In what sense can the Lie algebra of $GL_n(\mathbb{C})$ be "identified" with $M_n(\mathbb C)$? I would like to do Tao's exercise 6 (i) but before I can even attempt it I need to be clear about his terminology.
Exercise 6
Show that the Lie algebra $gl_n(\mathbb{C})$ of the general linear group $GL_n(\mathbb{C})$ can be identified with the space $M_n(\mathbb{C})$ of $n \times n$ complex matrices, with the Lie bracket $[A,B] := AB - BA$.
When he writes "identified", does he just mean there is an isomorphism? So the exercise would be to find an explicit isomorphism from $M_n(\mathbb{C})$ to $gl_n(\mathbb{C})$? Is that correct?
Many thanks for your help.
 A: My answer is a bit lacking, I omit calculations and take certain facts for granted.
Whenever a manifold $M$ is an open subset of some vector space $V$, there is, for every $m\in M$ a simple identification $T_mM\approx V$, which comes from the inclusion $\varphi :M\to V$ which is a smooth chart. Since $\mathrm{GL}_n(\mathbb{C})$ is an open subset of $\mathrm{M}_n(\mathbb{C})$, this gives you a natural identificaiton $$T_{Id}\mathrm{GL}_n(\mathbb{C})\approx \mathrm{M}_n(\mathbb{C}).$$ Explicitely, you map $A\in\mathrm{M}_n(\mathbb{C})$ to the class of paths $\lbrace t\in(-\epsilon,+\epsilon)\mapsto Id+tA\rbrace$ ($\epsilon$ depends on $A$, and is small enough so that $Id+tA$ is invertible). To compute the bracket you can use $[X,Y]_{Lie}=[X,Y]_{Vector~Fields}$ where $X,Y$ are in the Lie algebra and are identified (in the second bracket) to their left invariant extensions. The right hand side is explicitely computable if you know that the flow of $X\in\mathfrak{g}\approx\Gamma^{left~inv}(G)$ has $\mathrm{Fl}^t_X(g)=g\times\mathrm{Fl}^t_X(Id)$ and in the case of $\mathrm{GL}_n(\mathbb{C})$ you have $\mathrm{Fl}^t_X(Id)=\exp(tX)$ where $\exp$ is the usual matrix exponential. Then you only have to differentiate a path of matrices.
EDIT It's probably easier to go back to the definition of the Lie Bracket as $[X,Y]_{Lie}=ad(X,Y)=d_{Id}Ad(X)(Y)$, since for any $g\in\mathrm{GL}_n(\mathbb{C})$, $Ad(g):\mathrm{M}_n(\mathbb{C})\to\mathrm{M}_n(\mathbb{C})$ is conjugation by $g$.
A: "Identify" usually means "exhibit a nice isomorphism."  Making that rigorous in general is a bit tricky, but you know it when you see it. 
In this case, you could make it rigorous in the following way.  Note that both $\mathfrak{gl}_n(\mathbb{C})$ and $M_n(\mathbb{C})$ have a natural action of $\mathrm{GL}_n(\mathbb{C})$ (by the adjoint action and conjugation respectively), so a "nice" (or "natural" or "canonical") isomorphism should respect this action.
But Tao is almost certainly using "identify" informally.  He just means find an isomorphism, with the subtext that you shouldn't need to make any "arbitrary choices" along the way.
A: The group $G:=\text{GL}_n(\mathbb C)$ is an open subset of $M_n(\mathbb C)$. A vector field on $G$ is a map 
$$
X:G\to M_n(\mathbb C).
$$ 
We denote by $\ell(x)X$ the left action of $x\in G$ on such a vector field $X$, that is, we define the vector field $\ell(x)X$ by 
$$
(\ell(x)X)(y):=xX(x^{-1}y).
$$ 
Thus, $X$ is invariant under this action if and only if $X(x)=xX(1)$ for all $x$ in $G$. Such a vector field is said to be left invariant. In particular, for each $a\in M_n(\mathbb C)$ there is a unique left invariant vector field $\widetilde a$ such that $\widetilde a(1)=a$. We have $\widetilde a(x)=xa$ for all $x$ in $G$, and $\widetilde a$ is $C^\infty$. We must check 
$$
\left[\ \widetilde a\ ,\ \widetilde b\ \right]=\widetilde{[a,b]} 
$$ 
for all $a,b\in M_n(\mathbb C)$. 
Let $e_{ij}\in M_n(\mathbb C)$ be the matrix with a one in the $(i,j)$ position and zeros elsewhere. It suffices to verify that the above display holds for $a=e_{ij}$ and $b=e_{pq}$. 
We have 
$$
X_{ij}(x):=\widetilde{e_{ij}}(x)=x\ e_{ij}=\sum_{p,q}\ x_{pq}\ e_{pq}\ e_{ij}=\sum_p\ x_{pi}\ e_{pj}.
$$
If we write 
$$
X_{ij}=\sum_k\ x_{ki}\ \ \frac{\partial}{\partial x_{kj}}\quad,\quad 
X_{pq}=\sum_r\ x_{rp}\ \ \frac{\partial}{\partial x_{rq}}\quad, 
$$ 
then the Lie bracket $[X_{ij},X_{pq}]$ is just the commutator of the differential operators $X_{ij}$ and $X_{pq}$, and we get 
$$
[\widetilde e_{ij},\widetilde e_{pq}]=[X_{ij},X_{pq}]=\delta_{jp}\ X_{iq}-\delta_{qi}\ X_{pj}=\widetilde{[e_{ij},e_{pq}]}, 
$$ 
as was to be shown. 
