# Thread: Fixes Point

1. ## Fixes Point

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

2. Originally Posted by Chandru1
Let $T: \mathbb{R}^{3} \to \mathbb{R}^{3}$ be a linear transformation such that $det \ T=1$. If T is not the identity linear transformation and $T:S \to \mathbb{R}^{3}$ where $S=\{ (x,y,z) \in \mathbb{R}^{3} | x^{2}+y^{2}+z^{2}=1\}$.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 $\mathbb{R}^3$ to a vector in $\mathbb{R}^3$.

3. Originally Posted by dwsmith
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 $\mathbb{R}^3$ to a vector in $\mathbb{R}^3$.
We are given the determinant of a linear transformation $T$ - this has nothing to do with a column vector. In fact, if you were to write down the matrix for $T$, it would be $3\times 3$.

I'd like to point out a key fact. A linear transformation $T$ 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 $\det T=1$?

4. Originally Posted by Chandru1
Let $T: \mathbb{R}^{3} \to \mathbb{R}^{3}$ be a linear transformation such that $det \ T=1$. If T is not the identity linear transformation and $T:S \to \mathbb{R}^{3}$ where $S=\{ (x,y,z) \in \mathbb{R}^{3} | x^{2}+y^{2}+z^{2}=1\}$.Then prove that T fixes exactly 2 points of S.
Surely you mean that $T(S)=S$, otherwise this is false, as shows $T(a,b,c) = \left(\frac{a}{uv}, ub, vc\right)$, where $u, v, uv$ are real numbers distinct from $0,1$.

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

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