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Math Help - Eigenvalues and Eigenvectors

  1. #1
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    Eigenvalues and Eigenvectors

    I'm struggling with finding the eigenvectors for this paritcular question . I understand how to compute eigenvalues and hence the corresponding eigenvectors but my problem seems to be with reducing the matrix for it to be more easily solvable .

    My eigenvalue yields the 3x3 matrix ;
    7 8 -7
    -2 -1 2
    2 4 -2

    From which I must solve to find the eigenvector corresponding to the eigenvalue of -2 .
    I tried to reduce the matrix to echelon form but reach a stage where I can solve the equation but the matrix hasn't been reduced fully to echelon form .

    R1-3R3
    R1 + R2
    R3 + R1
    R3 + R1

    This yields the matrix -1 5 1
    -2 -1 2
    0 -6 0
    From this I can see that y will be 0 which allows me to solve x and z .
    This gives the eigenvector as (1,0,1) but this doesn't seem like I have completed the method correctly . If anyone could clarify what I've done as being correct or could help me with reducing to echelon form that would be great .
    Thank you .
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  2. #2
    MHF Contributor

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    It hard to understand what you are saying because you do not give the original matrix but I think you have a matrix which has an eienvalue of -2 and, after you subtract -2 from each diagonal element, you have the matrix
    \begin{bmatrix}7 & 8 & -7 \\ -2 & -1 & 2 \\ 2 & 4 & -2\end{bmatrix}.

    Now, you are seeking an eigenvector, a matrix, \begin{bmatrix}x \\ y \\ z \end{bmatrix} such that
    \begin{bmatrix}7 & 8 & -7 \\ -2 & -1 & 2 \\ 2 & 4 & -2\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}

    To do that you can row reduce the matrix (you don't need to list the final column of the augmented matrix since that is always 0).

    But you seem to be doing the row operations pretty much at random. I think you will find it simpler to always work on one column at a time, from the left, getting a "1" at the pivot and "0"s elsewhere. To reduce the first column, divide the row by 7, then add 2 times that new first row to the second row, then subract twice the new first row from the third. That gives
    \begin{bmatrix}1 & \frac{8}{7} & -1 \\ 0 & \frac{9}{7} & 0 \\ 0 & \frac{12}{7} & 0\end{bmatrix}

    Now, it is easy to see that we can divide the second row by 9/7, subrtact 8/7 times that new second row from the first, and subtract 12/7 times that new second row from the first to get
    \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}

    Now that, remember, is the same as saying
    \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x- z \\ y \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}

    That tells us that we must have x- z= 0 or z= x, and y= 0. That is, an eigenvector is of the form \begin{bmatrix} x \\ 0 \\ x\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}.
    Last edited by HallsofIvy; March 9th 2011 at 04:19 AM.
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