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Prove that the set of all $n \times n$ orthogonal matrices is a compact subset of $\mathbb{R}^{n^2}$.

I don't know how it can be done. Thanks.

Rodrigo de Azevedo
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Pedro
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1 Answers1

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Let $M^{n\times n}$ be the set of all matrices and $\mathcal{O}$ the subset of orthogonal matrices. Define $f\colon M\to M,\;A\mapsto A^\intercal A$. Then $\mathcal{O}=f^{-1}(I)$, where $I$ is the identity matrix. Since $f$ is continuous, $\mathcal{O}$ is closed as a preimage of a singleton. Since $\|Qx\|=\|x\|$ for each orthogonal matrix $Q$, $\mathcal{O}$ lies in the 1-ball, i.e. is bounded. Heine-Borel says that $\mathcal{O}$ is compact. Since all norms in finite-dimensional vectorspaces are equivalent it doesn't matter that we used the operatornorm

Bananach
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  • Is there other norm that I can use to conclude that the set is bounded? – Pedro Jul 19 '13 at 05:15
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    @Perguntador Take the Euclidean norm. Being orthogonal means the columns (as well as the rows) form an orthonormal basis of $\mathbb{R}^n$. That means the Euclidean norm of an orthogonal $n\times n$ matrix is $\sqrt{n}$. – Daniel Fischer Jul 19 '13 at 08:41
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    @Perguntador: To phrase Daniel Fischer's comment even more geometrically, note that each column of an $n \times n$ real orthogonal matrix is a unit vector in $\mathbb{R}^n$. Consequently, the set of all such matrices is a subset of the $n$-fold product $S^n \times \dots \times S^n \subset \mathbb{R}^{n \times n}$. – Andrew D. Hwang Jul 19 '13 at 11:09
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    Alternative way to prove that $\mathcal{O}$ is bounded. Define the norm as $\|A\|= (\sum_{1\leq i,j \leq n}{a_{ij}}^2)^{1/2}$ If Q is an orthogonal matrix then $Q^\intercal Q = I$ implies $\sum_{k=1}^n {q_{ik}}^2 = 1$ for $i =1,...,n$ Hence $\|Q\| = n^{1/2}$ for all Q in $\mathcal{O}$. – tsknakamura Sep 26 '13 at 13:21