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Math Help - Part of an eigenvector problem...

  1. #1
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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:

    \[<br />
\left( {\begin{array}{cc}<br />
 -1-\sqrt{3}i & 2i  \\<br />
 2i & 1-\sqrt{3}i  \\<br />
 \end{array} } \right)<br />
\]

    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:

    \[<br />
\left( {\begin{array}{cc}<br />
 -1-\sqrt{3}i & 2i  \\<br />
 2i & 1-\sqrt{3}i  \\<br />
 \end{array} } \right)<br />
\]

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


    \[<br />
\left( {\begin{array}{cc}<br />
 -1-\sqrt{3}i & 2i  \\<br />
 2i & 1-\sqrt{3}i  \\<br />
 \end{array} } \right)<br />
\] \begin{pmatrix}<br />
x_1+iy_1<br />
\\<br />
x_2+iy_2<br />
\end{pmatrix}=\begin{pmatrix}<br />
0<br />
\\<br />
0<br />
\end{pmatrix}
    Last edited by Also sprach Zarathustra; January 16th 2011 at 09:20 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    \textrm{r}(A)=1\Rightarrow \dim(\ker A)=2-1=1

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


    Fernando Revilla
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