Let $\mu$ be a top degree left invariant differential form on the Lie group $\operatorname{GL}(n,\mathbb{R})$. The divergence $\operatorname{div}(v)$ of a vector field $v$ is the function defined by the equation $$ \operatorname{div}(v) \mu = \mathcal{L}_v\mu$$ where $\mathcal{L}_v\mu$ denotes the Lie derivative of $\mu$ in the direction of $v$.
I want to compute such things. Say our vector field is given by $v(g) = g + g^{-1}$ where the right hand side is understood as an element of $\mathfrak{gl(2,\mathbb{R})}$ which is isomorphic to the tangent space at the point $g$.
How do I concretely compute it? I was thinking of using Cartan's formula for the Lie derivative together with taking the basis of left invariant vector fields given by the matrices $e_{ij} \in \mathfrak{gl(2,\mathbb{R})}$ for $1 \leq i,j \leq n$ and expressing $v$ as $\sum f_{ij}e_{ij}$ for some functions $f_{ij}$ (by a slight abuse of notation I denote both the element of the Lie algebra and the left invariant vector field by $e_{ij}$). My issue is that I am getting somewhat confused by all the isomorphisms and lost in the computations. At the point $g$, should I first move the matrix $g + g^{-1}$ to an element of $T_e\operatorname{GL}(2,\mathbb{R})$ via the differential of left multiplication by $g^{-1}$ or not?
If possible I'd like to have an explanation that does not rely too heavily on the fact that $\operatorname{GL}(n, \mathbb{R})$ is an open subset of a vector space and something which is more easily adaptable to any (Classical) Lie group, but beggars can't be choosers!