Show that the set of all orthogonal matrices in the set of all $n \times n$ matrices endowed with any norm topology is compact.
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2The column vectors of an orthogonal matrix are unit vectors. And there are $n$ column vectors. – NebulousReveal Jan 23 '11 at 21:28
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It would also be expeditious to use the operator norm. What is the operator norm of an orthogonal matrix? – hardmath Jan 23 '11 at 21:42
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Recall a compact subset of $R^{n \times n}$ is a set that is closed and bounded. One way to show closedness is to observe that the orthogonal matrices are the inverse image of the element $I$ under the continuous map $M \rightarrow MM^T$. Boundedness follows for example from the fact that each column or row is a vector of magnitude $1$.
Zarrax
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Each entry must be of absolute value at most 1, since the column it is in has magnitude 1, for example. – Zarrax Dec 12 '14 at 05:02
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1Why is the map continuous? Can you explain it a little bit more? I am thinking of a clear and concise way to prove it. (I have background in functional analysis.) – Oscar LIU Apr 22 '17 at 10:42
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The map is polynomial in the entries of $A$. Also, a function which is continuous in each coordinate (if the space has the product topology, which it does) is continuous. – Nerif Oct 24 '22 at 09:47