Let \(V\) be a finite-dimensional vector space over the field \(F\) and let \(S\) and \(T\) be linear operators on \(V\). We ask: When do there exist ordered bases \(\mathfrak{B}\) and \(\mathfrak{B}'\) for \(V\) such that \([S]_{\mathfrak{B}} = [T]_{\mathfrak{B}'}\)? Prove that such bases exist if and only if there is an invertible linear operator \(U\) on \(V\) such that \(T = USU^{-1}\). (Outline of proof: If \([S]_{\mathfrak{B}} = [T]_{\mathfrak{B}'}\), let \(U\) be the operator which carries \(\mathfrak{B}\) onto \(\mathfrak{B}'\) and show that \(T = USU^{-1}\). Conversely, if \(T = USU^{-1}\) for some invertible \(U\), let \(\mathfrak{B}\) be any ordered basis for \(V\) and let \(\mathfrak{B}'\) be its image under \(U\). Then show that \([S]_{\mathfrak{B}} = [T]_{\mathfrak{B}'}\).)
The printed outline has \(S=UTU^{-1}\) in its first direction while defining \(U\) to carry \(\mathfrak B\) onto \(\mathfrak B'\). For that choice of \(U\), the correct identity is \(T=USU^{-1}\), as in the theorem statement.
Solution
1. Equal matrices in suitable bases imply conjugacy.
Let \[ \mathfrak B=(v_1,\ldots,v_n), \qquad \mathfrak B'=(w_1,\ldots,w_n), \] and suppose \[ [S]_{\mathfrak B}=[T]_{\mathfrak B'}=(a_{ij}). \] Define \(U:V\to V\) by \(U(v_j)=w_j\) for every \(j\) and extend linearly. Since \(U\) sends a basis to a basis, it is invertible. By the definition of the matrix of a linear operator, \[ S(v_j)=\sum_{i=1}^n a_{ij}v_i, \qquad T(w_j)=\sum_{i=1}^n a_{ij}w_i \] for every \(j\) [@HK71]. Hence \[ US(v_j)=\sum_i a_{ij}U(v_i) =\sum_i a_{ij}w_i =T(w_j)=TU(v_j). \] Thus \(US=TU\) on a basis, hence on all of \(V\). Multiplying on the right by \(U^{-1}\) gives \[ T=USU^{-1}. \]
2. Conjugacy produces bases with the same matrix.
Conversely, suppose \(U\) is invertible and \(T=USU^{-1}\). Choose any ordered basis \[ \mathfrak B=(v_1,\ldots,v_n) \] and put \[ \mathfrak B'=(Uv_1,\ldots,Uv_n). \] This is an ordered basis because \(U\) is invertible. Write \[ S(v_j)=\sum_i a_{ij}v_i. \] Then \[ T(Uv_j)=USU^{-1}(Uv_j)=US(v_j) =\sum_i a_{ij}Uv_i. \] Therefore the \(j\)th column of \([T]_{\mathfrak B'}\) is the same as the \(j\)th column of \([S]_{\mathfrak B}\) for every \(j\). Hence \[ [T]_{\mathfrak B'}=[S]_{\mathfrak B}. \] This proves the converse and completes the equivalence.