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Math Help - How to find the corrosponding eigenvector?

  1. #1
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    How to find the corrosponding eigenvector?

    Im am nearly there to find the corrosponding eigenvector.

    I solved the equation (A - lambda * E)*x = 0, where x and 0 are vectors

    and the result is
    (0,1,0)
    (0,0,1)
    (0,0,0)

    how do I find the corrosponding vector? Maple says (1,0,0), but how do I reach that conclusion?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: How to find the corrosponding eigenvector?

    You have reduced the homogenous system to:
    \left( \begin{array}{ccc} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{array}\right)\left( \begin{array}{ccc} x\\y\\z \end{array} \right)= \left( \begin{array}{ccc} 0\\0\\0 \end{array} \right)
    That means the solution of the system is:
    \left \{ \begin{array}{ll} y=0 \\ z=0 \end{array} \right.

    But because we have a zero row we can choose (a parameter) x=t thus the eigenspace of the given eigenvalue is:
    \left \{ \left( \begin{array}{ccc} t\\0\\0 \end{array}\right) | t \in \mathbb{R} \right \}

    If you choose t=1 that will give you the eigenvector:
    \left(\begin{array}{ccc} 1\\0\\0 \end{array} \right)
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  3. #3
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    Re: How to find the corrosponding eigenvector?

    Thanks for a very good explanation, I wasnt totally clueless on that one. However, thats the case for

    lambda2 = i*sqrt(3) taken off the characteristic polynomial

    The matrix A is then given as;
    (6-i*sqrt3,0,1)
    (0,-i*sqrt3,-1)
    (0,3,-i*sqrt3)

    and the eigenvector, says maple, is (1/39(-6-i*sqrt(3) , i/sqrt(3) , 1)

    How do I reach that eigenvector?

    I rowreduce A to (1,0,0)(0,1,0)(0,0,1)
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: How to find the corrosponding eigenvector?

    I guess you have know:
    \left \{ \begin{array}{lll} (6-i\sqrt{3})x+z=0 \\ -i\sqrt{3}y-z=0 \\ 3y-i\sqrt{3}z=0 \end{array} \right.
    \Leftrightarrow \left \{ \begin{array}{lll} (6-i\sqrt{3})x+z=0 \\ -i\sqrt{3}y-z=0 \\ y=\frac{i\sqrt{3}z}{3} \end{array} \right.

    If we substitute y into the second equation we obtain:
    \left \{ \begin{array}{lll} (6-i\sqrt{3})x+z=0 \\ z=1 \\ y=\frac{i\sqrt{3}z}{3} \end{array} \right.

    Go further to find the x component of the eigenvector.
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