1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.


    A = 0

    A =\begin{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


    User Avatar
    Science Advisor
    Homework Helper

    (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

    (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


    User Avatar
    Science Advisor
    Homework Helper

    hi blockdummy! :smile:
    nooo …

    try again :wink:

    you can't divide by a vector!!​
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted