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Eigenvalues, eigenvectors, and eigenspaces

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Screen_shot_2012_02_26_at_3_59_09_PM.png
    3. The attempt at a solution
    T(1,0,0) = (3,-1,0)
    T(0,1,0) = (0,1,0)
    T(0,0,1) = (-1,2,4)

    Thus, we have the matrix,

    [itex]\left| \begin{array}{ccc}
    3 &0&-1 \\
    -1&1&2 \\
    0&0&4 \end{array} \right|[/itex]

    [itex]Δ_T (t) = det( \left| \begin{array}{ccc}
    3 &0&-1 \\
    -1&1&2 \\
    0&0&4 \end{array} \right| - tI)[/itex]

    I have this equaling: -(t-4)(t-3)(t-1), which is the characteristic polynomial. The roots are the eigenvalues, which are 4,3,1.

    To compute the eigenvectors:

    When t=4, we have,
    -x-z=0
    -x-3y+2z=0
    0z=0

    Which implies that eigenvectors are multiples of (-1,1,1).

    When t=3, we have,
    -z=0
    -x-2y=0
    z=0

    Which implies that eigenvectors are multiples of (-1,2,0)

    When t=1, we have,
    2x-z=0
    -x+2z=0
    3z=0

    Which implies that eigenvectors are multiples of (0,1,0).

    T is diagonizable because (-1,1,1),(-1,2,0),(0,1,0) are lin. indep.

    First of all, are these correct? Also, how does one determine eigenspaces? That's the part I feel that I really don't know.
     
  2. jcsd
  3. Feb 27, 2012 #2

    lanedance

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    to see if they're correct, test by multiplying your eignevectors by the matrix - do they behave as expected?
     
  4. Feb 27, 2012 #3

    lanedance

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    as for the eigenspaces, they will be the span of all eignevectors corresponding to a single eigenvalue

    what does the span of a single vector look like?
     
  5. Feb 27, 2012 #4
    Eigenspace are the eigenvectors. I obtained a different eigenvector for you second one. I don't believe I made a mistake but I could have.
     
  6. Feb 27, 2012 #5
    Okay. I have the equations

    -z=0
    -x-2y=0
    z=0

    So clearly z=0 and x=-2y. Thus, is the eigenvector multiples of (1,1/-2,0)? Or, equivalently, multiples of (-2,1,0)?

    So the span of a single vector is just a line, no? So, for example, considering the eigenvalue of 4, is the eigenspace just: span(-1,1,1)?

    For eigenvalue of 1: span(0,1,0)?

    If by behave as expected you mean give back the zero vector, then yes (after I made an adjustment to the second eigenvector calculation). If this is the expected result, why do I expect this matrix multiplication to give back a column of all zeros?
     
  7. Feb 27, 2012 #6

    lanedance

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    The line is correct, you coud parameterise to describe it if need be

    Are you multiplying by [itex] T [/itex] or [itex] (T - \mathbb{I} \lambda)[/itex]?

    Say your eignevector is [itex] u [/itex] multply by [itex] T [/itex] you should get the original vector multiplied by the corresponding eigenvalue [itex] Tu =\lambda_u u[/itex]
     
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