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Math Help - Eigenvectors

  1. #1
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    Eigenvectors

    I have the following matrix to solve for eigenvectors.
    \begin{vmatrix}-12.4171 & 5 & 1 \\ 5 & -11.4171 & 4 \\ 4 & 5 & -3.09496 \end{vmatrix}
    So far I have eliminated it to
    \begin{vmatrix}-12.4171 & 5 & 1 \\ 0 & -9.40375 & 4.40267 \\ 0 & 6.61068 & -3.09496 \end{vmatrix}
    I can not figure out how to eliminate the last row. Elimination of the value 6.61068 to zero would make x3=0.
    Last edited by pdfmath; September 25th 2010 at 02:26 PM.
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  2. #2
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    How did you eliminate it to that? Row reduction? You need first find the eigenvalues and row reduction does not help with that. Once you have found the eigenvalues, you can subtract an eigenvalue from the diagonal and row reduce that to find the eigenvectors.
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  3. #3
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    eigenvalues

    The eigenvalues were found. I am trying to find an eigenvector from one of the eigenvalues. The matrix there already has the first eigenvalue plugged in. I should have been more clear.
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  4. #4
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    Then, yes, of course, that matrix will row-reduce to a matrix having the last row 0- The equation (A- \lambda I)v= 0 must, by definition of "eigenvalue" have a non-zero solution and since 0 obviously is a solution, cannot have a unique solution. In fact, the set of all solutions, the "eigenspace", is a subspace and so it must have an infinite number of solutions.

    Doing exactly what you say reduces the matrix to
    \begin{bmatrix}-12.4171 & 5 & 1 \\ 0 & -9.40375 & 4.40267 \\ 0 & 0 & 0\end{bmatrix}
    which essentially means your equations reduce to -12.4171x+ 5y+ z= 0 and -9.40375y+ 4.40267z= 0.

    From the second equation, z= (9.40375/4.40267)y= 2.13592y.

    Putting that into the first equation we have -12.4171x+ 5y+ 2.13592y= -12.4171x+ 7.13592y= 0 so that x= (7.13592/12.4171)y= 0.57468y.

    That is, any eigenvector is of the form (0.57468y, y, 2.13592y)= (0.57468, 1, 2.13592)y where y can be any non-zero number (1 is simplest of course).
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  5. #5
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    Thank you that is exactly what i needed. The fact that this solution is not unique was key to my understanding. Just wasn't sure if you could get rid of the third row like that. Thanks again.
    Quote Originally Posted by HallsofIvy View Post
    Then, yes, of course, that matrix will row-reduce to a matrix having the last row 0- The equation (A- \lambda I)v= 0 must, by definition of "eigenvalue" have a non-zero solution and since 0 obviously is a solution, cannot have a unique solution. In fact, the set of all solutions, the "eigenspace", is a subspace and so it must have an infinite number of solutions.

    Doing exactly what you say reduces the matrix to
    \begin{bmatrix}-12.4171 & 5 & 1 \\ 0 & -9.40375 & 4.40267 \\ 0 & 0 & 0\end{bmatrix}
    which essentially means your equations reduce to -12.4171x+ 5y+ z= 0 and -9.40375y+ 4.40267z= 0.

    From the second equation, z= (9.40375/4.40267)y= 2.13592y.

    Putting that into the first equation we have -12.4171x+ 5y+ 2.13592y= -12.4171x+ 7.13592y= 0 so that x= (7.13592/12.4171)y= 0.57468y.

    That is, any eigenvector is of the form (0.57468y, y, 2.13592y)= (0.57468, 1, 2.13592)y where y can be any non-zero number (1 is simplest of course).
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