# Show that the trace class operators on a Hilbert space form an ideal

Let $(H, (\cdot, \cdot))$ be a separable Hilbert space over $\mathbb{L} = \mathbb{R}$ or $\mathbb{C}$. Suppose that $\{\phi_n\}_{n=1}^\infty$ is an orthonormal basis for $H$.

Let $\mathcal{B}(H)$ denote the bounded linear operators $H \to H$. For $A \in \mathcal{B}(H)$, let $|A|$ denote the positive, self-adjoint square root of $A^*A$.

We say that that $A \in \mathcal{B}(H)$ is of trace class if and only if

$$\sum_{n=1}^\infty (\phi_n, |A|\phi_n) < \infty.$$

Let $\mathscr{I}_1(H)$ denote the trace class on $H$. For $A \in \mathscr{I}_1(H)$, set $\operatorname{Tr}(A) = \sum_{n=1}^\infty (\phi_n, |A|\phi_n).$

I would like to prove that the trace class operators on $H$ form a two sided ideal. That is, I would like to show that, if $A \in \mathscr{I}_1(H)$ and $B \in \mathcal{B}(H)$, then $AB, BA \in \mathscr{I}_1(H).$

I am studying trace class operators from Reed and Simon's Method's of Modern Mathematical Physics, Vol.1, and this proposition above is stated in the text.

Some things that I know from my studies so far:

1. For $A \in \mathscr{I}_1(H)$, $\operatorname{Tr}(A)$ is independent of the orthonormal basis $\{\phi_n\}_{n=1}^\infty$ that we choose for $H$.

2. $\mathscr{I}_1(H)$ is vector subspace of $\mathcal{B}(H).$

3. If $A \in \mathscr{I}_1(H)$ and if $U \in \mathcal{B}(H)$ is unitary (i.e., if $\operatorname{Ran}U = H$ and $(Ux,Uy) = (x,y)$, all $x,y \in H$), then $UAU^{-1} \in \mathscr{I}_1(H)$.

I am hoping that I can prove that $\mathscr{I}_1(H)$ is an ideal without using too much more that (1)-(3), and perhaps some calculations. I do know some basics about adjoints and compact operators, and I am familiar with the construction of the polar decomposition for elements in $\mathcal{B}(H)$. Although I am not sure if any of these topics are relevant.

I am not familiar with the spectral theorem or the continuous functional calculus (I have yet to reach that chapter in Reed and Simon).

In fact, Reed and Simon offer a proof that $\mathscr{I}_1(H)$ forms an ideal, and their proof goes in two stages:

a) First, they prove that every $B \in \mathcal{B}(H)$ can be written as a linear combination of four unitary operators.

b) Then, they prove that, if $A \in \mathscr{I}_1(H)$ and $U$ is unitary, then $UA, AU \in \mathscr{I}_1(H)$.

Using a) and b), we can conclude that $\mathscr{I}_1(H)$ is a two-sided ideal, because it is already known that $\mathscr{I}_1(H)$ is a vector space.

The difficulty is, I do not understand the proofs that Reed and Simon give for a) and b). Based on this question I asked previoulsy, I am not even sure that a) is true in the case $\mathbb{L} = \mathbb{R}$. And in proving b), Reed and Simon claim that $|UA| = |A|$ and $|AU|= U^{-1}|A|U$, and I have had no luck proving these equalities either.

Any solutions or hints are greatly appreciated, whether it be an explication of Reed and Simon's proof, or a brand new argument. I have been stuck on this proposition for some time and I'd really like to get this figured out!

• You probably really want to think of this as a (non-commutative) version of the fact that $\ell^1\cdot\ell^{\infty}=\ell^1$. A proof along these lines is very easy, if (and perhaps only if) you have a few standard facts available: (1) $A$ is trace class precisely if the singular values are summable; (2) the singular values can be obtained from a min-max principle (this is in RS4, I believe). – user138530 Aug 15 '14 at 1:34
• @ChristianRemling---You comment that $A$ is trace class if the singular values are summable. We usually talk about singular values with regard to compact operators. Does this mean that a trace class operator is compact? I have been trying to prove this (directly from my definition above) by establishing that $A$ is a norm limit of finite rank operators. Do you think I might be on the right track? – JZS Aug 15 '14 at 16:55
• jtms88: Yes, this is correct: trace class is a subset of the compact operators, and the approximation argument you suggested will work. – user138530 Aug 15 '14 at 18:01

I'm really curious why you would want to consider bounded linear operators over a real space. Besides making everything more awkward, what is it there to gain?

Regarding your second question: $$|UA|=((UA)^*UA)^{1/2}=(A^*U^*UA)^{1/2}=(A^*A)^{1/2}=|A|.$$ And $$|AU|=((AU)^*AU)^{1/2}=(U^*A^*AU)^{1/2}=U^*(A^*A)^{1/2}U=U^*|A|U.$$ The last equality is justified by the fact that the positive square root is unique: from $(U^*|A|U)^2=U^*|A|UU^*|A|U=U^*|A|^2U=U^*A^*AU$ we deduce that $(U^*A^*AU)^{1/2}=U^*|A|U$.

• Real Hilbert spaces provide a way to do Tomita-Takesaki theory without unbounded operators – Iván Mauricio Burbano Oct 27 '17 at 15:27

for first question see page 27 of Gohberg, I. C, Krein, M. G. Introduction to the Theoryof Linear Non-Self-Adjoint Operators in Hilbert Space. For $A\in\mathscr{I}_1(H)$ and $B\in\mathcal{B}(H).$we have $s_{j}(AB)\leq{s_{j}{(A)}} \Vert{B}\Vert_\infty$ & $s_{j}(BA)\leq{s_{j}{(A)}} \Vert{B}\Vert_\infty$.