# Proof that n-Dimensional Rotations Leave A (n-2) Subspace Fixed

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.

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