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Thread: Finding Eigen Value and Eigen Vectors

  1. #1
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    Finding Eigen Value and Eigen Vectors

    Find the Eigen value and Eigen Vectors of the matrix
    $\displaystyle
    \begin{bmatrix}
    1 & -6 & -4 \\
    0 & 4 & 2 \\
    0 & -6 & -3 \\

    \end{bmatrix}
    $
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  2. #2
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    Quote Originally Posted by varunnayudu View Post
    Find the Eigen value and Eigen Vectors of the matrix
    $\displaystyle
    \begin{matrix}
    1 & -6 & -4 \\
    0 & 4 & 2 \\
    0 & -6 & -3 \\
    \end{matrix}
    $
    Where is the trouble here? There are two eigenvalues: $\displaystyle \lambda = 0$ and $\displaystyle \lambda = 1$ (repeated).

    Getting the eigenvectors does not look difficult.

    eg. A possible eigenvector (other forms are possible) asociated with $\displaystyle \lambda = 0$ is $\displaystyle \left(\begin{array}{c}
    -2 \\
    1 \\
    -2 \end{array}\right)
    $, found by solving

    $\displaystyle x - 6y - 4z = 0$ .... (1)

    $\displaystyle 4y + 2z = 0 \Rightarrow 2y + z = 0$ .... (2)

    $\displaystyle -6y - 3z = 0 \Rightarrow 2y + z = 0$ .... (3)

    Where are you stuck?
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  3. #3
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    Is this right?



    $\displaystyle \Rightarrow$$\displaystyle \begin{bmatrix}
    1-\lambda & -6 & -4 \\
    0 & 4- \lambda & 2 \\
    0 & -6 & -3 - \lambda
    \end{bmatrix}$
    $\displaystyle (1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0$
    $\displaystyle (1- \lambda)(- \lambda + \lambda ^2)$
    $\displaystyle \lambda = 0,1,1$

    for
    $\displaystyle \lambda = 0$ weget
    $\displaystyle
    \begin{bmatrix}
    1 & -6 & -4 \\
    0 & 4 & 2 \\
    0 & -6 & -3
    \end{bmatrix}
    $ $\displaystyle
    \begin{bmatrix}
    x \\
    y \\
    z
    \end{bmatrix}
    $=$\displaystyle
    \begin{bmatrix}
    0 \\
    0 \\
    0
    \end{bmatrix}
    $

    or$\displaystyle
    \begin{bmatrix}
    x-6y-4z=0 \\
    0x +4y+2z=0
    \end{bmatrix}
    $

    $\displaystyle
    \frac{x}{-12+16} = \frac{y}{2-0} = \frac{z}{4-0}
    $

    $\displaystyle \frac{x}{4} = \frac{y}{2} = \frac{z}{4}$

    Can u tell me hw u got 2 -1 2 instead of 2 1 2
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  4. #4
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    Quote Originally Posted by varunnayudu View Post


    $\displaystyle \Rightarrow$$\displaystyle \begin{bmatrix}
    1-\lambda & -6 & -4 \\
    0 & 4- \lambda & 2 \\
    0 & -6 & -3 - \lambda
    \end{bmatrix}$
    $\displaystyle (1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0$
    $\displaystyle (1- \lambda)(- \lambda + \lambda ^2)$
    $\displaystyle \lambda = 0,1,1$

    for
    $\displaystyle \lambda = 0$ weget
    $\displaystyle
    \begin{bmatrix}
    1 & -6 & -4 \\
    0 & 4 & 2 \\
    0 & -6 & -3
    \end{bmatrix}
    $ $\displaystyle
    \begin{bmatrix}
    x \\
    y \\
    z
    \end{bmatrix}
    $=$\displaystyle
    \begin{bmatrix}
    0 \\
    0 \\
    0
    \end{bmatrix}
    $

    or$\displaystyle
    \begin{bmatrix}
    x-6y-4z=0 \\
    0x +4y+2z=0
    \end{bmatrix}
    $

    $\displaystyle
    \frac{x}{-12+16} = \frac{y}{2-0} = \frac{z}{4-0}
    $

    $\displaystyle \frac{x}{4} = \frac{y}{2} = \frac{z}{4}$

    Can u tell me hw u got 2 -1 2 instead of 2 1 2
    x - 6y - 4z = 0 .... (1)

    2y + z = 0 .... (2)

    Let y = t. Then:

    From (2): z = -2y = -2t.

    From (1): x = 6y + 4z = 6t - 8t = -2t.

    Other forms of the solution are possible.
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