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Math Help - a little trouble finding the eigen vector here...

  1. #1
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    a little trouble finding the eigen vector here...

    So I have original matrix

    2 2 -2
    0 1 0
    0 0 1

    with eigen values 2, 1, 1

    i can find the eigen vector for the llamda 1 which is

    2
    0
    1

    and

    -2
    1
    0

    but I cannot find the eigen vector for value 2

    I get this matrix

    0 2 -2
    0 -1 0
    0 0 -1

    Please help.
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  2. #2
    MHF Contributor

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    Directly from the definition of "eigenvector": Av= \lambda v, if \begin{bmatrix} x \\ y \\ z\end{bmatrix} is an eigenvector of A corresponding to eigenvalue 2, then
    \begin{bmatrix}2 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= 2\begin{bmatrix} x \\ y \\ z\end{bmatrix}
    \begin{bmatrix}2x+ 2y- 2z \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}
    so we have the three equations 2x+ 2y- 2z= 2x, y= 2y, z= 2z. The last two give y= 0, z= 0, of course, and 2x= 2x+ 0- 0 is satisfied for any x.

    That is, of course, equivalent to saying \begin{bmatrix}0 & 2 & -2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}
    which gives the equations 2y- 2z= 0, y= 0, z= 0. That says nothing about x so x can be any number.
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  3. #3
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    Joined
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    So the eigen vector could be

    1
    0
    0

    when constructing the matrix made up of the eigen vectors but in columns, is there an order I must respect?

    The way I would go about it is to use llamda 1 (eigen value 1) for the first column, second for second and third for third. I get this matrix

    1 2 -2
    0 0 1
    0 1 0

    I think I can swap the 2nd and thrd column since their share the same eigen value. Is that true?
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