Fixes Point

Feb 2009
148
10
Chennai
Let \(\displaystyle T: \mathbb{R}^{3} \to \mathbb{R}^{3}\) be a linear transformation such that \(\displaystyle det \ T=1\). If T is not the identity linear transformation and \(\displaystyle T:S \to \mathbb{R}^{3}\) where \(\displaystyle S=\{ (x,y,z) \in \mathbb{R}^{3} | x^{2}+y^{2}+z^{2}=1\}\).Then prove that T fixes exactly 2 points of S.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Let \(\displaystyle T: \mathbb{R}^{3} \to \mathbb{R}^{3}\) be a linear transformation such that \(\displaystyle det \ T=1\). If T is not the identity linear transformation and \(\displaystyle T:S \to \mathbb{R}^{3}\) where \(\displaystyle S=\{ (x,y,z) \in \mathbb{R}^{3} | x^{2}+y^{2}+z^{2}=1\}\).Then prove that T fixes exactly 2 points of S.
I am confused here but maybe another member can clear it up. How can we have a det of a 3x1 column vector since T sends a vector in \(\displaystyle \mathbb{R}^3\) to a vector in \(\displaystyle \mathbb{R}^3\).
 
Nov 2009
485
184
I am confused here but maybe another member can clear it up. How can we have a det of a 3x1 column vector since T sends a vector in \(\displaystyle \mathbb{R}^3\) to a vector in \(\displaystyle \mathbb{R}^3\).
We are given the determinant of a linear transformation \(\displaystyle T\) - this has nothing to do with a column vector. In fact, if you were to write down the matrix for \(\displaystyle T\), it would be \(\displaystyle 3\times 3\).


I'd like to point out a key fact. A linear transformation \(\displaystyle T\) has a nonzero fixed point if and only if it has eigenvalue 1. Is it possible to show that this is the case, given that \(\displaystyle \det T=1\)?
 

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
Let \(\displaystyle T: \mathbb{R}^{3} \to \mathbb{R}^{3}\) be a linear transformation such that \(\displaystyle det \ T=1\). If T is not the identity linear transformation and \(\displaystyle T:S \to \mathbb{R}^{3}\) where \(\displaystyle S=\{ (x,y,z) \in \mathbb{R}^{3} | x^{2}+y^{2}+z^{2}=1\}\).Then prove that T fixes exactly 2 points of S.
Surely you mean that \(\displaystyle T(S)=S\), otherwise this is false, as shows \(\displaystyle T(a,b,c) = \left(\frac{a}{uv}, ub, vc\right)\), where \(\displaystyle u, v, uv\) are real numbers distinct from \(\displaystyle 0,1\).
 
Last edited:

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
Surely you mean that \(\displaystyle T(S)=S\), otherwise this is false, as \(\displaystyle T(a,b,c) = \left(\frac{a}{uv}, ub, vc\right)\), where \(\displaystyle u, v\) are real numbers distinct from \(\displaystyle 0,1\).
Here's a nice but somewhat technical way to prove this : let \(\displaystyle \sigma : \hat{\mathbb{C}}\rightarrow S\) be the usual stereographic projection; then \(\displaystyle f=\sigma T \sigma^{-1} : \hat{\mathbb{C}} \to \hat{\mathbb{C}}\) is an automorphism of \(\displaystyle \hat{\mathbb{C}}\), i.e. a Möbius transformation, which has one or two fixed points; but the fixed points of \(\displaystyle T\) come in pairs, so the case where \(\displaystyle f\) has only one fixed point is excluded.

(If we had \(\displaystyle \mbox{det }T=-1\), then all we could say is that \(\displaystyle f\) is antiholomorphic, i.e. \(\displaystyle f(\overline z)\) is conformal; for instance, reflection about a suitable plane passing through the origin in \(\displaystyle \mathbb{R}^3\) corresponds to \(\displaystyle z\mapsto \overline z\), which is not conformal, and which has infinitely many fixed points, all of which lie on a circle.)
 
Last edited: