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Prove true or false eigenvalue question

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove true or false.

    If A^2+A=0 then λ=1 may not be an eigenvalue.


    2. Relevant equations

    To find the eigenvalues of A I find the solutions to det(λ-A).

    The definition of an eigenvalue from my understanding, AX = λX.

    A(A+I) = 0


    3. The attempt at a solution

    I'm unable to find the connection between the restriction of A^2+A=0 and its effects on the eigenvalues.

    Since A^2 + A =0 , I've thought of some matrices where this is true.

    Scenarios:

    A = 0

    A =\begin{pmatrix}
    -1\end{pmatrix}

    A = \begin{pmatrix}
    -1 & 0\\
    0 & -1\end{pmatrix}

    A = \begin{pmatrix}
    -1 & -1\\
    0 & 0\end{pmatrix}


    It seems that when you have -1 on the diagonal the first equation holds true and the eigenvalues are -1 or 0. With that being said, this statement appears true, but I have no idea how to find scenarios which could be counterexamples nor do I have a formal proof if the statement is indeed true.

    Clarification 1: Since I've only been able to find matrices with -1 (after reduction) on the diagonal to satisfy A^2 + A = 0, it appears that the eigenvalue will be always negative because when I find the characteristic polynomial its always in the form of (λ+1)^n which gives me a negative eigenvalue.

    Any help would be appreciated.
     
    Last edited: Nov 11, 2013
  2. jcsd
  3. Nov 11, 2013 #2

    tiny-tim

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    Science Advisor
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    (try using the X2 button just above the Reply box

    hi blockdummy! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)

    if 1 is an eigenvalue and x is the eigenvector, then (A2 + A)x = … ? :wink:
     
  4. Nov 11, 2013 #3
    Hmm...

    (A2 + A)x = (1)x

    Is it correct to assume that both sides have the same eigenvector?

    If so the x's cancel and... (A2 + A) = 1 which is a contradiction. This doesn't seem correct since the only way this wouldn't be a contradiction would be if λ=0.
     
  5. Nov 11, 2013 #4

    tiny-tim

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    hi blockdummy! :smile:
    nooo …

    try again :wink:
    how??

    you can't divide by a vector!!​
     
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