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Math Help - Find eigenvalues and eigenvector

  1. #1
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    Find eigenvalues and eigenvector

    I am to find the eigenvalues and corresponding eigenvectors of the following matrix:

    [3 -1 -1]
    [-12 0 5]
    [4 -2 -1]

    Here is what I have so far:
    det(I - A) = det [-3 1 1], [12 -5], [-4 2 +1] = (-3)()(+1)


    I know that this is not completely correct, but not sure how to finish. Thanks in advance.
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  2. #2
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    Re: Find eigenvalues and eigenvector

    your determinant does not look correct. i get you should have:

    (λ-3)(λ)(λ+1) + (1)(-5)(-4) + (1)(12)(2) - (-4)(λ)(1) - (-5)(2)(λ-3) - (λ+1)(1)(12)

    = λ^3 - 2λ^2 - 3λ + 20 + 24 + 4λ + 10λ - 30 - 12λ - 12

    = λ^3 - 2λ^2 - λ + 2
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  3. #3
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    Re: Find eigenvalues and eigenvector

    Quote Originally Posted by page929 View Post
    I am to find the eigenvalues and corresponding eigenvectors of the following matrix:

    [3 -1 -1]
    [-12 0 5]
    [4 -2 -1]

    Here is what I have so far:
    det(I - A) = det [-3 1 1], [12 -5], [-4 2 +1] = (-3)()(+1)


    I know that this is not completely correct, but not sure how to finish. Thanks in advance.

    A=\begin{pmatrix} 3& -1 & -1\\  -12& 0 & 5\\  4& -2 & -1\end{pmatrix}

    First we will find the characteristic polynomial- \Delta (t)

    \Delta (t)=det(tI-A)=det\begin{pmatrix} t-3& -1 & -1\\ -12& t & 5\\  4& -2 & t+1\end{pmatrix}=t^3-2t^2-t+2=(t-2)(t-1)(t+1)

    Thus, our eigenvalues are:

    \lambda_1=2

    \lambda_2=1

    \lambda_3=-1


    Now, I'll find eigenvector for eigenvalue - \lambda_1=2 (And you will do the rest...)

    M:= \begin{pmatrix} 3& -1 & -1\\ -12& 0 & 5\\ 4& -2 & -1\end{pmatrix}-I\lambda_1= \begin{pmatrix} 3& -1 & -1\\  -12& 0 & 5\\  4& -2 & -1\end{pmatrix}-2I=\begin{pmatrix} 3-2& -1 & -1\\  -12& 0-2 & 5\\ 4& -2 & -1-2\end{pmatrix}=\begin{pmatrix}1& -1 & -1\\  -12& -2 & 5\\ 4& -2 & -3\end{pmatrix}

    The eigenvectors that belongs to eigenvalue \lambda_1=2 creating the solution for homogeneous system MX=0, in other words:

    \begin{pmatrix}1& -1 & -1\\ -12& -2 & 5\\  4& -2 & -3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0

    Now, you just need to solve it...

    The solution (x_1,y_1,z_1) is your eigenvector that belongs to eigenvalue \lambda_1=2 .
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  4. #4
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    Re: Find eigenvalues and eigenvector

    λ=2: (λI-A)=0
    [(-1,1,1), (12,2,-5), (-4,2,3)] [x1, x2, x3] = 0
    -x1 + x2 + x3 = 0
    x1 = 2
    x2 = 1
    x3 = 1

    λ=1: (λI-A)=0
    [(-2, 1, 1), (12, 1, -5), (-4, 2, 2)] [x1, x2, x3] = 0
    -2x1 + x2 + x3 = 0
    x1 = 1
    x2 = 1
    x3 = 1

    λ=-1: (λI-A)=0
    [(-4, 1, 1), (12, -1, -5), (-4, 2, 0)] [x1. x2. x3] = 0
    -4x1 + x2 + x3 = 0
    x1 = 1
    x2 = 2
    x3 = 2

    Are these eigenvector correct?
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  5. #5
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    Re: Find eigenvalues and eigenvector

    I alos need help on finding the eigenvector and want some clarity.

    Are you suppose to just multiply the 3x3 and the 3x1 matrix to get a 3x1 matrix or are you suppose to take the parametric vector form?

    Also shouldn't the solution be (x1, y1, z1)^T?

    ANd what is x1, y1, and z1? Are they our respective eigenvalues?

    And what do you do when there is a repeated eigenvalue? Does that mean there will be two of the same eigenvectors?
    Last edited by Reefer; November 26th 2011 at 04:21 PM.
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