Differential of a linear map between matrix spaces 
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?
 A: 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.
A: Maybe you got confused as me when i first read it. So maybe it's better to denote the point $X$ in $L_X$ as a more suggestive notation $L_{X_0}$, which means that $X_0$ is fix.  
Suppose that $(X^i_j)$ denote the standard global coordinate for $\text{GL}(n, \mathbb{R})$ and let $X_0=(X_0)^i_j \in \text{GL}(n, \mathbb{R})$ be a chosen point where we want to compute $A^\text{L}|_{X_0}$. The matrix representation of the map $L_{X_0} : \text{GL}(n, \mathbb{R}) \to \text{GL}(n, \mathbb{R})$, $X \mapsto X_0X$ in this coordinates is
$$
(X^i_j) \mapsto (L_{X_0}X)^i_j = (X_0)^i_k X^k_j.
$$
So 
\begin{align*}
A^{L}|_{X_0} &= d(L_{X_0})_{I_n}\bigg( A^i_j \frac{\partial}{\partial X^i_j}\bigg|_{I_n} \bigg) = A^i_j \, \bigg(\frac{\partial (L_{X_0})^k_m}{\partial X^i_j}\bigg)_{I_n} \, \frac{\partial}{\partial X^k_m}\bigg|_{X_0} \\ &= A^i_j \, \bigg(\frac{\partial (X_0X)^k_m}{\partial X^i_j}\bigg)_{I_n} \frac{\partial}{\partial X^k_m}\bigg|_{X_0} \\
&= A^i_j \, \bigg(\frac{\partial ((X_0)^k_l X^l_m)}{\partial X^i_j}\bigg)_{I_n} \, \frac{\partial}{\partial X^k_m}\bigg|_{X_0}  \\&= A^i_j (X_0)^k_l \delta_{il} \delta_{jm} \, \frac{\partial}{\partial X^k_m}\bigg|_{X_0}\\
&=   (X_0)^i_j  A^j_k \frac{\partial}{\partial X^i_k}\bigg|_{X_0}.
\end{align*}
