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Thread: eigenvectors, vectors..confused

  1. #1
    Dec 2008

    eigenvectors, vectors..confused

    v=[7,3,6] and w=[3,2,6] how would i calculate (A-(llambda)I)v and (A-(llambda)I)w. for any old 3x3 matrix A. im confused what llambda is supposed to be.
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  2. #2
    Super Member Gamma's Avatar
    Dec 2008
    Iowa City, IA


    An eigenvalue $\displaystyle \lambda_1$ corresponding to an eigenvector $\displaystyle v_1$ satisfies the following relationship.

    $\displaystyle Av_1=\lambda_1v_1$
    To see this from the equation you supplied, just multiply it out and move the negative part to the other side and you will get what I wrote above.

    If you are given the eigenvector v you just multiply it by A (Av), see what you get and if it is an eigenvector, there will be some scalar multiple lambda that you can multiply your vector by to get the same thing, as in the above equation.

    The equation in the form you supplied is key for actually solving what these eigenvalues and vectors are. You will subtract lambda from the diagonal and take the determinant. Set this equal to 0 and this is your characteristic polynomial. The roots of this polynomial will be your eigenvalues. Then to find the corresponding eigenvectors you simply subtract those roots from the diagonal and solve the equation for v.
    $\displaystyle (A-I\lambda)v=0$
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