[proofplan]
We identify $M(n,\mathbb F)$ with the coordinate space of matrix entries, using real coordinates when $\mathbb F=\mathbb C$. The determinant is a polynomial function of these coordinates, so the nonvanishing locus of the determinant is open and gives $GL(n,\mathbb F)$ an open submanifold structure. Matrix multiplication is polynomial in the entries, while inversion is given by the adjugate formula with nonzero denominator $\det A$. These coordinate formulas prove that both group operations are smooth.
[/proofplan]
[step:Model the matrix space by real coordinates]
Let $M(n,\mathbb F)$ denote the [vector space](/page/Vector%20Space) of $n\times n$ matrices with entries in $\mathbb F$. If $\mathbb F=\mathbb R$, identify $M(n,\mathbb R)$ with $\mathbb R^{n^2}$ by the coordinate map
\begin{align*}
A\mapsto (A_{ij})_{1\le i,j\le n}.
\end{align*}
If $\mathbb F=\mathbb C$, identify $M(n,\mathbb C)$ with $\mathbb R^{2n^2}$ by the coordinate map
\begin{align*}
A\mapsto (\operatorname{Re}A_{ij},\operatorname{Im}A_{ij})_{1\le i,j\le n}.
\end{align*}
All smoothness assertions below are taken with respect to these real coordinate models. Since an open subset of a finite-dimensional real vector space has its standard smooth manifold structure, it remains to prove that $GL(n,\mathbb F)$ is open and that the two group operations are smooth in these coordinates.
[/step]
[step:Express the general linear group as the nonvanishing locus of the determinant]
Define the determinant map
\begin{align*}
\det:M(n,\mathbb F)\to\mathbb F.
\end{align*}
In matrix entries it is given by
\begin{align*}
\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n} A_{i,\sigma(i)},
\end{align*}
where $S_n$ is the [symmetric group](/page/Symmetric%20Group) on $\{1,\dots,n\}$ and $\operatorname{sgn}(\sigma)\in\{-1,1\}$ is the sign of the permutation $\sigma$. Thus $\det$ is a polynomial in the real coordinates introduced above: in the complex case, multiplication in $\mathbb C$ is polynomial in real and imaginary parts. Therefore $\det$ is continuous as a real-valued map when $\mathbb F=\mathbb R$ and as a map to $\mathbb R^2$ when $\mathbb F=\mathbb C$.
The set $\mathbb F\setminus\{0\}$ is open in $\mathbb F$ with its usual Euclidean topology, and
\begin{align*}
GL(n,\mathbb F)=\det^{-1}(\mathbb F\setminus\{0\}).
\end{align*}
Hence $GL(n,\mathbb F)$ is open in $M(n,\mathbb F)$.
[guided]
The point of this step is to turn invertibility into an open condition. The determinant map
\begin{align*}
\det:M(n,\mathbb F)\to\mathbb F
\end{align*}
is defined by the finite formula
\begin{align*}
\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^{n} A_{i,\sigma(i)}.
\end{align*}
Every term in this sum is a product of matrix entries, and there are only finitely many permutations $\sigma\in S_n$. Therefore, in the real case, $\det$ is a polynomial map from $\mathbb R^{n^2}$ to $\mathbb R$. In the complex case, after writing each entry as $A_{ij}=x_{ij}+iy_{ij}$, complex addition and multiplication are polynomial operations in the real variables $x_{ij}$ and $y_{ij}$. Thus $\det$ is a polynomial map from $\mathbb R^{2n^2}$ to $\mathbb R^2$.
Polynomial maps between finite-dimensional real coordinate spaces are continuous. Since $\mathbb F\setminus\{0\}$ is open in $\mathbb F$, the inverse image
\begin{align*}
\det^{-1}(\mathbb F\setminus\{0\})
\end{align*}
is open in $M(n,\mathbb F)$. Finally, a matrix over $\mathbb F$ is invertible exactly when its determinant is nonzero, so this inverse image is precisely $GL(n,\mathbb F)$. Hence $GL(n,\mathbb F)$ is an open subset of the ambient [matrix space](/page/Matrix%20Space).
[/guided]
[/step]
[step:Give the open subset its inherited smooth manifold structure]
Because $GL(n,\mathbb F)$ is open in the finite-dimensional real vector space $M(n,\mathbb F)$, it inherits the smooth manifold structure whose charts are restrictions of the coordinate charts on $M(n,\mathbb F)$. Thus $GL(n,\mathbb R)$ is a smooth manifold of real dimension $n^2$, and $GL(n,\mathbb C)$ is a smooth manifold of real dimension $2n^2$.
[/step]
[step:Show that matrix multiplication is smooth in entries]
Define the multiplication map
\begin{align*}
m:GL(n,\mathbb F)\times GL(n,\mathbb F)\to GL(n,\mathbb F)
\end{align*}
by
\begin{align*}
m(A,B)=AB.
\end{align*}
For each pair of indices $1\le i,j\le n$, the $(i,j)$ entry of $AB$ is
\begin{align*}
(AB)_{ij}=\sum_{k=1}^{n} A_{ik}B_{kj}.
\end{align*}
This is a polynomial function of the entries of $A$ and $B$, and in the complex case it is polynomial in their real and imaginary parts. Hence $m$ is smooth as a map between open subsets of finite-dimensional real coordinate spaces.
[/step]
[step:Show that matrix inversion is smooth by the adjugate formula]
Define the inversion map
\begin{align*}
\iota:GL(n,\mathbb F)\to GL(n,\mathbb F)
\end{align*}
by
\begin{align*}
\iota(A)=A^{-1}.
\end{align*}
Define the adjugate map
\begin{align*}
\operatorname{adj}:M(n,\mathbb F)\to M(n,\mathbb F)
\end{align*}
by the rule that its $(i,j)$ entry is the signed determinant of the $(n-1)\times(n-1)$ submatrix obtained by deleting the $j$th row and $i$th column of $A$. For $A\in GL(n,\mathbb F)$, the adjugate formula gives
\begin{align*}
A^{-1}=\frac{\operatorname{adj}(A)}{\det A}.
\end{align*}
Thus each entry of $\operatorname{adj}(A)$ is a polynomial in the entries of $A$. Since $\det A\neq 0$ on $GL(n,\mathbb F)$, each real coordinate function of $A^{-1}$ is smooth: in the case $\mathbb F=\mathbb R$ it is a quotient of smooth polynomial functions, and in the case $\mathbb F=\mathbb C$ it is obtained by multiplying by the complex conjugate and dividing by $|\det A|^2$, which is again a smooth real-valued denominator that does not vanish on $GL(n,\mathbb C)$. Therefore $\iota$ is smooth.
[/step]
[step:Conclude that the smooth manifold is a Lie group]
The set $GL(n,\mathbb F)$ is a group under matrix multiplication, with identity matrix $I_n$ and inverse operation $A\mapsto A^{-1}$. The preceding steps show that $GL(n,\mathbb F)$ is a smooth manifold and that the multiplication and inversion maps
\begin{align*}
m:GL(n,\mathbb F)\times GL(n,\mathbb F)\to GL(n,\mathbb F)
\end{align*}
and
\begin{align*}
\iota:GL(n,\mathbb F)\to GL(n,\mathbb F)
\end{align*}
are smooth. Therefore $GL(n,\mathbb F)$ is a real Lie group.
[/step]
