Let \(V\) be the space of complex \(n \times n\) matrices with inner product \((A \mid B) = \operatorname{tr}(AB^*)\). For each \(M\) in \(V\), let \(T_M\) be the linear operator defined by \(T_M(A) = MA\). Show that \(T_M\) is unitary if and only if \(M\) is a unitary matrix.
Solution
1. Adjoint and composition of the left-multiplication operator:
1.1. For any \(A, B \in V = M_n(\mathbb{C})\), compute the inner product:\[ (T_M(A) \mid B) = \operatorname{tr}\big((MA) B^*\big) = \operatorname{tr}(M A B^*) = \operatorname{tr}\big(A B^* M\big) = \operatorname{tr}\big(A (M^* B)^*\big) = (A \mid T_{M^*}(B)). \]
cyclic property of the trace \(\operatorname{tr}(XY) = \operatorname{tr}(YX)\) with \(X = M\) and \(Y = AB^*\), and \((M^* B)^* = B^* M\).
1.2. Thus the adjoint of \(T_M\) on the Hilbert space \(V\) is \(T_M^* = T_{M^*}\).
definition of adjoint operator.
1.3. The composition \(T_M^* T_M\) on \(V\) satisfies:\[ (T_M^* T_M)(A) = T_{M^*}(MA) = (M^* M) A = T_{M^* M}(A) \quad \text{for all } A \in V. \]
associativity of matrix multiplication.
2. Equivalence of unitarity:
2.1. Forward direction (\(\implies\)): Assume \(T_M\) is a unitary operator, so \(T_M^* T_M = \operatorname{id}_V\).Evaluate this operator equality on the identity matrix \(I_n \in V\):\[ I_n = (T_M^* T_M)(I_n) = (M^* M) I_n = M^* M. \]Thus \(M^* M = I_n\), which means \(M\) is a unitary matrix.
evaluating the identity operator at \(I_n\).
2.2. Reverse direction (\(\impliedby\)): Assume \(M\) is a unitary matrix, so \(M^* M = M M^* = I_n\).For all \(A \in V\):\[ (T_M^* T_M)(A) = (M^* M) A = I_n A = A \implies T_M^* T_M = \operatorname{id}_V, \]\[ (T_M T_M^*)(A) = (M M^*) A = I_n A = A \implies T_M T_M^* = \operatorname{id}_V. \]Thus \(T_M\) is a unitary operator on \(V\).
1 and definition of unitary operator.
3. Conclusion:\(T_M\) is unitary if and only if \(M\) is a unitary matrix. Q.E.D.
2.