Clearly, if $A\colon V\to V$ is an endomorphism on a finite-dimensional vector space and $V$ is the direct sum of two vector spaces $V_1$ and $V_2$ such that $A(V_i)\subset V_i$, then
$$
\mathrm{Tr}\,A=\mathrm{Tr}\,A|_{V_1}+\mathrm{Tr}\,A|_{V_2}.
$$
I was wondering if this is also true in the infinite-dimensional case, possibly under certain restricting circumstances. It seems very plausible, but I couldn't find any reference online.
Trace of operator defined on direct sum of two vector spaces
functional-analysisoperator-theorytrace
Best Answer
With respect to the decomposition we can write $A$ in block form $$ A = \begin{pmatrix} A_{00} & A_{01} \\ A_{10} & A_{11} \end{pmatrix} $$ where $A_{00} \in B(V_1)$, $A_{01} \in B(V_2, V_1)$, $A_{10} \in B(V_1,V_2)$, and $A_{11} \in B(V_2)$. That is $A \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} A_{00} x + A_{01} y \\ A_{10} x + A_{11} y\end{pmatrix}$ Then supposing $V_1$ and $V_2$ are closed subspaces of some Hilbert space we can define and orthonormal basis for $V_1$, $\{e_{1,i}\}_i$ and an orthonormal basis for $V_2$, $\{e_{2,i}\}_i$ such that $\{(e_{1,i},0)\}_i \cup \{(0,e_{2,i})\}_i$ form an orthonormal basis for $V_1 \oplus V_2$. From which you can show that $\mathrm{tr}(A) = \mathrm{tr}(A_{00}) + \mathrm{tr}(A_{11})$.