I found a nice identity involving the trace of some commutators of real symmetric matrices:
$$
\operatorname*{Tr}\, [A,B][A,[C,B]] = 0.
$$
Here $A$,$B$, and $C$ are real symmetric $n\times n$ matrices, $[A,B]=AB-BA$, and $\operatorname*{Tr}$ is the trace. I am simply asking for a nice proof of this identity, which hopefully generalizes to prove
$$
\operatorname*{Tr}\, [A,[A,B]][A,[A,[C,B]]] = 0.
$$
So far I only have a messy proof of the first identity, and an empirical verification of the second (for some random matrices).
A summary of the messy proof
The proof I found is just to expand the commutators out, use the cyclic property of the trace, and rearrange the terms to obtain the identity
$$
\operatorname*{Tr}\, [A,B][A,[C,B]]
= 2\operatorname*{Tr}\,ABA [C,B] + \operatorname*{Tr} A^2 [B^2,C].
$$
Then I use the fact that the matrices $A$,$B$, and $C$ are symmetric to see that $ABA$ and $A^2$ are symmetric while $[C,B]$ and $[B^2,C]$ are antisymmetric. Then we use the fact that the trace of the product of a symmetric matrix with an antisymmetric matrix is zero.
Best Answer
Since $[A,B]^T=[B^T,A^T]$, the commutator of two symmetric or antisymmetric matrices is antisymmetric, and the commutator of a symmetric and an antisymmetric matrix is symmetric.
As a result, if $A,B,C$ are symmetric, then $[A,B]$ is antisymmetric, $[A,[C,B]]$ is symmetric, and the third one $[A,[A,[C,B]]]$ is again antisymmetric. Thus, you can use the fact the trace of a product of symmetric and antisymmetric matrix is zero without having to expand commutators and use the trace's cyclic property.