# Trace-class operators between different Hilbert spaces

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 $$\Tr(AT) = \Tr(TA)$$?

Notice that $$|AT|^2 = T^*A^*AT \leq \|A\|^2T^*T = \|A\|^2|T|^2,$$ so, by Proposition 1.3.8 in  (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.
 Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London - New York -San Francisco: Academic Press. X, 416 p. $60.00 (1979). ZBL0416.46043. • I was not aware the operator monotonicity of square-root. Thanks for pointing out it. I'm also curious whether$\mathrm{Tr}(AT) = \mathrm{Tr}(TA)$holds when$H_3 = H_1$. Do you know some textbooks/papers about it? Mar 22, 2021 at 2:55 • It you consider all operators in sight as acting on$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.