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Thread: eigenvector question

  1. #1
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    eigenvector question

    Could you please refresh my memory in eigenvalues and tell me how did it get these values in equations E4 and E5?
    Thanks in advance

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  2. #2
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    Re: eigenvector question

    A linear transformation, A, from vector space V to V, has eigenvalue \lambda if and only if there exist a non-zero vector, v, such that [tex]Av= \lambda v[/b]. That can be written as Av- \lambda v= (A- \lambda I)v= 0 where I is the identity transformation, that maps every vector to itself. Written as matrices the identity transformation is the identity matrix that has 1s on the main diagonal, 0 everywhere else.

    (A- \lambda I)v= 0 has the "trivial" solution v= 0. In order that there be a non-trivial, not zero, solution, there the matrix A- \lambda I must not be invertible which, in turn, means that determinant of A- \lambda I must be 0.

    The specific example you give refers to "eq. 5.5a" which you do not show so it is not clear what "A" and " \lambda are but matrix A- \lambda I are:
    [tex]\begin{bmatrix}-10\omega^2+ 35 & -5 \\ -5 & -\omega^2+ 5\end{bmatrix}
    and we want to find \omega so that the determinant of that matrix is 0.

    For a 2 by 2 matrix, \begin{bmatrix}a & b \\ c & d \end{bmatrix} is ad- bc.
    Here that is (-10\omega^2+ 35)(-\omega^2+ 5)- 25= -10\omega^24- 85\omega^2+ 175- 25= -10\omega^4- 85\omega^2+ 150 and we want to find \omega so that is 0. The first thing I would do is divide through by -10 and let \omega^2 to get the quadratic equation x^2+ 8.5x- 15= 0. Use the quadratic formula to solve that: x= \frac{-8.5\pm\sqrt{72.25+60}}{2}= -4.25\pm 6.75 so that has the two values 6.75- 4.25= 2.50 and -6.75- 4.25= -11. From x= \omega^2= 2.5 and x= \omega^2= -11. \omega has four solutions, 2 real, \sqrt{2.5} and -\sqrt{2.5}, and two imaginary solutions i\srqt{11} and -i\sqrt{11}.

    (Please check my arithmetic!)
    Last edited by HallsofIvy; Jul 28th 2017 at 10:35 AM.
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  3. #3
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    Re: eigenvector question

    Thank you for your reply.
    Could you please explain me how did it get the values 1,2 and 1,-5?
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  4. #4
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    Re: eigenvector question

    There are an infinite number of possible vectors that satisfy the conditions of being an eigenvector. The author chose 1 for the top number and calculated what the second number needed to be. Any scalar multiple of each vector will also be a valid eigenvector
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  5. #5
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    Re: eigenvector question

    I'm sorry I don't understand
    Could you give me an example how did it get the values 1,2 and 1,-5?
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