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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 $M:=\mathbb{R}^{n\times n}$ 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 vector spaces are equivalent it doesn't matter that we used the operator norm.

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
  • Another norm would be for example $||A||:=\sum_{i,j=1}^n|A_{ij}|$. For all $O\in\mathcal O$ one can see immediately $O_{ij}^2\leq\sum_{k=1}^nO_{kj}^2 = (O^TO)_{jj} = 1$ implying that $||O||\leq n^2$. – Mathematics enthusiast Jul 10 '23 at 00:20
  • why is $f$ continuous ? Is this obvious ? I do not seem to see this immediately ... – Snowflake Mar 26 '24 at 18:39
  • Because all components are polynomials (quadratics to be precise) of the coordinates – Bananach Mar 27 '24 at 10:13