[proofplan] We identify the tangent space at the identity with the ambient [matrix space](/page/Matrix%20Space) because $GL(n,\mathbb C)$ is open in $M(n,\mathbb C)$. Given a smooth one-parameter subgroup $\gamma$, set $X=\gamma'(0)$ and compare $\gamma$ with the exponential one-parameter subgroup $t\mapsto\exp(tX)$. The uniqueness part of the exponential one-parameter subgroup theorem then forces equality for all real $t$, and differentiating at $0$ proves uniqueness of the matrix. [/proofplan]

[step:Identify the infinitesimal velocity as a matrix]Since $GL(n,\mathbb C)$ is open in the real [vector space](/page/Vector%20Space) $M(n,\mathbb C)$, the tangent space $T_I GL(n,\mathbb C)$ is identified with $M(n,\mathbb C)$ by the differential of the open inclusion \begin{align*} j:GL(n,\mathbb C)\to M(n,\mathbb C). \end{align*} Define \begin{align*} X:=\gamma'(0)\in M(n,\mathbb C) \end{align*} under this identification. Because $\gamma$ is a [group homomorphism](/page/Group%20Homomorphism) from $(\mathbb R,+)$ to $GL(n,\mathbb C)$, it satisfies \begin{align*} \gamma(t+s)=\gamma(t)\gamma(s) \end{align*} for all $s,t\in\mathbb R$, and \begin{align*} \gamma(0)=I. \end{align*}[/step]

[guided]The derivative $\gamma'(0)$ is initially a tangent vector in $T_I GL(n,\mathbb C)$, not merely a symbol. Since $GL(n,\mathbb C)$ is an open subset of the real vector space $M(n,\mathbb C)$, the open inclusion \begin{align*} j:GL(n,\mathbb C)\to M(n,\mathbb C) \end{align*} identifies $T_I GL(n,\mathbb C)$ with $M(n,\mathbb C)$. Through this identification we define \begin{align*} X:=\gamma'(0)\in M(n,\mathbb C). \end{align*} The phrase one-parameter subgroup means that $\gamma$ is a group homomorphism from the additive Lie group $(\mathbb R,+)$ to the matrix group $GL(n,\mathbb C)$. Therefore, for every $s,t\in\mathbb R$, \begin{align*} \gamma(t+s)=\gamma(t)\gamma(s). \end{align*} Taking $s=0$ gives \begin{align*} \gamma(t)=\gamma(t)\gamma(0). \end{align*} Multiplying on the left by $\gamma(t)^{-1}$ gives \begin{align*} \gamma(0)=I. \end{align*} Thus $X$ is exactly the infinitesimal velocity of the subgroup at the identity matrix.[/guided]

[step:Compare $\gamma$ with the exponential subgroup having the same initial velocity] By [citetheorem:8783], for this matrix $X\in M(n,\mathbb C)$ the map \begin{align*} \eta:\mathbb R\to GL(n,\mathbb C),\qquad \eta(t)=\exp(tX), \end{align*} is the unique smooth one-parameter subgroup satisfying \begin{align*} \eta'(0)=X. \end{align*} The map $\gamma$ is also a smooth one-parameter subgroup and, by the definition of $X$, satisfies \begin{align*} \gamma'(0)=X. \end{align*} The uniqueness assertion in [citetheorem:8783] therefore gives \begin{align*} \gamma(t)=\eta(t)=\exp(tX) \end{align*} for every $t\in\mathbb R$. [/step]

[step:Differentiate at the identity to prove uniqueness of the matrix] Suppose $Y\in M(n,\mathbb C)$ also satisfies \begin{align*} \gamma(t)=\exp(tY) \end{align*} for every $t\in\mathbb R$. Differentiating this identity at $t=0$ in the ambient vector space $M(n,\mathbb C)$ gives \begin{align*} \gamma'(0)=\left.\frac{d}{dt}\right|_{t=0}\exp(tY). \end{align*} From the matrix exponential series, \begin{align*} \exp(tY)=I+tY+\sum_{m=2}^{\infty}\frac{t^mY^m}{m!}, \end{align*} so the derivative at $t=0$ is $Y$. Hence \begin{align*} X=\gamma'(0)=Y. \end{align*} Thus the matrix representing the one-parameter subgroup is unique, and it is precisely $\gamma'(0)$. [/step]