# Fixes Point

#### Chandru1

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
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 $$\displaystyle \mathbb{R}^3$$.

#### roninpro

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
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 $$\displaystyle 0,1$$.

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#### Bruno J.

MHF Hall of Honor
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.)

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