Corollary 3.46. Suppose $G \subset \operatorname{GL}(n, \mathbb{C})$ is a matrix Lie group with Lie algebra $\mathfrak{g}$. Then a matrix $X$ is in $\mathfrak{g}$ if and only if there exists a smooth curve $\gamma$ in $\operatorname{M}_n(\mathbb{C})$ with $\gamma(t) \in G$ for all $t$ and such that $\gamma(0) = I$ and $\mathrm{d}\gamma/\mathrm{d}t|_{t=0} = X$. Thus, $\mathfrak{g}$ is the tangent space at the identity to $G$.
Proof. If $X$ is in $\mathfrak{g}$, then we may take $\gamma(t) = e^{tX}$ and then $\gamma(0) = I$ and $\mathrm{d}\gamma/\mathrm{d}t|_{t=0} = X$. In the other direction, suppose that $\gamma(t)$ is a smooth curve in $G$ with $\gamma(0) = I$. For all sufficiently small $t$, we can write $\gamma(t) = e^{\delta(t)}$, where $\delta$ is a smooth curve in $\mathfrak{g}$. Now, the derivative of $\delta(t)$ at $t = 0$ is the same as the derivative of $t \mapsto t \delta'(0)$ at $t = 0$. Thus, by the chain rule, we have $$ \gamma'(0) = \left. \frac{\mathrm{d}}{\mathrm{d}t} e^{\delta(t)} \right|_{t=0} = \left. \frac{\mathrm{d}}{\mathrm{d}t} e^{t \delta'(0)} \right|_{t=0} = \delta'(0). $$ Since $\delta(t)$ belongs to $\mathfrak{g}$ for all sufficiently small $t$, we conclude (as in the proof of Theorem 3.20) that $\delta'(0) = \gamma'(0)$ belongs to $\mathfrak{g}$.
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The conclusion say $\delta \in \mathfrak{g} \implies \delta'(0) \in \mathfrak{g}$. So why does the derivative belong to $\frak{g}$ too?
Since Lie algebra is a linear space, we suppose $\{v_1,v_2,\cdots,v_n\}$ is a basis of $\mathfrak{g}$.
$\delta(t)\in \mathfrak{g}$, so we have $\delta(t)=\sum_{j=1}^nf_j(t)v_j$.
It's obvious that $\delta'(t)|_{t=0}=\sum_{j=1}^nf_j'(0)v_j\in \mathfrak{g}$.