Define $$X_A(x):=A^i_{\ j}x^j$$ where $A$ is a matrix. Why is there a minus sign in the following formula? $$[X_A,X_B]=-X_{[A,B]}$$
Edit: perhaps the question is not well posed, since what I really meant was something like $$X_A(x):=A^i_{\ j}x^j\frac{\partial}{\partial x^i}$$ and this indeed yelds the above commutation relation.
Now, I have a new question, in some sense related to the above: it regards left (or right) multiplication in a group. Define $$L_X(A):=AX$$ and prove that $$[L_X,L_Y]=L_{[X,Y]}:$$ again the sign bothers me, and here I would like to ask some good reference about this stuff, or to point out where $A,X$ live: do $A$ stay in the group, while $X$ in the tangent space, or what? (sorry for the confusion)
Let $V$ be a finite dim. vector space over $\mathbb K$ and $X_A\in End_{\mathbb K}(V)$, defined as in the question, for any $n\times n$ matrix $A\in M_n(\mathbb K)$ with $n=dim V$. We can choose $\mathbb K=\mathbb R$ or $\mathbb C$ for simplicity.
If no non-trivial metric on $V$ is specified, then we lower and/or raise indices with no extra signs.
Then the $i$-th component $(([[X_A,X_B]])(x))_i$ of $([[X_A,X_B]])(x)\in V$ is
$$(([[X_A,X_B]])(x))_i:=(X_A(X_B(x))-X_B(X_A(x)))_i=(A_{ij}B_{jk}-B_{ij}A_{jk})x_k,$$
and
$$(X_{[A,B]}(x))_i:=[A,B]_{ik}x_k=(A_{ij}B_{jk}-B_{ij}A_{jk})x_k. $$
So the linear map $\varphi:(M_{n}(\mathbb K),[\cdot,\cdot])\rightarrow (End_{\mathbb K}(V),[[,\cdot,\cdot]])$ with $A\mapsto \varphi(A):=X_A$ is a Lie algebra morphism.
If we consider this second setting, we can identify the $X_A$'s with tangent vectors at $p$, where $p$ is an arbitrary point with coordinates $(x_1,\dots,x_n)$ of a given real Riemannian manifold $M$ with Euclidean metric (we can raise/lower indices with no extra signs, as supposed in the above setting). In other words,
$$X_A(x)\in T_{p}M,$$
for all $A\in M_{n}(\mathbb R).$
By definition, the coefficients of the $X_A$'s are monomials of degree $1$ in the coordinates $x_j$ at $p$ and
$$\mathcal L_{X_A}(f)(x):=X_A(x)(f):=A_{ij}x_j\frac{\partial f}{\partial x_i},$$
for all $f\in C^{\infty}(U)$, with $U$ open set containing $p$. Then
$$(X_A(x)\circ X_B(x))(f):=X_A(x)(X_B(x)(f))=A_{ij}B_{ki}x_j\frac{\partial f}{\partial x_k}+A_{ij}B_{kl}x_jx_l\frac{\partial^2 f}{\partial x_i x_k},$$ $$(X_B(x)\circ X_A(x))(f):=X_B(x)(X_A(x)(f))=B_{kl}A_{ik}x_l\frac{\partial f}{\partial x_i}+B_{kl}A_{ij}x_lx_j\frac{\partial^2 f}{\partial x_k x_i}, $$ $$X_{[A,B]}(x)(f):=[A,B]_{kj}x_j\frac{\partial f}{\partial x_k}= (A_{ki}B_{ij}-B_{ki}A_{ij})x_j\frac{\partial f}{\partial x_k},$$
with $\frac{\partial x_i}{\partial x_k}=\delta_{ik}.$
The relation
$$[X_A(x),X_B(x)]=-X_{[A,B]}(x)~~~~(1)$$
holds, as the terms quadratic in the coordinates $x_{\bullet}$ appearing in the l.h.s. of $(1)$ cancel out.