# connected component of $\rm{GL_n}(\mathbb{R})$ [duplicate]

Possible Duplicate:
How many connected components does $\mathrm{GL}_n(\mathbb R)$ have?

I know $\rm{GL_n}(\mathbb{R})$ is not connected and connected components are $C_1 =\{A:\rm{det}\ A>0\}$ and $C_2=\{A:\rm{det} \ A<0\}$.

Given that $C_1= \rm{det}^{-1}(0,\infty)$ and $C_2=\rm{det}^{-1}(-\infty,0)$, $\rm{det}:M_n(\mathbb{R})\to \mathbb{R}$.

But how can one prove that $C_1$ and $C_2$ are connected in $\rm{M_n}(\mathbb{R})$?

Thank you.

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Think of $M_n(\mathbb{R})$ as $\mathbb{R}^{n\times n}$. –  M.B. Jul 17 '12 at 9:45
@Davide: Sorry, I hadn't refreshed to see your duplicate before I posted the answer. I've now voted to close as duplicate. –  joriki Jul 17 '12 at 9:56
could any one explain me in a little detail about the point (ii) of the fisrt answer of the question? –  Bunuelian Trick Jul 18 '12 at 12:47
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## marked as duplicate by Davide Giraudo, joriki, Zhen Lin, t.b., Amitesh DattaJul 17 '12 at 11:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

Given two invertible matrices, consider their singular value decompositions. We can continuously change their (non-zero) singular values to $1$. Each orthogonal matrix is either a rotation or can be written as a rotation times a reflection in, say, the first component. The reflections cancel if there are two of them. There remains a single reflection if the original matrix has negative determinant, so if the two matrices have determinants of the same sign, they now either both have or both don't have one reflection. It remains only to continuously transform the rotations to the identity. Now both matrices have been continuously transformed either to the identity or to the same reflection. Thus each component is path-connected, and hence connected.

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