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Math Help - Antisymmetric matrix.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Antisymmetric matrix.

    Let A be real antisymmetric matrix with size of nxn, when n is odd.

    Prove that 0 is eigenvalue of A.
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  2. #2
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let A be real antisymmetric matrix with size of nxn, when n is odd.

    Prove that 0 is eigenvalue of A.
    Hint: Show that \det(A) = -\det(A), so that \det(A) has to be 0.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Thanks, but how I continue from your hint please...?

    p(x)=det(Ix-A) =?
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  4. #4
    Senior Member roninpro's Avatar
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    The idea is that if a matrix has determinant 0, then it has 0 as an eigenvalue. Try to prove it if you can't see it.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    0 is an eigenvalue iff there exist vector v!=0 so A(v)=0v=0...

    Thank you all...
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  6. #6
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    The determinant of a matrix is the product of its eigenvalues. If the determinant is 0, then one of the eigenvalues must be 0.
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  7. #7
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    http://www.mathhelpforum.com/math-he...rexamples.html

    Post 3 shows that the det is the product of the eigen values.
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