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Math Help - Determine eigenvector

  1. #1
    Inf
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    Determine eigenvector

    Hey guys,

    i am supposed to determine the eigenvalues and eigenvectors of:
    \begin{pmatrix}1 & 1 \\1 & 0\\ \end{pmatrix}

    I calculated the eigenvalues with the formula:

    det(\lambda I - A) = 0

    to be \lambda_1 = \frac{1}{2}(1 + \sqrt 5) and \lambda_2 = \frac{1}{2}(1 - \sqrt 5)

    To get the eigenvectors i need to solve the following equation, right?

    \begin{pmatrix}\lambda - 1 & -1 \\-1 & \lambda\\ \end{pmatrix}\cdot\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}

    Now i get x = 0 and y = 0, but thats just not right says Wolfram Alpha and the exercise text.

    What is wrong?

    Thanks, Inf
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Inf View Post
    To get the eigenvectors i need to solve the following equation, right?

    \begin{pmatrix}\lambda - 1 & -1 \\-1 & \lambda\\ \end{pmatrix}\cdot\begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}

    No, you have to solve the systems (A-\lambda_iI)\begin{pmatrix}x \\ y \end{pmatrix}=\begin{pmatrix}0\\ 0 \end{pmatrix} \quad (i=1,2)
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  3. #3
    Member alexgeek's Avatar
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    To find eigenvectors you solve simultaneous equations for:
    \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \begin{matrix} x \\ y \end{matrix} = \lambda \begin{matrix} x \\ y \ \end{matrix}
    Last edited by topsquark; May 18th 2011 at 04:26 PM. Reason: Changed to [tex] tags
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  4. #4
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    if λ is an eigenvalue for A, then Av = λv, for some eigenvector v. which means that

    (A - λI)v = 0. this is the system you have to solve (for each value of λ). so for λ1 = (1/2)(1 + √5) this would be:


    [(1/2)(1 - √5) ...........1........][x]....[0]
    [.......1.......... (-1/2)(1 + √5)][y] = [0], which leads to:

    y = (-1/2)(1 - √5)x.

    if we choose (as apparently wolfram|alpha) did, y = 1, this makes x = (1/2)(1 + √5).

    you should be able to compute the 2nd eigenvector, now.
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  5. #5
    Inf
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    Thank you guys!

    I thought both equations have to be solved at the same time.

    Now I did it like you told me and get this:

    p_1 = \begin{pmatrix}\frac{1 +\sqrt 5}{2}\\1\\\end{pmatrix}
    p_2 = \begin{pmatrix}\frac{1 -\sqrt 5}{2}\\1\\\end{pmatrix}
    P = \begin{pmatrix}\frac{1 +\sqrt 5}{2} & \frac{1 -\sqrt 5}{2}\\1 &1\\\end{pmatrix}

    And this perfectly solves the equation: P^{-1} \cdot A \cdot P = D

    Thank you guys!
    Greetings, Inf
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