[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*}
[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]
[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]
