Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 8 '13 at 9:14
3  
@MattN. It has determinant $-1$. –  Cocopuffs Apr 8 '13 at 9:21
2  
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 8 '13 at 9:25
    
@MattN. That's a reflection. It fixes an entire great circle of the unit sphere. –  Chris Eagle Apr 8 '13 at 9:36
    
Duh. Thanks. Had a moment of idiocy there. –  Rudy the Reindeer Apr 8 '13 at 9:44

2 Answers 2

up vote 1 down vote accepted
  1. 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.
  2. 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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.