This bit of text comes from Lee's Introduction to Smooth Manifolds.
I don't see why (8.15) holds. Note first of all that Lee assumes the Einstein summation convention, while I will not in my formulation. I would think that we have
$$
A^L\vert_X=\sum_{i=1}^n\sum_{j=1}^n X^i_j A^i_j\frac\partial{\partial X^i_j}\bigg\vert_X,
$$
instead of
$$
A^L\vert_X=\sum_{k=1}^n\sum_{i=1}^n\sum_{j=1}^n X^i_j A^j_k\frac\partial{\partial X^i_k}\bigg\vert_X,
$$
since I think it holds that
$$
d(L_X)_{I_n}\left(\frac\partial{\partial X^i_j}\bigg\vert_{I_n}\right)=X^i_j\frac\partial{\partial X^i_j}\bigg\vert_X.
$$
I argued this using the coordinate reprsentation of the differential, which is given for an arbitrary smooth map $F\colon M\to N$ by
$$
dF_p\left(\frac\partial{\partial x^i}\bigg\vert_p\right)=\frac{\partial\hat F^i}{\partial x^j}\bigg\vert_{\hat p}\frac\partial{\partial y^j}\bigg\vert_{F(p)},
$$
where $(x^i)$ are local coordinates for some open $U\ni p$, and $(y^i)$ are local coordinates for some open $V\ni F(p)$.
Hence, if we take $(E^i_j)$ as our basis for $\operatorname M_n(\mathbb R)$, then $E^i_j$ is mapped by $L_X$ to $X^j_i$. And therefore
$$
\frac{\partial(L_X)^i_j}{\partial x^k_l}=\delta_{ik}\delta_{jl} X^j_i.
$$
Note that also here, I don't assume Einstein summation convention.
So I don't see why (8.15) holds… could someone clarify?
Best Answer
Note that when you left multiply by $X$, you now have the matrix whose $ik$-entry is $X^i_j A^j_k$. If we want its $ij$-entry, we should change letters around and write $X^i_\ell A^\ell_j$. This then becomes the coefficient of $\partial/\partial X^i_j$.
Bottom line, the matrix $A$ at the identity element left-translates to the matrix $XA$ at the point $X$. Writing that in terms of the standard basis is precisely what Lee has done.