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Math Help - Finding eigenvectors for a matrix

  1. #1
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    Finding eigenvectors for a matrix

    Need some help with this, the book is written in a very confusing way. I'm following the examples exactly and not getting the correct result.

    The matrix A = [2,-1 ; -1,2]

    I found the two eigenvalues, to be λ = 1 and λ = 3

    Now when I do the A - λI calculation and reduce the matrix I get

    for λ = 1

    x1 + x2 = 0 --> x1 = -x2

    however, the solution in the book says the answer is

    [1,1]

    for λ = 3

    -x1 + x2 = 0 --> x1 = x2

    but again, the solution says

    [-1, 1]



    Can someone please clarify or point out what I am doing wrong?

    Thanks.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Finding eigenvectors for a matrix

    So you solved the characteristic equation to get  \lambda = 3 and  \lambda = 1 . I got the same eigen values.

    Now to find the eigen vectors

     \begin{bmatrix}2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix} in which you get to solve a homogenous equation  \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} . Which yields infinetly many solutions of the form  \begin{bmatrix} -p \\ p \end{bmatrix} So one eigen vector for the eigen value  \lambda = 3 would be p = 1,  \begin{bmatrix} -1 \\ 1 \end{bmatrix}

    For  \lambda = 1 i got eigen vectors using the same processing above, infinitely many solutions of the form \begin{bmatrix} p \\ p \end{bmatrix} . So p = 1,  \begin{bmatrix} 1 \\ 1 \end{bmatrix} is a eigen vector for eigen value  \lamba = 1
    Last edited by jakncoke; November 25th 2012 at 05:14 PM.
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    Re: Finding eigenvectors for a matrix

    Quote Originally Posted by jakncoke View Post
    So you solved the characteristic equation to get  \lambda = 3 and  \lambda = 1 . I got the same eigen values.

    Now to find the eigen vectors

     \begin{bmatrix}2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix} in which you get to solve a homogenous equation  \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} . Which yields infinetly many solutions of the form  \begin{bmatrix} -p \\ p \end{bmatrix} So one eigen vector for the eigen value  \lambda = 3 would be p = 1,  \begin{bmatrix} -1 \\ 1 \end{bmatrix}

    For  \lambda = 1 i got eigen vectors using the same processing above, infinitely many solutions of the form \begin{bmatrix} p \\ p \end{bmatrix} . So p = 1,  \begin{bmatrix} 1 \\ 1 \end{bmatrix} is a eigen vector for eigen value  \lamba = 1
    I appologize, the matrix A is actually [2,1 ; 1,2]

    could you check it once again with the right matrix
    Last edited by zdravkoBG; November 25th 2012 at 05:35 PM. Reason: Typo
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Finding eigenvectors for a matrix

    Quote Originally Posted by zdravkoBG View Post
    I appologize, the matrix A is actually [2,1 ; 1,2]

    could you check it once again with the right matrix
    with this matrix the eigen vectors are merely exchanged.

    for  \lambda = 3 i got eigen vectors of the form  \begin{bmatrix} p \\ p \end{bmatrix} so for p = 1  \begin{bmatrix} 1 \\ 1 \end{bmatrix} is the one of the eigen vector for  \lambda = 3

    for  \lambda = 1 i got eigen vectors of the form  \begin{bmatrix} -p \\ p \end{bmatrix} so for p = 1  \begin{bmatrix} -1 \\ 1 \end{bmatrix} is the one of the eigen vector for  \lambda = 1
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    Re: Finding eigenvectors for a matrix

    Quote Originally Posted by jakncoke View Post
    with this matrix the eigen vectors are merely exchanged.

    for  \lambda = 3 i got eigen vectors of the form  \begin{bmatrix} p \\ p \end{bmatrix} so for p = 1  \begin{bmatrix} 1 \\ 1 \end{bmatrix} is the one of the eigen vector for  \lambda = 3

    for  \lambda = 1 i got eigen vectors of the form  \begin{bmatrix} -p \\ p \end{bmatrix} so for p = 1  \begin{bmatrix} -1 \\ 1 \end{bmatrix} is the one of the eigen vector for  \lambda = 1
    That's exactly what I thought the correct answer was. But in the back of the book with the answers, it states that for eigenvalue 3, the vector is [-1,1] and that for eigenvalue 1, the vector is [1,1].
    Could they have just switched them by accident? Because I am getting the same results as you are.
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: Finding eigenvectors for a matrix

    Yea, even books make mistakes
    Thanks from zdravkoBG
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