Commutators and scalar product. I have trouble understanding a fairly simple equality:
$\langle x, [A,B]x\rangle = \langle Ax, Bx \rangle  + \langle Bx, Ax \rangle$ where $x$ is a vector in $\mathbb{R}^n$ and $[A,B]$ is symmetric with $A,B \in M_{nn}(\mathbb{R}) $ and $A^T = A, B = -B^T$.
I can only decompose the expression in terms of $ABx$ and $BAx$ but I dont know how to obtain $Ax$ and $Bx$ terms.
 A: This formula is not correct: If we take $$A = B = \begin{pmatrix} 1 & 0\\
0 & 1\end{pmatrix}$$
we have that $[A,B] = 0$ but we find that for all $x \in \mathbb{R}^2$ that 
$$0 = \langle x, 0\rangle = 2 \langle x, x \rangle = 2\| x \|^2$$
which is only possible if $x = 0$. This gives a counterexample. 
Moreover, note that $[A,B] = AB - BA$, so you would need at least a minus in your right hand side (this would make the previous counterexmaple work, since we would find that $\langle x, 0 \rangle = \langle x, x \rangle - \langle x, x \rangle = 0$). 

$\textbf{EDIT}$: 
With the edited question (so adding the constraints that $A = A^T$ and $B = - B^T$, we find the following:
\begin{align}
\langle x, [A,B]x \rangle &= x^T (AB - BA)x\\
                          &= x^TABx - x^TBAx\\
                          &= (A^Tx)^T(Bx) - (B^Tx)^T(Ax)\\
                          &= (Ax)^T(Bx) + (Bx)^T(Ax)\\
                          &= \langle Ax, Bx \rangle + \langle Bx, Ax \rangle
\end{align}
where we used in equation two the linearity of the inner product, in equation three we used that $(AB)^T = B^TA^T$ and in the fourht equation, we used the assumption on $A$ and $B$ and $(\lambda B)^T = \lambda B^T$ for any scalar $\lambda$.


Note that you do not need that $[A,B]$ since this immediately follows from your assumptions on $A,B$ (convince yourself) :)
