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Good evening

I am looking at the general linear group $GL_n(\Bbb R)$ which is a subset of $\Bbb R^{n^2}$. I want to prove now that the identity component of $GL_n(\Bbb R)$ is $GL_n^+(\Bbb R)$ but I got stuck the show this. Can anybody helps me out? Thanks!

  • What is that you denote by $GL_n^+$? – R. Alexandre Mar 23 '20 at 18:47
  • I am sry, I mean with $GL_n^+(\Bbb R)$ all matrices with positive determinant. So this means $GL_n^+(\Bbb R) = {M(n \times n, \Bbb R): det(M)>0}$ –  Mar 23 '20 at 18:51
  • There is one easy inclusion given by the fact that a path can't change the sign of the determinant. Now, for the other inclusion, you can try to show that this is true locally and hence globally (think of how a connected group can be generated by a neighborhood of the origin) – R. Alexandre Mar 23 '20 at 18:55
  • Sorry but why can a path no change the sign of the determinant? –  Mar 23 '20 at 18:58
  • Hi Dexter, versions of this question have been asked before on SE. There are several proofs here although to finish these arguments just note that $I$ belongs to the component with positive determinant. Hopefully the answer below or one of those answers your question! – Andres Mejia Mar 23 '20 at 19:29
  • @Dexter Suppose there is a path $\gamma$ from $A$ to $B$ where $det(A) > 0$ and $det(B)<0$. Now consider $det\circ \gamma\colon [0,1] \to \mathbb{R}$. This path necessarily passes through $0$ (by the Intermediate Value Theorem!) or in other words $\gamma$ must pass through a matrix with determinant $0$. – William Mar 23 '20 at 19:30

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Certainly the identity component is contained in $GL_n^+$. To see this, note that the image of the continuous map $\operatorname{det}:GL_n \to \mathbb{R}$ has two connected components, namely the positive real numbers and the negative real numbers. Since the determinant of the identity is positive, the whole identity component must have positive determinant. To show that you get all of $GL_n^+$, you can try to write down a path from an arbitrary matrix in here to the identity through other positive determinant matrices. To solve this, you usually use some sort of matrix factorization technique. For instance, you can try looking at QR factorization.

Ethan Dlugie
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