Isometries of the trace class
Bernard Russo · Proceedings of the American Mathematical Society · 1969
REMARK 1. The theorem provides a partial answer to [3, Remark 1, p. 231 ]. PROOF. The adjoint CV is a linear isometry of ? onto ? so by results of Kadison [2, Theorem 7, Corollary 11] has the form V(A) = Ua(A) where a and U are as described in the statement of the theorem. It is elementary that 1?(T) =T(TU) where V'=a. The proof will be complete if it is shown that a is the adjoint of a-' (restricted to 3). By the folk result [1, pp. 256, 9] it is sufficient to check this in the following two cases: (i) a(A) = VA V-1 with V a fixed unitary operator; then (T, a(A)) = (T, VA V-1)= (V-1TV, A)= (a-cI(T), A), (ii) after the choice of an orthonormal basis, a (A) is the transposed matrix of A; then (T, a (A)) =Tr(Ta (A)) =Tr(a (T)A) = (a-'(T), A). REMARK 2. A previous version of the above proof exploited a knowledge of the extreme points of the unit sphere of 3. These were determined to be the partial isometries with initial (hence final) domain one-dimensional.