \{f_1,f_2,\dots,f_n\}\) of the dual space \(V^*\) is defined with the property \(f_i(v_j) = \delta_{ij}\). Step 3: The double-dual basis \(\beta^{**} = \{f_1^*,f_2^*,\dots,f_n^*\}\) of the double-dual space \(V^{**}\) is defined with the property \(f_i^*(f_j) = \delta^{*}_{ij}\). Step 4: Upon applying the function \(\psi\) to each basis element \(\beta\), we obtain \(\psi(v_i) = f_i^{**}\) for all \(i\). It follows that \((\psi(v_i))(f_j) = f_i^{**}(f_j)\), which simplifies to \(\delta_{ij}\) as per the definition of the dual basis, verifying \(\psi(\beta) = \beta^{**}\).