Q:

The Lie exponential map coincides with the matrix exponential?

From the wikipedia:

If $G$ is a matrix Lie group, then the exponential map coincides with the matrix exponential.

Is there any proof of this? I searched and did not find any. I failed to prove it myself.

A:

Let $\mathfrak g$ be the Lie algebra of $G$. It turns out that$$\mathfrak{g}=\left\{X\in\mathfrak{gl}(n,\mathbb{R})\,\middle|\,(\forall t\in\mathbb{R}):\exp(tX)\in G\right\}.$$Now, let $X\in\mathfrak{g}$. Consider the map$$\begin{array}{rccc}\gamma_X\colon&\mathbb R&\longrightarrow&G\\&t&\mapsto&\exp(tX).\end{array}$$Then $\gamma_X$ is a Lie group homomorphism from $(\mathbb{R},+)$ into $G$ such that ${\gamma_X}'(0)=X$. Therefore, by definition, $\exp_G(X)=\gamma_X(1)=\exp(X)$.

