Let $G$ be a matrix Lie group and $\frak g$ be the corresponding Lie algebra. Furthermore, $X:I\rightarrow G$ is a smooth path in $G$.
I'd like to understand this statement better:
$$X^{-1} \cdot \frac{d X}{dt} \in \frak g.$$
Is there a proof for this from first principles? (e.g relying on matrix algebra, matrix exponential etc.)
Are there any illustrative, insightful examples?
The basic point here is that left multiplication by $X^{-1}$ is a linear map, and thus its derivative also is left multiplication by $X^{-1}$. Now take a curve $c:I\to G$ with $c(0)=X$, so $c'(0)$ is tangent to $G$ at $X$. Then $X^{-1}\cdot c(t)$ is a curve in $G$ mapping $0$ to the unit matrix. Hence $\frac{d}{dt}|_{t=0}(X^{-1}\cdot c(t))$ is an element of $\mathfrak g$. But by linearity, this derivative is just $X^{-1}\cdot c'(0)$. Now you can do this in any point of a curve, which proves the claim.