If \(B\) is an arbitrary member of the inner product space in Exercise 7, show that \(L_B\) is unitarily equivalent to \(R_{B^t}\).
Solution
Define \[ U:V\longrightarrow V, \qquad U(A)=A^t. \]
1. The transpose map \(U\) is unitary.
The matrix units \(E_{ij}\) form an orthonormal basis of \(V\) for the inner product \[ (A\mid C)=\operatorname{tr}(AC^*). \] Since \[ U(E_{ij})=E_{ji}, \] \(U\) merely permutes this orthonormal basis. Hence it preserves the inner product and is unitary. Also \[ U^2=I, \] so \(U^{-1}=U\).
2. Conjugating \(L_B\) by \(U\) gives \(R_{B^t}\).
For every \(A\in V\), \[ \begin{aligned} (UL_BU^{-1})(A) &=U\bigl(B A^t\bigr)\\ &=(BA^t)^t\\ &=AB^t\\ &=R_{B^t}(A). \end{aligned} \] Therefore \[ \boxed{UL_BU^{-1}=R_{B^t}.} \] Since \(U\) is unitary, \(L_B\) and \(R_{B^t}\) are unitarily equivalent.