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Math Help - diagonalizable 2x2 matrix

  1. #1
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    diagonalizable 2x2 matrix

    I need to find all real x values such that:
    (x,1
    1,0) is diagonalizable. I believe that x can be any real number however I am not sure how to check this.

    Anyone have an idea?
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  2. #2
    Senior Member roninpro's Avatar
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    Did you try looking for eigenvalues first?
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    Quote Originally Posted by hansard View Post
    I need to find all real x values such that:
    (x,1
    1,0) is diagonalizable. I believe that x can be any real number however I am not sure how to check this.

    Anyone have an idea?
    You need to identify eigenvalues that produce an eigenspace of 2.

    The det=-1 and tr=x; therefore, your characteristic polynomial is \lambda^2-x\lambda-1
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  4. #4
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    Quote Originally Posted by hansard View Post
    I need to find all real x values such that:
    (x,1
    1,0) is diagonalizable. I believe that x can be any real number however I am not sure how to check this.

    Anyone have an idea?
    \sqrt{b^2-4ac}\geq 0 We know a=1, c=-1

    \sqrt{b^2+4}\geq 0 So we are good in that aspect but are there values of x that will produce eigenvalues of multiplicity 2?
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    Quote Originally Posted by dwsmith View Post
    You need to identify eigenvalues that produce an eigenspace of 2.

    The det=-1 and tr=x; therefore, your characteristic polynomial is \lambda^2-x\lambda-1

    every number that I've tried so far, (eg. 1, -1, 5, -10) produces a quadratic characteristic eqn. that factors into 2 eigenvalues, so the eigenspace is 2.
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  6. #6
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    Quote Originally Posted by hansard View Post
    every number that I've tried so far, (eg. 1, -1, 5, -10) produces a quadratic characteristic eqn. that factors into 2 eigenvalues, so the eigenspace is 2.
    You can run into problems only if \lambda_1=\lambda_2 because then there is a chance of having only 1 eigenvector.
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    Quote Originally Posted by dwsmith View Post
    \sqrt{b^2-4ac}\geq 0 We know a=1, c=-1

    \sqrt{b^2+4}\geq 0 So we are good in that aspect but are there values of x that will produce eigenvalues of multiplicity 2?
    I couldn't find any eigenvalues with multiplicity 2. I think x can be any real number but I'm still not sure how to prove it.
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  8. #8
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    Quote Originally Posted by hansard View Post
    I couldn't find any eigenvalues with multiplicity 2. I think x can be any real number but I'm still not sure how to prove it.
    If x\in\mathbb{R}, then \sqrt{b^2-4ac}\geq 0.

    Now show that the quadratic cant be of the form (\lambda\pm y)^2 where y\in\mathbb{R}.
    Last edited by dwsmith; May 24th 2010 at 04:26 PM.
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