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Trying to understand Eigenvectors

  1. May 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the Eigenvalues of A=

    4 0 1
    -2 1 0
    -2 0 1

    Then find the eigenvectors corresponding to each of the eigenvalues.



    2. Relevant equations



    3. The attempt at a solution

    I found the Characteristic Polynomial of the matrix, computed the Eigenvalues which are 1,2,3.

    What I'm trying to get my head around is the concept of the eigenvectors.

    First of all I attempted to find the eigenvector(s) for λ=1. So I constructed the matrix (A-Iλ), row-reduced and got the matrix:

    1 0 0
    0 0 1
    0 0 0

    This matrix corresponds to the set of linear eqns (A-Iλ)x, and x must be non-zero. So normally I'd just read the solutions from this matrix and tell myself

    x1=0
    and
    x3=0

    I did this in maple and it gave me the value (0,1,0) as the eigenvector corresponding to λ=1, but x2 doesn't equal zero in any of these rows. Can someone explain this to me?
     
  2. jcsd
  3. May 4, 2014 #2

    CAF123

    User Avatar
    Gold Member

    There is no constraint on x2, so it is arbitrary. 1 is chosen so that the eigenvector is properly normalized. The eigenspace corresponding to eigenvalue ##\lambda =1## is then all vectors of the form ##\langle 0, k, 0\rangle##.
     
  4. May 4, 2014 #3
    Ok, so what you're saying is that x1 has to be zero and x3 has to be zero but x2 can be anything?

    I think I get it, I think I'll just need to practice more questions like this. It's a bit subtle.
     
  5. May 4, 2014 #4

    Mark44

    Staff: Mentor

    It might be helpful to put some words with the work you did.

    When you were finding the eigenvectors for λ = 1, you ended up with this matrix:
    $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

    Your work started with the matrix equation (A - 1I)x = 0. By row-reduction, you got to the matrix I show above.

    This matrix represents the linear system
    x1 + 0x2 + 0x3 = 0
    0x1 + 0x2 + x3 = 0
    0x1 + 0x2 + 0x3 = 0

    Or more simply,
    x1 = 0
    x3 = 0

    Since there are no conditions or restrictions on x2, you can substitute any real value for x2. This means that any vector of the form <0, k, 0> is an eigenvector for λ = 1. For convenience's sake, k = 1 is chosen, making the eigenvector <0, 1, 0>. Any multiple of that vector would also be an eigenvector for this eigenvalue.
     
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