Is the trace invariant under conjugation of isometry

Question

$\DeclareMathOperator{\tr}{tr}$

Let $H$ be a Hilbert space and $U$ be an isometry. Then which of the following are true?

1. If $T$ is trace-class, then $\tr(U TU^*)=\tr T$
2. If $T$ is trace-class, then $\tr(|U TU^*|)=\tr |T|$ where $|T|=\sqrt{T^*T}$
3. $T\mapsto U TU^*$ is a linear isometry on the space of Hilbert-Schmidt operators

I know that the statements are all true if $U$ were unitary, but how about the weakened case of isometry.

Answer 1

Let $H$ be separable, and $(e_k)$ be an orthonormal (Schauder) basis. Then $$ tr(UTU^*) = \sum_k (|T|U^*e_k,U^*e_k). $$ We can write $$ U^*e_k = \sum_i (U^*e_k,e_i)e_i. $$ Plugging this in, then sum over $k$ first, we find $$ tr(UTU^*) = \sum_{i,j,k} (|T|e_i,e_j)(U^*e_k, e_i)(e_j,U^*e_k)\\ =\sum_{i,j,k} (|T|e_i,e_j)(Ue_j,(Ue_i,e_k)e_k)\\ =\sum_{i,j} (|T|e_i,e_j)(Ue_j,Ue_i). $$ Since $U$ is an isometry, it preserves the inner product, and $(Ue_j,Ue_i)=(e_j,e_i)$. So $$ tr(UTU^*)=\sum_{i,j} (|T|e_i,e_j)(e_j,e_i) = \sum_i (|T|e_i,e_i) = tr(T). $$ This proves (1). (2) follows from (1) since $tr(T)=tr(|T|)$.

For (3), we find by (1) $$ \|UTU^*\|_{HS}^2 = tr(UT^*TU^*) = tr(T^*T) = \|T\|_{HS}^2. $$