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Math Help - No real eigenvalues

  1. #1
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    Question No real eigenvalues

    Show that the following matrix has no real eignvalues, and thus no eigenvectors. Interpret your result geometrically.
    [0 -1]
    [-1 0]
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  2. #2
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    a^2-1=0 where a=eigen value.
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  3. #3
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    I am still not sure what this means.
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  4. #4
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    i get 1 and -1 as the eigenvalues for your matrix and these are real eigenvalues.
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  5. #5
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    Are you sure you have written the matrix correctly?
    Charikar is right. The matrix has real eigenvalues and eigenvectors. The matrix is a reflection - the eigenvectors are the bisectors of the axes.
    Your question would make sense if the matrix had been
    [0 -1]
    [1 0]
    as that would have been a rotation, which does not have real eigenvectors.
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  6. #6
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    Yes I figured it out and got the same thing as Charikar. I did have the wrong matrix. I get x^2+1 as the determinant. I don't know where to go from there at all.
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  7. #7
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    Quote Originally Posted by lm6485 View Post
    Yes I figured it out and got the same thing as Charikar. I did have the wrong matrix. I get x^2+1 as the determinant. I don't know where to go from there at all.
    If det(A-\lambda I) = f(\lambda) for some function f, then the eigenvalues of A are the roots of f(\lambda). If it only has solutions in F = \mathbb{C} - \mathbb{R} then its eigenvalues are not real, therefore it has no eigenvectors as your question described.

    In this case, f(\lambda) = \lambda^2 + 1.

    Can you find the roots of f?
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  8. #8
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    sqrt of -1?
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  9. #9
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    Yes, And therefore the eigenvalues are ?
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