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Let $M_n(\mathbb{R})$ be the set of real $n\times n$ matrices. I've proved that the map $\left \|\cdot \right \| \mapsto \left \| A \right \| :=\sqrt{\text{tr}(A^tA)}$ is a norm. Then I defined the metric as $d(A,B):=\left \| A-B \right \|$. Now, I want to prove that the orthogonal group $O_n(\mathbb{R})$ is an open set in these metric space:

Let $P\in M_n(\mathbb{R})$. We have to find an $\varepsilon>0$ such that $B(P,\varepsilon)\subset O_n(\mathbb{R})$. Then

\begin{equation*} \begin{split} B(P,\varepsilon)&=\left \{ Q\in O_n({\mathbb{R}})\ |\ \sqrt{\text{tr}(Q^tQ)-2\text{tr}(Q^tP)+\text{tr}(P^tP)}<\varepsilon \right \} \\ &=\left \{ Q\in O_n({\mathbb{R}})\ |\ \sqrt{\text{tr}(I_n)-2\text{tr}(Q^tP)+\text{tr}(I_n)}<\varepsilon \right \}\\ &=\left \{ Q\in O_n({\mathbb{R}})\ |\ \sqrt{n-2\text{tr}(Q^tP)+n}<\varepsilon \right \} \\ &= \left \{ Q\in O_n({\mathbb{R}})\ |\ \sqrt{2}\cdot\sqrt{1-\text{tr}(Q^tP)}<\varepsilon \right \} \end{split} \end{equation*} And now?

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  • $\begingroup$ it is quite obvious that in the neighborhood (for your norm) of an orthogonal matrix there is always a non-orthogonal matrix, no ? for example $P+\epsilon I$ is not orthogonal in general. but in general, if $M$ is inversible then $M+\epsilon A$ is inversible when $\epsilon$ is small enough, hence the set of inversible matrix is an open set. $\endgroup$ – reuns Apr 21 '16 at 19:16
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No. You have to look at a set of arbritray matrices in a neighbourhood of a given $P$ and show that any matrix in that set is orthogonal.

You won't succeed with this, though, since that set is actually not open but compact.

(It's bounded and closed as the counterimage of a point under a continuous map)

(Edit: Actually it is a closed smooth submanifold of lower dimension than $M_n$, so it does not even contain an interior point).

(2nd Edit: maybe you are mixing this up. The General Linear Group $GL_n$ of invertible matrices is a (dense) open subset of $M_n$).

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  • $\begingroup$ Right, thank you! Can you advise me some literature of this norm and something about manifolds? $\endgroup$ – user296708 Apr 21 '16 at 17:54
  • $\begingroup$ @Victor This norm is rather common but not very spectacular, it's just the one derived from the standard scalar product in $\mathbb{R}^{n\times n}$, written down using other terms. Any textbook on linear algebra should do. Introductory chapters on manifolds are found in many textbooks on analysis, if you want to have a deep dive you should consult textbooks on differential topology or differential geometry. I'm out of the business since 15 years, so I don't know which textbooks are preferred today. I learned these topics with Hirsch's book on DT and Spivak's comprehensive introduction to DG. $\endgroup$ – Thomas Apr 21 '16 at 18:02

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