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Math Help - Need some clarification on eigenvalues/eigenvectors

  1. #1
    Junior Member
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    Need some clarification on eigenvalues/eigenvectors

    i have a quick question about eigenvalue/eigenvectors

    so say i have a matrix

    -3 0
    2 -1

    with eigen values -1 and -3 (i'm pretty sure they are right)

    so for the first eigenvalue (-1) i get a matrix

    -2 0
    2 0

    and for the second (-3)

    0 0
    2 2

    well i not realy sure how to get the eigen vector from these?...wondering if someone could help me...this would really help, thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by action259 View Post
    i have a quick question about eigenvalue/eigenvectors

    so say i have a matrix

    -3 0
    2 -1

    with eigen values -1 and -3 (i'm pretty sure they are right)

    so for the first eigenvalue (-1) i get a matrix

    -2 0
    2 0

    and for the second (-3)

    0 0
    2 2

    well i not realy sure how to get the eigen vector from these?...wondering if someone could help me...this would really help, thanks
    You have the eigenvalues right, but what is this matrix associated with the eigenvalue? I've never heard of it. Anyway, to get the eigenvectors:

    \lambda = -1:

    Let the eigenvector be \left ( \begin{array}{c} a \\ b \end{array} \right )

    So:
    \left ( \begin{array}{cc} -3 & 0 \\ 2 & -1 \end{array} \right ) \left ( \begin{array}{c} a \\ b \end{array} \right ) = (-1) \left ( \begin{array}{c} a \\ b \end{array} \right )

    \left ( \begin{array}{c} -3a \\ 2a - b \end{array} \right ) = \left ( \begin{array}{c} -a \\ -b \end{array} \right )

    Thus -3a = -a or a = 0, and 2a - b = -b which says that 0 - b = -b, leaving b undetermined. So the eigenvector for \lambda = -1 is of the form
    \left ( \begin{array}{c} 0 \\ b \end{array} \right )
    where b is undetermined.

    You can use the same method to get the eigenvector for \lambda = -3.

    -Dan
    Last edited by topsquark; November 1st 2006 at 04:05 PM. Reason: Typo
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  3. #3
    Senior Member
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    The characteristic polynomial is your friend; you'll use it a lot. That is,

    Ax = (lambda)x --> (A - (lamdba)(I))x = 0, where I is the identity matrix.
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