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Math Help - Help with a proof - eigenvalues

  1. #1
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    Help with a proof - eigenvalues

    let \varphi :V\rightarrow V be a linear transformation.
    let u be eigenvector of \varphi with the eigenvalue \lambda
    let v be eigenvector of \varphi with the eigenvalue \mu
    I need to prove that if u+v is an eigenvector of \varphi, then \lambda=\mu

    this is what I tried:
    \varphi(u+v)=a(u+v)=au+av
    on the other hand:
    \varphi(u)+\varphi(v)=\lambda u+\mu v

    but how can I conclude that a=\lambda and a=\mu?
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  2. #2
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    Re: Help with a proof - eigenvalues

    φ(u+v) = λu + μv = λ(u+v)
    (λ- λ)u + (μ- λ)v = 0
    u and v are linearly independent so λ = λ , μ = λ
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  3. #3
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    Re: Help with a proof - eigenvalues

    Why are they linear independent?
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  4. #4
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    Re: Help with a proof - eigenvalues

    Quote Originally Posted by Stormey View Post
    Why are they linear independent?
    Suppose \lambda \ne \mu and u and v are linearly dependent.

    Then there is some t \ne 0 such that v=tu
    In that case \phi v = \phi (tu) = t (\phi u) = t (\lambda u) = \lambda (tu) = \lambda v, which is a contradiction (why?).

    In other words, either \lambda = \mu, or u and v are linearly independent.
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    Re: Help with a proof - eigenvalues

    Quote Originally Posted by Stormey View Post
    Why are they linear independent?
    If λ =μ (starting assumption) the eigen vectors are linearly independent. You'll have to look it up in a linear algebra book or on-line.

    = means not equal
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    Re: Help with a proof - eigenvalues

    Thanks for the help.

    I know that vectors of different eigenvalues are linear independent, you don't need to convince me about that.
    the thing is: it's not said that  \lambda \neq \mu.
    it's only said that u+v is a eigenvector.
    are you saying that I need to assume that \lambda \neq \mu and show that it leads to a contradiction?
    (just want to wrap my head around it)
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Help with a proof - eigenvalues

    Quote Originally Posted by Stormey View Post
    Thanks for the help.

    I know that vectors of different eigenvalues are linear independent, you don't need to convince me about that.
    the thing is: it's not said that  \lambda \neq \mu.
    it's only said that u+v is a eigenvector.
    are you saying that I need to assume that \lambda \neq \mu and show that it leads to a contradiction?
    (just want to wrap my head around it)
    This is where a proof by contradiction comes in.

    Suppose u+v is an eigenvector of \phi, but \lambda \ne \mu ...

    If it leads to a contradiction, then we must conclude that if u+v is an eigenvector of \phi, then \lambda = \mu.
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  8. #8
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    Re: Help with a proof - eigenvalues

    Great.
    Thank you.
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