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Math Help - Fixes Point

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    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.
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    Quote Originally Posted by Chandru1 View Post
    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.
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    Senior Member roninpro's Avatar
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    Quote Originally Posted by dwsmith View Post
    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?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    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.
    Last edited by Bruno J.; May 28th 2010 at 11:54 AM.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    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.)
    Last edited by Bruno J.; May 28th 2010 at 09:44 AM.
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