Let \(V\) and \(W\) be finite-dimensional inner product spaces having the same dimension. Let \(U\) be an isomorphism of \(V\) onto \(W\). Show that:
The mapping \(T \to UTU^{-1}\) is an isomorphism of the vector space \(L(V, V)\) onto the vector space \(L(W, W)\).
\(\operatorname{trace}(UTU^{-1}) = \operatorname{trace}(T)\) for each \(T\) in \(L(V, V)\).
\(UT_{\alpha,\beta}U^{-1} = T_{U\alpha, U\beta}\) (\(T_{\alpha,\beta}\) defined in Exercise 10).
\((UTU^{-1})^* = UT^*U^{-1}\).
Solution
1. Part (a): Vector space isomorphism \(\Phi: L(V, V) \to L(W, W)\):
1.1. Define \(\Phi(T) = U T U^{-1}\) for \(T \in L(V, V)\).For any \(T_1, T_2 \in L(V, V)\) and \(c \in K\):\[ \Phi(c T_1 + T_2) = U(c T_1 + T_2)U^{-1} = c U T_1 U^{-1} + U T_2 U^{-1} = c \Phi(T_1) + \Phi(T_2). \]Thus \(\Phi\) is a linear map.
1.2. Define \(\Psi: L(W, W) \to L(V, V)\) by \(\Psi(S) = U^{-1} S U\).Then \(\Psi(\Phi(T)) = U^{-1}(U T U^{-1})U = T\) and \(\Phi(\Psi(S)) = U(U^{-1} S U)U^{-1} = S\), so \(\Psi = \Phi^{-1}\).Thus \(\Phi\) is a vector space isomorphism.
2. Part (b): Invariance of the trace:
2.1. By the cyclic property of the trace functional on linear operators \(\operatorname{tr}(AB) = \operatorname{tr}(BA)\):\[ \operatorname{tr}\left(U T U^{-1}\right) = \operatorname{tr}\left((UT) U^{-1}\right) = \operatorname{tr}\left(U^{-1} (UT)\right) = \operatorname{tr}\left(\operatorname{id}_V T\right) = \operatorname{tr}(T). \]
3. Part (c): Action on rank-one operators \(T_{\alpha, \beta}\):
3.1. The operator \(T_{\alpha, \beta} \in L(V, V)\) is defined by \(T_{\alpha, \beta}(v) = \langle v, \beta \rangle_V \alpha\).For any vector \(w \in W\):\[ \left(U T_{\alpha, \beta} U^{-1}\right)(w) = U\left(T_{\alpha, \beta}\left(U^{-1}w\right)\right) = U\left( \langle U^{-1}w, \beta \rangle_V \alpha \right) = \langle U^{-1}w, \beta \rangle_V (U\alpha). \]
3.2. Since \(U: V \to W\) is an inner product space isomorphism (unitary map), it preserves inner products:\[ \langle U^{-1}w, \beta \rangle_V = \langle U(U^{-1}w), U\beta \rangle_W = \langle w, U\beta \rangle_W. \]
3.3. Substituting this into 1:\[ \left(U T_{\alpha, \beta} U^{-1}\right)(w) = \langle w, U\beta \rangle_W (U\alpha) = T_{U\alpha, U\beta}(w). \]Since this holds for all \(w \in W\), \(U T_{\alpha, \beta} U^{-1} = T_{U\alpha, U\beta}\).
4. Part (d): Adjoint identity \((UTU^{-1})^* = UT^*U^{-1}\):
4.1. Because \(U\) is an inner product space isomorphism, \(U\) is unitary, which means \(U^* = U^{-1}\) and \((U^{-1})^* = (U^*)^* = U\).
4.2. Applying the property of adjoints on composite operators \((ABC)^* = C^* B^* A^*\):\[ \left(U T U^{-1}\right)^* = \left(U^{-1}\right)^* T^* U^* = U T^* U^{-1}. \]
5. Conclusion:Parts (a), (b), (c), and (d) are proven. Q.E.D.