Looking at many of the questions on this site and literature, there are many references made to the fact that rotations in ${\Bbb R}^n$ leave a $(n-2)$ dimensional subspace fixed.

I have looked for a proof of this statement, and couldn't find any.

This is as close as I got:

Let $R$ be a rotation in ${\Bbb R}^n$. If for some non-zero vector $v$, $Rv = v$, then there exists $(n-3)$ vectors $w_1, w_2, ...$ mutually orthogonal to themselves and $v$, such that $Rw=w$

Based on the question Rotation in 4D? But I failed to find a proof as I couldn't prove that the Eigenspace corresponding to the '$+1$' eigenvectors was $n-2$ dimensional.

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    $\begingroup$ What’s your definition of a rotation? $\endgroup$ – amd Dec 12 '17 at 19:16
  • $\begingroup$ In 4D rotation it can happen that there is no fixed vectors. $\endgroup$ – Widawensen Dec 13 '17 at 10:15
  • $\begingroup$ @amd a rotation is an orthogonal matrix R with det(R) = 1 $\endgroup$ – Adam Kelly Dec 13 '17 at 13:21
  • $\begingroup$ @Widawensen how is this? $\endgroup$ – Adam Kelly Dec 13 '17 at 13:26
  • $\begingroup$ @AdamKelly See en.wikipedia.org/wiki/… for details $\endgroup$ – Widawensen Dec 14 '17 at 7:11

The proposition is false. Take, for example, $R=\operatorname{diag}(-1,-1,-1,-1,1,\dots)$. Its determinant is $1$, so it’s a rotation, but the eigenspace of $1$ is obviously $(n-4)$-dimensional. (If you don’t like using what seems like a reflection, replace the first four rows with a pair of Givens rotations.) Indeed, a rotation in $\mathbb R^{2m}$ for $m\ge2$ need not have any non-trivial fixed points at all: a simple example is $-I$.

Now, you could define a simple rotation, similar to a Givens rotation, that only has a single plane of rotation, but then the proposition pretty much is the definition of the rotation itself.

There are some interesting things that one can say about planes of rotation and their relationship to the eigenvalues of a rotation, though. This Wikipedia article is a pretty good gloss on the subject.


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