consider a bounded linear $2\times 2$ matrix-valued operator $A: L^2(R)\otimes L^2(R)\rightarrow L^2(R)\otimes L^2(R)$, i.e. where each component $A_{i,j}$ ($1\leq i,j\leq 2$) is a bounded linear scalar operator acting on square-integrable functions on $R$ relative to the Lebesgue measure. If the scalar operators $A_{i,j}$ are trace-class, is it possible to conclude that $A$ is trace-class? If $A$ is trace-class too, how does the trace of $A$ relate to the traces of $A_{i,j}$?
Thank you very much for suggestions!
Yes, this is the case. Let me define $|A| := \sqrt{A^*A}$, $P_{i,j}$ the operators that restricts to the $(i,j)^{\text{th}}$ space, so that for example $P_{1,1}A= \left(\begin{matrix}A_{1,1} &0 \\0 & 0\end{matrix}\right)$, and I will use the same notation for the operator that takes an operator acting on $L^2$ and gives an operator on $L^2\otimes L^2$, so for example $P_{1,1}B = \left(\begin{matrix}B &0 \\0 & 0\end{matrix}\right)$.
First, if $B$ is a positive operator acting on $L^2$, then $P_{1,1}B$ is positive as well and we can use the fact that if $(\psi_k)_{k\in\Bbb N}$ is a Hilbert basis of $L^2$, then $(\psi_k\otimes \psi_j)_{k\in\Bbb N}$ is a basis of $L^2\otimes L^2$ to see that $$ \mathrm{Tr}(P_{1,1}B) = \sum_{j,k} \langle\psi_j\otimes\psi_k, P_{1,1}B\,\psi_j\otimes\psi_k\rangle \\= \sum_{j} \langle\psi_j, B\psi_j\rangle = \mathrm{Tr}(B) $$ and one operator is trace class if and only if the other is trace class. On another hand, it is not difficult to see that $$ \left|P_{1,1}A\right| = \left|\left(\begin{matrix} A_{1,1} &0 \\0 & 0\end{matrix}\right)\right| = \left(\begin{matrix} |A_{1,1}| &0 \\0 & 0\end{matrix}\right) = P_{1,1}|A_{1,1}| $$ since they are positive operators with the same square. Similarly $\left|P_{2,2}A\right| = P_{2,2}|A|$. For the other parts, notice that $$ \left|P_{1,2}A\right|^2 = \left(\begin{matrix} 0&0 \\A_{1,2}^* & 0\end{matrix}\right)\left(\begin{matrix} 0&A_{1,2} \\0 & 0\end{matrix}\right) \\ = \left(\begin{matrix} 0&0 \\0 & |A_{1,2}|^2\end{matrix}\right) = P_{2,2} |A_{1,2}|^2 $$ to deduce that $\left|P_{1,2}A\right| = P_{2,2} |A_{1,2}|$, and similarly, $\left|P_{2,1}A\right| = P_{1,1} |A_{2,1}|$. Hence since $A = P_{1,1}A + P_{1,2}A + P_{2,1}A + P_{2,2}A$, it follows by the triangle inequality for the trace norm and the previous computations that $$ \mathrm{Tr}(|A|) \leq \mathrm{Tr}(|P_{1,1}A|) + \mathrm{Tr}(|P_{1,2}A|) + \mathrm{Tr}(|P_{2,1}A|) + \mathrm{Tr}(|P_{2,2}A|) \\ \leq \mathrm{Tr}(P_{1,1}|A_{1,1}|) + \mathrm{Tr}(P_{2,2}|A_{1,2}|) + \mathrm{Tr}(P_{1,1} |A_{2,1}|) + \mathrm{Tr}(P_{2,2}|A_{2,2}|) \\ \leq \mathrm{Tr}(|A_{1,1}|) + \mathrm{Tr}(|A_{1,2}|) + \mathrm{Tr}(|A_{2,1}|) + \mathrm{Tr}(|A_{2,2}|) $$ so $A$ is trace class if its blocks are trace class.
In this case, to compute the trace, you can use again the definition of the trace using the tensor product basis, and it gives $$ \mathrm{Tr}(A) = \mathrm{Tr}(A_{1,1}) + \mathrm{Tr}(A_{2,2}). $$