Proving $\chi$ fixes two points in the unit sphere.

Question

Let $\chi:\mathbb{R}^{3}\to\mathbb{R}^{3}$ be an orthogonal transformation such that $\det(\chi)=1$ and $\chi$ is not the identity linear transformation. Let $S \subset \mathbb{R}^{3}$, be the unit sphere. Then how do we prove that $\chi$ fixes only two points of $S$?

Any ideas of proceeding for the solution?

Do you mean "fixes no more than $2$ points"? Cause the matrix with rows $(010, 100, 001)$ is a matrix with det = 1 fixing only one point in the unit sphere... — Rudy the Reindeer, Apr 08 '13 at 09:14
Idea: such a $\chi$ is a rotation in some 2D subspace $V$ (a plane) of $\mathbb{R}^3$. The fixed points are the intersection of the normal to that plane with $S$. — Cocopuffs, Apr 08 '13 at 09:25
@MattN. That's a reflection. It fixes an entire great circle of the unit sphere. — Chris Eagle, Apr 08 '13 at 09:36

score 2 · Answer 1 · answered Apr 08 '13 at 09:28

2

An orthogonal transformation with $\det(1)$ is a rotation. The only fixed points are the one on the rotation axis. And the intersection of the rotation axis and the unit sphere are exacty two points.

answered Apr 08 '13 at 09:28

Dominic Michaelis

19,935

Branimir Ćaćić · Accepted Answer · 2013-04-08T10:30:02.247

First, convince yourself that $\operatorname{det}(\chi - I_3) = 0$ (hint: $\operatorname{det}(\chi - I_3) = \operatorname{det}(\chi(I_3 - \chi^T)) = \ldots$). Thus, $\dim \ker(\chi - I_3) \geq 1$, and hence $\ker(\chi - I_3)$ contains a $1$-dimensional subspace, i.e., a line through the origin, which intersects the unit sphere in two points $v_1$ and $v_2$. Thus, $\chi$ fixes at least two points.
Now, suppose by contradiction that $\chi$ fixes a third point $v_3 \in S^2$, which therefore cannot line on the same line through the origin as $v_1$ and $v_2 = -v_1$. Then $\{v_1,v_2\}$ span a $2$-dimensional subspace of $\ker(\chi - I_3)$ (why?), so that $\dim \ker(\chi - I_3) \geq 2$; if $\dim \ker(\chi - I_3) = 3$ then $\chi = I_3$, which is a contradiction, so suppose instead that $\dim \ker(\chi - I_3) = 2$. Now, let $x \in \ker(\chi - I_3)^\perp$ be non-zero, so that $x$ spans $\ker(\chi - I_3)^\perp$; check that$\chi x \in \ker(\chi - I_3)^T$ (hint: for $v \in \ker(\chi - I_3)$, $v = \chi^T \chi v = \chi^T v$). Hence, $\chi x = \lambda x$ for some $\lambda \in \mathbb{R}$, so that $\chi$ is diagonalisable with eigenvalues $1$ (with multiplicity $2$) and $\lambda$ (with multiplicity $1$). Why does this then give a contradiction?

What will come out of this is that $\ker(\chi - I_3)$ is indeed $1$-dimensional, giving the axis of rotation for $\chi$; one can then show that $\chi$ restricted to $\ker(\chi - I_3)^\perp \cong \mathbb{R}^2$ is indeed a non-trivial rotation.

Proving $\chi$ fixes two points in the unit sphere.

2 Answers2