TITLE: How many connected components does $\mathrm{GL}_n(\mathbb R)$ have?
QUESTION [76 upvotes]: I've noticed that $\mathrm{GL}_n(\mathbb R)$ is not a connected space, because if it were $\det(\mathrm{GL}_n(\mathbb R))$ (where $\det$ is the function ascribing to each $n\times n$ matrix its determinant) would be a connected space too, since $\det$ is a continuous function. But $\det(\mathrm{GL}_n(\mathbb R))=\mathbb R\setminus\{0\},$ so not connected.
I started thinking if I could prove that $\det^{-1}((-\infty,0))$ and $\det^{-1}((0,\infty))$ are connected. But I don't know how to prove that. I'm reading my notes from the topology course I took last year and I see nothing about proving connectedness...

REPLY [2 votes]: We will show that $G:=Gl_n({\bf R})^+$, set of matrices of positive
determinant, is path connected.
If $A=[a_1\cdots a_n]\in G$, consider $H:={\bf R}^{n-1}$ which is
spanned by $\{ a_i\}_{i=2}^n$ Here there exists some $v_1=e_i$ or
$-e_i$ s.t.
$$ \angle (v_1,a_1) < \pi $$ where $e_i$ is a canonical basis. Clearly there exists
a path, which do not change sign of determinant, so that we have $$
A_1:=[ v_1\ a_2\cdots a_n]  $$
By repeating this process we have $A_n\in SO(n)$. That is if $SO(n)$
is connected, then we are done
Note the fact that $SO(n)/SO(n-1)=S^{n-1}$. Since $SO(2)$ is
connected, by induction, we complete the proof.