[proofplan] We prove the equality by two inclusions inside the ambient real [vector space](/page/Vector%20Space) $M(n,\mathbb C)$. If $X\in\mathfrak g$, the curve $t\mapsto\exp(tX)$ lies in $G$ and has ambient derivative $X$ at $t=0$, so $X$ is tangent to $G$ at the identity. Conversely, if $V\in T_I G$, left translations show that the ambient vector field $A\mapsto AV$ is tangent to $G$; solving the corresponding ODE in the embedded submanifold and comparing with the ambient matrix ODE gives $\exp(tV)\in G$ for small $t$, and the exponential law extends this to all real $t$. The dimension statement then follows because equality identifies $\mathfrak g$ with the tangent space of the manifold $G$ at $I$. [/proofplan]

[step:Identify tangent vectors through the ambient inclusion] Let \begin{align*} j:G\to M(n,\mathbb C) \end{align*} denote the inclusion map. Since $G$ is an embedded submanifold of the real vector space $M(n,\mathbb C)$, the differential \begin{align*} dj_I:T_I G\to T_I M(n,\mathbb C) \end{align*} is injective. We identify $T_I M(n,\mathbb C)$ with $M(n,\mathbb C)$ by the standard translation identification for a finite-dimensional real vector space. Thus we regard $T_I G$ as the real linear subspace $dj_I(T_I G)\subset M(n,\mathbb C)$. [/step]

[step:Send each one-parameter subgroup generator to a tangent vector] Let $X\in\mathfrak g$. Define the curve \begin{align*} \gamma:\mathbb R\to G,\quad t\mapsto \exp(tX). \end{align*} This is well-defined by the definition of $\mathfrak g$. The ambient curve $j\circ\gamma:\mathbb R\to M(n,\mathbb C)$ is \begin{align*} (j\circ\gamma)(t)=\exp(tX). \end{align*} By the [absolute convergence of the matrix exponential series](/theorems/8777) [citetheorem:8777], \begin{align*} \exp(tX)=I+tX+\sum_{k=2}^{\infty}\frac{t^kX^k}{k!}. \end{align*} Hence \begin{align*} \lim_{t\to 0}\frac{\exp(tX)-I}{t}=X. \end{align*} Therefore the tangent vector $\gamma'(0)\in T_I G$ has ambient representative $X$. Thus \begin{align*} \mathfrak g\subset T_I G. \end{align*} [/step]

[step:Use left translations to build a tangent vector field from a tangent vector at the identity]Let $V\in T_I G$, viewed as an element of $M(n,\mathbb C)$ through the identification above. For each $A\in G$, define left translation by $A$ as the smooth map \begin{align*} L_A:G\to G,\quad B\mapsto AB. \end{align*} Its inverse is $L_{A^{-1}}$, so $L_A$ is a diffeomorphism. The ambient extension of $L_A$ is the real [linear map](/page/Linear%20Map) \begin{align*} \Lambda_A:M(n,\mathbb C)\to M(n,\mathbb C),\quad B\mapsto AB. \end{align*} Therefore, under the ambient identification of tangent spaces, \begin{align*} d(L_A)_I(V)=AV. \end{align*} Since $d(L_A)_I$ maps $T_I G$ into $T_A G$, it follows that \begin{align*} AV\in T_A G \end{align*} for every $A\in G$.[/step]

[guided]We need to turn the single tangent vector $V\in T_I G$ into a tangent direction at every other point of $G$. Throughout this guided argument, $V$ is viewed as its ambient representative in $M(n,\mathbb C)$ through the inclusion-induced identification of $T_I G$ with a real linear subspace of $M(n,\mathbb C)$. The group structure provides exactly this transport. For a fixed matrix $A\in G$, define \begin{align*} L_A:G\to G,\quad B\mapsto AB. \end{align*} This map is smooth because multiplication on the matrix Lie group is smooth, and its inverse is \begin{align*} L_{A^{-1}}:G\to G,\quad B\mapsto A^{-1}B. \end{align*} Thus $L_A$ is a diffeomorphism of $G$. Now compare this intrinsic map with its ambient linear extension \begin{align*} \Lambda_A:M(n,\mathbb C)\to M(n,\mathbb C),\quad B\mapsto AB. \end{align*} Because $\Lambda_A$ is linear, its derivative at every point is itself. In particular, when tangent vectors are represented inside $M(n,\mathbb C)$ through the inclusion $j:G\hookrightarrow M(n,\mathbb C)$, the differential of $L_A$ at the identity sends the ambient representative $V$ to \begin{align*} d(L_A)_I(V)=AV. \end{align*} The important point is that $d(L_A)_I$ is a differential between tangent spaces of $G$: \begin{align*} d(L_A)_I:T_I G\to T_A G. \end{align*} Since $V\in T_I G$, the image $AV$ must lie in $T_A G$. Therefore the assignment $A\mapsto AV$ defines a tangent vector at every point $A\in G$.[/guided]

[step:Solve the tangent ODE inside $G$ and compare it with the matrix exponential] Let $TG:=\bigsqcup_{A\in G}T_A G$ denote the tangent bundle of $G$. Define the vector field \begin{align*} W:G\to TG, \end{align*} by sending each $A\in G$ to the tangent vector $AV\in T_A G$. Since $G$ is an embedded submanifold of the finite-dimensional real vector space $M(n,\mathbb C)$, $G$ is a finite-dimensional smooth manifold. The map $A\mapsto AV$ is a smooth real-linear map from $M(n,\mathbb C)$ to $M(n,\mathbb C)$, and its restriction to $G$ is smooth. The previous step shows that this restricted map has value in $T_A G$ at each $A\in G$, so $W$ is a smooth tangent vector field on the smooth manifold $G$. By the local existence theorem for smooth ordinary differential equations on finite-dimensional smooth manifolds, applied to the smooth vector field $W$ with initial point $I\in G$, there exist $\varepsilon>0$ and a smooth curve \begin{align*} \beta:(-\varepsilon,\varepsilon)\to G \end{align*} such that \begin{align*} \beta(0)=I \end{align*} and \begin{align*} \beta'(t)=\beta(t)V \end{align*} for all $t\in(-\varepsilon,\varepsilon)$. Viewed as a curve in the ambient vector space $M(n,\mathbb C)$, the same curve solves the matrix [initial value problem](/page/Initial%20Value%20Problem) \begin{align*} Y'(t)=Y(t)V,\qquad Y(0)=I. \end{align*} The curve \begin{align*} \alpha:\mathbb R\to M(n,\mathbb C),\quad t\mapsto \exp(tV) \end{align*} also solves this initial value problem. Indeed, by [citetheorem:8770], $\alpha$ is a smooth one-parameter subgroup of $GL(n,\mathbb C)$. The expansion at $0$ obtained from the absolutely convergent matrix exponential series [citetheorem:8777] gives $\alpha'(0)=V$. For each $t\in\mathbb R$, the homomorphism identity gives $\alpha(t+h)=\alpha(t)\alpha(h)$, so differentiating with respect to $h$ at $h=0$ yields \begin{align*} \alpha'(t)=\alpha(t)V=\exp(tV)V. \end{align*} Also $\alpha(0)=I$. The right-hand side $Y\mapsto YV$ is a linear, hence smooth and locally Lipschitz, vector field on the finite-dimensional real vector space $M(n,\mathbb C)$. Therefore uniqueness for linear ordinary differential equations in finite-dimensional real vector spaces applies, and \begin{align*} \beta(t)=\exp(tV) \end{align*} for every $t\in(-\varepsilon,\varepsilon)$. Since $\beta(t)\in G$, we have \begin{align*} \exp(tV)\in G \end{align*} for every $t\in(-\varepsilon,\varepsilon)$. [/step]

[step:Extend local membership in $G$ to all real times] Let $t\in\mathbb R$. Choose $m\in\mathbb N$ such that $|t|/m<\varepsilon$. From the previous step, \begin{align*} \exp((t/m)V)\in G. \end{align*} Since $G$ is a subgroup of $GL(n,\mathbb C)$, the $m$-fold product of this element with itself lies in $G$. The exponential law for one matrix [citetheorem:8778], applied repeatedly to the single matrix $(t/m)V$, gives \begin{align*} \left(\exp((t/m)V)\right)^m=\exp(tV). \end{align*} Therefore $\exp(tV)\in G$ for every $t\in\mathbb R$. Hence $V\in\mathfrak g$, and so \begin{align*} T_I G\subset\mathfrak g. \end{align*} [/step]

[step:Conclude equality of the Lie algebra and the tangent space] The two inclusions give \begin{align*} \mathfrak g=T_I G \end{align*} as real linear subspaces of $M(n,\mathbb C)$. Since the tangent space of a smooth manifold at any point has real dimension equal to the manifold dimension, we obtain \begin{align*} \dim_{\mathbb R}\mathfrak g=\dim_{\mathbb R}T_I G=\dim_{\mathbb R}G. \end{align*} This proves both the tangent-space identification and the dimension statement. [/step]