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Math Help - 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
    <br />
\begin{bmatrix}<br />
1 & -6 & -4 \\<br />
0 & 4 & 2 \\<br />
0 & -6 & -3 \\ <br /> <br />
\end{bmatrix}<br />
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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
    <br />
\begin{matrix}<br />
1 & -6 & -4 \\<br />
0 & 4 & 2 \\<br />
0 & -6 & -3 \\ <br />
\end{matrix}<br />
    Where is the trouble here? There are two eigenvalues: \lambda = 0 and \lambda = 1 (repeated).

    Getting the eigenvectors does not look difficult.

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

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

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

    -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?



    \Rightarrow \begin{bmatrix}<br />
1-\lambda & -6 & -4 \\<br />
0 & 4- \lambda & 2 \\<br />
0 & -6 & -3 - \lambda <br />
\end{bmatrix}
    (1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0
    (1- \lambda)(- \lambda + \lambda ^2)
    \lambda = 0,1,1

    for
    \lambda = 0 weget
     <br />
\begin{bmatrix}<br />
1 & -6 & -4 \\<br />
0 & 4 & 2 \\<br />
0 & -6 & -3 <br />
\end{bmatrix}<br />
 <br />
\begin{bmatrix}<br />
x \\<br />
y \\<br />
z<br />
\end{bmatrix}<br />
=  <br />
\begin{bmatrix}<br />
0 \\<br />
0 \\<br />
0<br />
\end{bmatrix}<br />

    or  <br />
\begin{bmatrix}<br />
x-6y-4z=0 \\<br />
0x +4y+2z=0<br />
\end{bmatrix}<br />

    <br />
\frac{x}{-12+16} = \frac{y}{2-0} = \frac{z}{4-0}<br />

    \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


    \Rightarrow \begin{bmatrix}<br />
1-\lambda & -6 & -4 \\<br />
0 & 4- \lambda & 2 \\<br />
0 & -6 & -3 - \lambda <br />
\end{bmatrix}
    (1- \lambda){(4- \lambda)(-3- \lambda)+12} + 0 +0
    (1- \lambda)(- \lambda + \lambda ^2)
    \lambda = 0,1,1

    for
    \lambda = 0 weget
     <br />
\begin{bmatrix}<br />
1 & -6 & -4 \\<br />
0 & 4 & 2 \\<br />
0 & -6 & -3 <br />
\end{bmatrix}<br />
 <br />
\begin{bmatrix}<br />
x \\<br />
y \\<br />
z<br />
\end{bmatrix}<br />
=  <br />
\begin{bmatrix}<br />
0 \\<br />
0 \\<br />
0<br />
\end{bmatrix}<br />

    or  <br />
\begin{bmatrix}<br />
x-6y-4z=0 \\<br />
0x +4y+2z=0<br />
\end{bmatrix}<br />

    <br />
\frac{x}{-12+16} = \frac{y}{2-0} = \frac{z}{4-0}<br />

    \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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