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Thread: Help with a proof - eigenvalues

  1. #1
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    Help with a proof - eigenvalues

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

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

    but how can I conclude that $\displaystyle a=\lambda$ and $\displaystyle a=\mu$?
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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 $\displaystyle \lambda \ne \mu$ and $\displaystyle u$ and $\displaystyle v$ are linearly dependent.

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

    In other words, either $\displaystyle \lambda = \mu$, or $\displaystyle u$ and $\displaystyle v$ 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$\displaystyle \lambda \neq \mu$.
    it's only said that $\displaystyle u+v$ is a eigenvector.
    are you saying that I need to assume that $\displaystyle \lambda \neq \mu$ and show that it leads to a contradiction?
    (just want to wrap my head around it)
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    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$\displaystyle \lambda \neq \mu$.
    it's only said that $\displaystyle u+v$ is a eigenvector.
    are you saying that I need to assume that $\displaystyle \lambda \neq \mu$ 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 $\displaystyle u+v$ is an eigenvector of $\displaystyle \phi$, but $\displaystyle \lambda \ne \mu$ ...

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

    Great.
    Thank you.
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