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Thread: Part of an eigenvector problem...

  1. #1
    Senior Member
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    Part of an eigenvector problem...

    Upon solving an eigenvector problem, I've been summoned to find the kernel of the following matrix:

    $\displaystyle \[
    \left( {\begin{array}{cc}
    -1-\sqrt{3}i & 2i \\
    2i & 1-\sqrt{3}i \\
    \end{array} } \right)
    \]$

    Perhaps I'm just sleep deprived, but I'm having trouble getting this into row-echelon form. Any advice?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Glitch View Post
    Upon solving an eigenvector problem, I've been summoned to find the kernel of the following matrix:

    $\displaystyle \[
    \left( {\begin{array}{cc}
    -1-\sqrt{3}i & 2i \\
    2i & 1-\sqrt{3}i \\
    \end{array} } \right)
    \]$

    Perhaps I'm just sleep deprived, but I'm having trouble getting this into row-echelon form. Any advice?
    Try to solve...:


    $\displaystyle \[
    \left( {\begin{array}{cc}
    -1-\sqrt{3}i & 2i \\
    2i & 1-\sqrt{3}i \\
    \end{array} } \right)
    \]$$\displaystyle \begin{pmatrix}
    x_1+iy_1
    \\
    x_2+iy_2
    \end{pmatrix}=\begin{pmatrix}
    0
    \\
    0
    \end{pmatrix}$
    Last edited by Also sprach Zarathustra; Jan 16th 2011 at 08:20 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    $\displaystyle \textrm{r}(A)=1\Rightarrow \dim(\ker A)=2-1=1$

    So, use only the second equation (for example) and do $\displaystyle x_2=1$ (for example) .You'll obtain a basis of $\displaystyle \ker A$ .


    Fernando Revilla
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