Invertible Matrices as a Manifold

Asked Apr 10 '15 at 05:13

Active Apr 10 '15 at 05:34

Viewed 358 times

I've been trying to prove that the set of all invertible $n \times n$ matrices is a differentiable manifold. My attempt is as follows:

Define a map $\alpha : X \to XX^{-1} - I$
I take the inverse: $\alpha^{-1}(0)$ which would map to the set of invertible $n \times n$ matrices. I believe this suffices to show it is a manifold, but I am stumped in finding it's dimension. How can I approach this?

edited Apr 10 '15 at 05:34

asked Apr 10 '15 at 05:13

Stefan G.

What is your domain of $\alpha$? And this map is just constant.... – Apr 10 '15 at 05:35
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Hint: first try to prove that $\Bbb{R} \backslash {0}$ is a manifold without treating the two components separately. (This is the $n=1$ case of your proposition.) – Sammy Black Apr 10 '15 at 05:41
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This has already been done here. – Mnifldz Apr 10 '15 at 05:43
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Sorry about the duplicate, and thanks for your hints. It turns out that with a bit of sleep I was able to get it :) – Stefan G. Apr 10 '15 at 14:26

Invertible Matrices as a Manifold

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