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Homework Help: Eigenvalue/Eigenvector problem. Check my solution please.

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose A is an n x n matrix with the property that A2=A
    a. Show that if λ is an eigenvalue of A, then λ=0 or λ=1
    b. Prove that A is diagonalizable.

    2. Relevant equations
    Av=λv (v : eigenvector)

    3. The attempt at a solution

    solution for a.
    A2v=A(Av)=A(λv)=λ(Av)=λλv. also, Av=λv, therefore, λ2v=λv
    => (λ2-λ)v=0. So, λ=0 or λ=1.

    I want you to check if this sounds right. If you see any errors, let me know. Now I'm working on the b part, but im pretty much stuck. I will post up when I find something. Thanks in advance
  2. jcsd
  3. Dec 5, 2011 #2


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    Looks good so far!
  4. Dec 5, 2011 #3
    Thanks for checking.

    I've been trying the second part, but have not find any satisfying answers.
    from the question, it looks like A is diagonalizable.

    But A is n-dimensional, but has only 2 distinct eigenvalues. then, A is not diagonaliable. am I right?
  5. Dec 5, 2011 #4


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    No, that's not true. The number of eigenvalues is irrelevant. What is relevant is the number of (independent) eigenvectors.

    A linear transformation, A, from V to itself, is "diagonalizable" if and only if there exist a basis for V consisting of eigenvectors of A. For an n by n matrix, that means if and only if there exist n independent eigenvectors.

    I don't know whether you can use this or not but if A is not diagonalizable, it can be put in Jordan Normal Form- there exist a part of the matrix of the form
    [tex]\begin{array}{cc} a & 1 \\ 0 & a\end{array}[/tex]
    where a is either 0 or 1 and they are on the diagonal.

    Show that [itex]A^2[/itex] cannot have that form.

    You should be able to show that the corresponding place on [itex]A^2[/itex] will be
    [tex]\begin{array}{cc} a^2 & 2a \\ 0 & a^2\end{array}[/tex]
    contradicting [itex]A^2= A[/itex].
    Last edited by a moderator: Dec 5, 2011
  6. Dec 5, 2011 #5
    Thanks HallsofIvy.

    I've already submited the homework, but I'm just curious now.

    the matrix you showed above is 2 x 2.

    what will be Jordan Normal Form in n x n matrix?

    upper triangle with 1s?
  7. Dec 5, 2011 #6

    Ray Vickson

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    If J is the Jordan form then A = P^(-1)JP, and J = PAP^(-1). Thus J^2 = PAP^(-1)PAP(-1) = PA^2P^(-1) = PAP^(-1) [because A^2=A], so J^2 = J. The only type of Jordan form in which this can happen is the diagonal one.

  8. Dec 5, 2011 #7


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    As HallsofIvy said, this is not true. For example, the matrix
    \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}has only two distinct eigenvalues, but you can find three independent eigenvectors, e.g. (1,0,0), (0,1,0), and (0,0,1). And obviously, it's diagonalizable because it's already diagonal.

    Jordan normal form looks like, for example,
    \begin{pmatrix} \lambda_1 & 1 & 0 & 0 \\ 0 & \lambda_1 & 1 & 0 \\ 0 & 0 & \lambda_1 & 0 \\ 0 & 0 & 0 & \lambda_2 \end{pmatrix}In each block corresponding to an eigenvalue, you have 1s just above the diagonal.
  9. Dec 6, 2011 #8
    Thank you for your replies.

    I have never studied the jordan normal form (at least not in my class.)

    But it looks like it has some role in eigenvalue/vector subject.

    Time to dig in!
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