In https://arxiv.org/pdf/1505.04503.pdf it was mentioned that "the linear span of the idempotents is a lie ideal".
I wonder where I can find a proof for that.
Let $A$ be a Banach algebra and let $\chi$ be the linear span of the idempotents. We want to show that $ax-xa \in \chi$ for all $x\in \chi, a\in A$. From here, I'm not sure how to use the properties of idempotents to show that.
Any references or suggestions will be appreciated.
The commutator $[a,x]$ is linear in $x$, so it suffices to consider the case where $x$ itself is idempotent. Now note that $$[a,x]=[a,x^2]=x[a,x]+[a,x]x$$ so it suffices to show that $x[a,x]$ is in the span of the idempotents (since then so is $[a,x]x$ by symmetry). But observe that $$x[a,x]x=xax-xax=0$$ so $$(x+x[a,x])^2=x+x[a,x]$$ and thus $x[a,x]$ is a linear combination of the idempotents $x$ and $x+x[a,x]$.
The following perspective may help illuminate what's going on here (and is in fact how I came up with the argument above). Since $x$ is idempotent, we can decompose $A$ as a direct sum $$xAx\oplus xA(1-x)\oplus (1-x)Ax\oplus (1-x)A(1-x).$$ Moreover, if we represent elements of $A$ as $2\times 2$ matrices using this direct sum decomposition, we can multiply them using matrix multiplication, with $x$ corresponding to $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ (where I write the top left entry as $1$ since $x$ is in fact the unit of the algebra $xAx$, and also acts as a unit on $xA(1-x)$ and $(1-x)Ax$). We can then calculate that the commutator of $x$ with a matrix $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ is $\begin{pmatrix}0 & -b \\ c & 0\end{pmatrix}$. But now notice that any matrix of the form $\begin{pmatrix} 1 & 0 \\ c & 0\end{pmatrix}$ is idempotent, and so by taking linear combinations we can get any matrix $\begin{pmatrix} 0 & 0 \\ c & 0\end{pmatrix}$. Similarly, we can get any matrix $\begin{pmatrix} 0 & b \\ 0 & 0\end{pmatrix}$ as a linear combination of idempotents, and thus our commutator is a linear combination of idempotents.