1. ## Fixes Point

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.

2. Originally Posted by 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.
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 $\displaystyle \mathbb{R}^3$.

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 $\displaystyle \mathbb{R}^3$ 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 $\displaystyle \det T=1$?

4. Originally Posted by 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.
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 $\displaystyle 0,1$.

5. Originally Posted by Bruno J.
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.)