I’ve noticed that GLn(R) is not a connected space, because if it were det(GLn(R)) (where det is the function ascribing to each n×n matrix its determinant) would be a connected space too, since det is a continuous function. But det(GLn(R))=R∖{0}, so not connected.

I started thinking if I could prove that det−1((−∞,0)) and det−1((0,∞)) 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…

**Answer**

Your suspicion is correct, GLn has two components, and det may be used to show there are at least two of them. The other direction is slightly more involved and requires linear algebra rather than topology. Here is a sketch of how to do this:

i) If b is any vector, let Rb denote the reflection through the hyperplane perpendicular to b. These are all reflections. Any two reflections Ra,Rb with a,b linear independent can be joined by a path consisting of reflections, namely Rta+(1−t)b,t∈[0,1].

ii) Any X∈O+(n) (orthogonal matrices with positive determinant) is the product of an even number of reflections. Since matrix multiplication is continuous O(n)×O(n)→O(n) and by i) you can join any product RaRb with RaRa=Id it follows that O+(n) is connected.

iii) det shows O(n) is not connected.

iv) O−(n)=RO+(n) for any reflection R. Hence O−(n) is connected.

v) Any X∈GLn is the product AO of a positive matrix A and O∈O(n) (polar decomposition). Now you only need to show that the positive matrices are connected, which can be shown again using convex combination with Id. This proves the claim.

**Attribution***Source : Link , Question Author : Bartek , Answer Author : Community*