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Find eigenvalues and eigenvectors of weird matrix

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    find eigenvalues and eigenvectors for the following matrix

    |a 1 0|
    |1 a 1|
    |0 1 a|

    2. Relevant equations


    3. The attempt at a solution
    I'm trying to find eigenvalues, in doing so I've come to a dead end at 1 + (a^3 - lambda a^2 -2a^2 lambda + 2a lambda^2 + lambda^2 a - lambda^3 - a + lambda)

    This is because i have to put in a - lamnda across the board, then it gets tricky when trying to find eigenvalues with all these variables, can someone please help in this bizarre question.
     
  2. jcsd
  3. Oct 3, 2015 #2

    vela

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    The polynomial you came up with isn't correct. I suggest you resist the temptation to multiply everything out immediately. You should find that ##(a-\lambda)## is a factor.
     
  4. Oct 3, 2015 #3
    Ok so before i multiplied everything out, it came to:

    det|1 (a - lamnda)| + (a - lamnda)*det|(a - lamnda) 1 | = 0
    |0 1 | | 1 (a - lamnda)|

    which becomes 1 + (a - lamnda)((a - lamnda)(a - lamnda) - 1)

    I think i see now, you're right with the a - lamnda

    Thus a represents an eigenvalue, and is the only one then?
     
  5. Oct 3, 2015 #4
    The

    0 1 | | 1 (a - lamnda)|

    should be shifted accordingly to match the top row
     
  6. Oct 3, 2015 #5

    vela

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    I don't think your expansion is correct. To get matrices to appear correctly, use LaTeX. It's pretty easy to learn. If you reply to this post, you can see an example.
    $$\begin{vmatrix}
    a-\lambda & 1 & 0 \\
    1 & a-\lambda & 1 \\
    0 & 1 & a-\lambda
    \end{vmatrix}$$
     
  7. Oct 3, 2015 #6

    Ray Vickson

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    If you set ##b = a - \lambda## the determinant of ##A - \lambda I## becomes ##b^3 -2b##, so equating this to zero gives roots ##b = 0## and ##b = \pm \sqrt{2}##.
     
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