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Math Help - [SOLVED] Eigenvalues

  1. #1
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    [SOLVED] Eigenvalues

    A nonzero vector can't correspond to two different eigenvalues of A.

    True, can only correspond if the eigenvalue is of multiplicity > 1.

    Correct?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by dwsmith View Post
    A nonzero vector can't correspond to two different eigenvalues of A.

    True, can only correspond if the eigenvalue is of multiplicity > 1.

    Correct?
    What do you mean by saying that a "non-zero vector x corresponds to an eigenvalue \lambda of A"? Du you mean that Ax=\lambda x holds (i.e. that x is an eigenvector of A for the eigenvalue \lambda)?

    And is what you wan't to show that, from x being non-zero and that Ax = \lambda_1 x and Ax=\lambda_2 x, it follows that \lambda_1=\lambda_2?
    If so, this has got nothing to do with multiplicities of those eigenvalues:
    Ax=\lambda_1 x=\lambda_2x\Leftrightarrow (\lambda_1-\lambda_2)x=0\Leftrightarrow \lambda_1-\lambda_2=0\Leftrightarrow \lambda_1=\lambda_2.
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  3. #3
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    Quote Originally Posted by Failure View Post
    What do you mean by saying that a "non-zero vector x corresponds to an eigenvalue \lambda of A"? Du you mean that Ax=\lambda x holds (i.e. that x is an eigenvector of A for the eigenvalue \lambda)?

    And is what you wan't to show that, from x being non-zero and that Ax = \lambda_1 x and Ax=\lambda_2 x, it follows that \lambda_1=\lambda_2?
    If so, this has got nothing to do with multiplicities of those eigenvalues:
    Ax=\lambda_1 x=\lambda_2x\Leftrightarrow (\lambda_1-\lambda_2)x=0\Leftrightarrow \lambda_1-\lambda_2=0\Leftrightarrow \lambda_1=\lambda_2.

    I mean exactly what it says. These questions from test I have taking this semester and I am just redoing them for the final. You getting all the information verbatim as it appears.
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  4. #4
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    Did you understand what Failure (he really needs to change that!) is saying? He showed you exactly how to answer your question.
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