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Math Help - Eigenvectors correspondig to eigenvalues

  1. #1
    Junior Member Greg98's Avatar
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    Eigenvectors correspondig to eigenvalues

    Hello,
    the task is following:
    Let's assume that vector \bar v is eigenvector corresponding to eigenvalue \lambda_{A} of matrix A. Also \bar v corresponds to eigenvalue \lambda_{B} of matrix B. Now the questions:

    I Is the \bar v eigenvector of matrix A+B?
    II Is the \bar v eigenvector of matrix AB?

    If answer is yes to atleast one question, it must be shown, to which eigenvalue vector \bar v corresponds.

    I think I should use following theorem:
    "If \bar v is eigenvector corresponding to eigenvalue of matrix A, then \bar v is is eigenvector corresponding to A^k's \lambda^k".

    But somehow, I can't get anywhere, though I inspected eigenvalues and -vectors definitions closely. Also it's no problem to me calculate those vectors and values in normal cases. So, any help is appreciated. Thank you!
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  2. #2
    A Plied Mathematician
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    Try computing the quantities

    (A+B)\bar{v} and AB\bar{v}. What do you get?
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  3. #3
    Junior Member Greg98's Avatar
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    Hmm. I got:
    (A+B)\bar{v}
    =\bar{v}A+\bar{v}B
    =\bar{v}\lambda_{A}+\bar{v}\lambda_{B}

    So, I think \bar v corresponds to \lambda_{A} and \lambda_{B}, in this case. If I calculated it right. Second:

    \bar{v}(AB)
    =(\bar{v}A)B
    =A(\bar{v}B)
    =(\bar{v}\lambda_{A})\lambda_{B}
    =\lambda_{A}(\bar{v}\lambda_{B})

    Which means that, it isn't eigenvector of AB. Does this make any sense or Am I completely lost? Thanks for your help again! It's been very useful.
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  4. #4
    A Plied Mathematician
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    I am so not used to seeing leftward-looking operators. I can see that you're on the right track, though. Try this on for size:

    (A+B)\bar{v}=A\bar{v}+B\bar{v}=\lambda_{A}\bar{v}+  \lambda_{B}\bar{v}=(\lambda_{A}+\lambda_{B})\bar{v  }. What does that tell you?

    And, we also have

    AB\bar{v}=A(\lambda_{B}\bar{v})=\lambda_{B}A\bar{v  }=\lambda_{B}\lambda_{A}\bar{v}=(\lambda_{A}\lambd  a_{B})\bar{v}.

    What can you conclude?
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  5. #5
    Junior Member Greg98's Avatar
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    Revisiting following basic definition:
    "If A\bar{v}=\lambda\bar{v}, then \lambda is eigenvalue of A and \bar{v} eigenvector corresponding to \lambda",

    I would like to conclude, that \bar{v} is eigenvector of both (which is obviously wrong). This also should say something to me, but...
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  6. #6
    A Plied Mathematician
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    Right. You need to have the nonzero condition on the eigenvector in there as well. Why can't \bar{v} be an eigenvector of both A+B and AB? Haven't I basically just proven that it is?
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  7. #7
    Junior Member Greg98's Avatar
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    Yeah, you have it correct. You've been very kind again, so word of thanks for you. I tend to make wrong guesses quite often, so I wasn't serious. \bar{v} can be eigenvector of both, Q.E.D.
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  8. #8
    A Plied Mathematician
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    You're welcome. Have a good one!
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