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Thread: Finding eigenvectors

  1. #1
    Senior Member I-Think's Avatar
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    Finding eigenvectors

    The matrix A, given by
    $\displaystyle A=\left(\begin{array}{ccc}7&-4&6\\2&2&2\\-3&4&-2\end{array}\right)$

    has eigenvalues 1,2,4. Find the set of corresponding vectors.

    Every time I perform these operations I only get my answers as 0's. I'll geve an example.

    $\displaystyle x(A-\delta{I})=0$ taking $\displaystyle \delta$ as 1

    $\displaystyle \left(\begin{array}{ccc}6&-4&6\\2&1&2\\-3&4&-3\end{array}\right)$$\displaystyle \left(\begin{array}{c}x\\y\\z\end{array}\right)=\l eft(\begin{array}{c}0\\0\\0\end{array}\right)$

    Solving
    $\displaystyle 6x-4y+6z=0$
    $\displaystyle 2x+y+2z=0$
    $\displaystyle -3x+4y-3z=0$

    $\displaystyle -x+5y-z=0$
    $\displaystyle x=5y-z$

    $\displaystyle 30y-6z-4y+6z=0$
    $\displaystyle y=0$
    And so on. Could someone please direct me as to what I am doing wrong?
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  2. #2
    Super Member Gamma's Avatar
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    Just by inspection, you can see that an eigenvector that will work is <1,0,-1>.
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  3. #3
    Super Member
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    You're solving the equations wrong (note that the coefficients of x and z will always vanish together if we try to sum up one equation to another) , the solutions for that eigenvalue should be x=-z
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