Let $H$ be a (separable) Hilbert space and $B(H)$ be the space of bounded linear operators defined on $H$. Suppose $\{ e_i \}$ be an orthonormal basis of $H$, we say $T \in B(H)$ is a trace-class operator if

$$ \|T\|_1 := \sum_i \langle |T| e_i, e_i \rangle < \infty, $$

where $|T|$ is the positive operator such that $|T|^2 = T^*T$ and $T^*$ is the adjoint operator of $T$. If $T \in B(H)$ is a trace-class operator, we can define its trace as

$$ \newcommand{\Tr}{\mathrm{Tr}} \Tr(T) := \sum_i \langle Te_i, e_i \rangle. $$

From Wikipedia, or Conway's book A course in functional analysis, we know that trace-class operators have some nice properties:

  • If $T \in B(H)$ is a trace-class operator, then for any $A \in B(H)$, we know both $AT$, $TA$ are trace-class, and $\Tr(AT) = \Tr(TA)$, and $\|AT\|_1 \leq \|A\| \|T\|_1$.

However, I'm curious about the definition of trace-class operators between two different Hilbert spaces $H_1$ and $H_2$. Let $B(H_1, H_2)$ be the space of bounded linear operators from $H_1$ to $H_2$.

It seems the above definition can be generalized to this case as follows: We say an operator $T \in B(H_1, H_2)$ is trace-class if $\sqrt{T^*T} : H_1 \to H_1$ is trace-class and define its trace-norm as $\|T\|_1 := \| \sqrt{T^*T} \|_1$. This generalization looks reasonable to me, but I am not sure whether there exist some subtleties.

I am trying to prove the aforementioned properties of trace-class operators under such definition. Suppose $H_3$ is another Hilbert space, and $A \in B(H_2, H_3)$. To show that $AT$ is trace-class, by definition it suffices to show that $\sqrt{T^* A^* AT}$ is trace-class. But I don't know how to proceed.

I guess that the definition of trace-class operators can be generalized and the aforementioned properties also hold. I would appreciate it if you could provide some references about these.

(BTW: It seems that the Hilbert-Schmidt operator can be easily generalized to different Hilbert spaces.)

Update: Thanks for the answer of Ruy. There is another question about these trace-class operators: When $A \in B(H_2, H_1)$, can we conclude that $\Tr(AT) = \Tr(TA)$?


1 Answer 1


Notice that $$ |AT|^2 = T^*A^*AT \leq \|A\|^2T^*T = \|A\|^2|T|^2, $$ so, by Proposition 1.3.8 in [1] (square-root is an operator monotone function), one has that $$ |AT| \leq \|A\||T|, $$ and it follows that $AT$ is trace class according to the OP's definition.

[1] Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London - New York -San Francisco: Academic Press. X, 416 p. $ 60.00 (1979). ZBL0416.46043.

  • $\begingroup$ I was not aware the operator monotonicity of square-root. Thanks for pointing out it. I'm also curious whether $\mathrm{Tr}(AT) = \mathrm{Tr}(TA)$ holds when $H_3 = H_1$. Do you know some textbooks/papers about it? $\endgroup$
    – Barth2131
    Mar 22, 2021 at 2:55
  • $\begingroup$ It you consider all operators in sight as acting on $H_1\oplus H_2$, then $$ \pmatrix{0 & A \cr 0 & 0}\pmatrix{0 & 0 \cr T & 0}=\pmatrix{AT & 0 \cr 0 & 0} $$ while, $$ \pmatrix{0 & 0 \cr T & 0}\pmatrix{0 & A \cr 0 & 0}=\pmatrix{0 & 0 \cr 0 & TA} $$ so, yes, $\text{tr}(AT)=\text{tr}(TA)$. Kadison and Ringrose has a lot about traces in the second volume. $\endgroup$
    – Ruy
    Mar 22, 2021 at 3:52

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