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Math Help - Special orthogonal group

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    Special orthogonal group

    Let P be a matrix in SO_3(\mathbb{C}).
    Prove that if X ia an eigenvector with eigenvalue 1 and if P \ne I, then X^t X\ne 0.
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    Quote Originally Posted by KaKa View Post
    Let P be a matrix in SO_3(\mathbb{C}).
    Prove that if X ia an eigenvector with eigenvalue 1 and if P \ne I, then X^t X\ne 0.

    Perhaps I'm misunderstanding something here, but isn't X^tX just the inner (scalar) product of the vector X with itself, also known as ||X||^2 ?
    If so then it's obvious it can't be zero, and this much is true for ANY eigenvector of ANY operator/matrix, since ||X||=0\Longleftrightarrow X=0 , and zero cannot be an eigenvector...

    Tonio
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    Quote Originally Posted by tonio View Post
    Perhaps I'm misunderstanding something here, but isn't X^tX just the inner (scalar) product of the vector X with itself, also known as ||X||^2 ?
    If so then it's obvious it can't be zero, and this much is true for ANY eigenvector of ANY operator/matrix, since ||X||=0\Longleftrightarrow X=0 , and zero cannot be an eigenvector...

    Tonio
    Base field is \mathbb{C}!!!
    For example, if P=I, then any non zero column vector X can be eigenvector.
    If X=\begin{bmatrix}1\\i\\0\end{bmatrix},then X^tX=1-1+0=0
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    Quote Originally Posted by KaKa View Post
    Base field is \mathbb{C}!!!
    For example, if P=I, then any non zero column vector X can be eigenvector.
    If X=\begin{bmatrix}1\\i\\0\end{bmatrix},then X^tX=1-1+0=0

    Yeah, I totally missed the complex thing: I'm used to call a complex matrix unitary...but then you have a counterexample: the unity matrix I is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors X for which X^tX=0...! Thus, the claim is false

    Tonio
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    Quote Originally Posted by tonio View Post
    Yeah, I totally missed the complex thing: I'm used to call a complex matrix unitary...but then you have a counterexample: the unity matrix I is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors X for which X^tX=0...! Thus, the claim is false

    Tonio
    Quote Originally Posted by KaKa View Post
    Let P be a matrix in SO_3(\mathbb{C}).
    Prove that if X ia an eigenvector with eigenvalue 1 and if \bold{P \ne I}, then X^t X\ne 0.
    Last edited by Defunkt; December 1st 2009 at 05:55 PM.
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    Quote Originally Posted by KaKa View Post
    Let P be a matrix in SO_3(\mathbb{C}).
    Prove that if X ia an eigenvector with eigenvalue 1 and if P \ne I, then X^t X\ne 0.
    If P\in SO_3(\mathbb{C}) then the product of the three eigenvalues of P is \det (P) = 1. If one of the eigenvalues is 1 then the other two must be \lambda and \lambda^{-1}, where \lambda\in\mathbb{C} and \lambda \ne 1 because P\ne I.

    Let X,\;Y,\;Z be eigenvectors for the three eigenvalues. Then PX=X and so X = P^{-1}X = P^{\textsc t}X (because P is orthogonal). Thus X^{\textsc t}Y = (P^{\textsc t}X)^{\textsc t}Y = X^{\textsc t}PY = \lambda X^{\textsc t}Y. Since \lambda\ne1, it follows that X^{\textsc t}Y = 0. Similarly X^{\textsc t}Z = 0.

    If in addition X^{\textsc t}X = 0 then X^{\textsc t}W = 0 for all W\in\mathbb{C}^3 (because \{X,Y,Z\} is a basis for \mathbb{C}^3). In particular, X^{\textsc t}\overline X = 0, where \overline X is the vector whose coordinates are the complex conjugates of those of X. But that implies that X=0: contradiction. Therefore X^{\textsc t}X \ne 0.
    Last edited by Opalg; December 3rd 2009 at 03:26 AM. Reason: corrected error
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