I am looking to prove that the set of $2\times 2$ matrices with positive determinant is connected. I understand that the set of invertible $2\times 2$ matrices is however disconnected since the determinant can not equal zero.
I know That my aim is to prove that $GL^+ (2,\mathbb{R})$ cannot be written as a disjoint Union of two open and non-empty subsets. How do I go about this?
Thanks
The Gram-Schmidt process allows us to make a deformation retract from $GL^+(2,\mathbb{R})$ to $SO(2)$. Since $SO(2) \simeq S^1$, it follows that $SO(2)$ is path-connected and thus $GL^+(2,\mathbb{R})$ must also be.
This argument can be generalized to show that $GL^+(n,\mathbb{R})$ is connected, using the fact that $SO(n)$ is connected (which can be inferred, for instance, from the fact that the spheres are connected, but this is a separate argument).
In case you want an explicit path :
let $M \in GL^+(2,\mathbb{R}), M = \pmatrix{ a & b\\ c & d}$
My goal is to prove that any matrix of $GL^+(2,\mathbb{R})$ can be connected to $I$, i.e. that $\forall M \in GL^+(2,\mathbb{R}), \exists f(t) \in C^0, \forall t \in [0;1] f(t)\in GL^+(2,\mathbb{R}), f(0) = M, f(1)=I$
We know that $\det(M)=ad-cb>0$
So $ad > 0$ or $cb < 0$
We now have two case to consider :
let $M' = \pmatrix{ a & 0\\ 0 & d}$ and $f(t)=tM+(1-t)M'$
$M' \in GL^+(2,\mathbb{R})$ and $f(t) \in GL^+(2,\mathbb{R}) \forall t \in [0;1]$
1.1) If $a>0$ then $d>0$ and $M'$ is easily connected to $I$ through $g(t)=tM'+(1-t)I$
1.2) If $a<0$ then $d<0$ then let $\epsilon >0$ and $M_\epsilon=\pmatrix{ a & \epsilon\\ -\epsilon & d}$ which is connected to $M'$, $M_\epsilon$ can be connected to $I$ using the instruction in the section 2
let $M'=\pmatrix{ 0 & b\\ c & 0}\in GL^+(2,\mathbb{R})$ which is connected to $M$ through $f(t) = tM+(1-t)M'$
then consider $M''=\pmatrix{ 1 & b\\ c & 1}\in GL^+(2,\mathbb{R})$ which is connected to $M'$ through $g(t) = tM'+(1-t)M''$
In the end $M''$ is connected to $I$ through $h(t)=tI+(1-t)M''$
If $A$ is connected to $B$ and $B$ is connected to $C$ then $A$ is connected to $C$. Which means that any matrix can be connected to $I$. All in all, $GL^+(2,\mathbb{R})$ is connected
If $det(M)= det(N)$, then $M= SN$ for some $S\in SL(2,\mathbb{R})$.
In particular, given $M\in GL^+(2,\mathbb{R})$ with determinant $D$, $M = S \pmatrix{ 1 & 0\\ 0 & D}$ for some $S\in SL(2,\mathbb{R})$. As $D>0$, there is a path $\gamma$ in $\mathbb{R}_+^*$ connecting $D$ and $1$. Hence $t\mapsto S \pmatrix{ 1 & 0\\ 0 & \gamma(t)}$ is a path in $GL^+(2,\mathbb{R})$ connecting $M$ and $S$.
Now $SL(2,\mathbb{R})$ is spanned by transvection matrices $I_2 + \alpha E_{i,j}$ for $i\neq j$. Let's denote with $T_{i,j}(\alpha)$ such a matrix. Then clearly, $t\mapsto T_{i,j}(t\alpha)$ is a path in $SL(2,\mathbb{R})$ connecting $I_2$ and $T_{i,j}(\alpha)$. So any product of those matrices is connected to $I_2$ in $SL(2,\mathbb{R})$.
In particular, any matrix in $SL(2,\mathbb{R})$ is connected in $SL(2,\mathbb{R})(\subset GL^+(2,\mathbb{R}))$ to $I_2$. It's the case for $S$. Therefore $M$ is also connected to $I_2$. Therefore $GL^+(2,\mathbb{R})$ is path-connected, hence connected.
Of course this generalizes to $GL^+(n,\mathbb{R})$
