Differential manifold of dimension $m^2$

Question

How could I prove that $\mathrm{GL}(m, \mathbb{R})$ is a differential manifold of dimension $m^2$?

$\mathrm{GL}(m, \mathbb{R})$ is the set of all non-singular $m\times m$ matrices in $\mathbb{R}$

Thank you

Answer 1

It can clearly be represented to be the complement in $\mathbb R^{m^2}$ of the zeros of the "determinant" polynomial. So it is an open subset of $\mathbb R^{m^2}$, and thus can inherit the trivial differential structure on $\mathbb R^{m^2}$.

This space is obviously not connected, since there is no path from the matrices with positive determine to the matrices with negative determinants.