# Part of an eigenvector problem...

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• Jan 16th 2011, 03:31 AM
Glitch
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?
• Jan 16th 2011, 03:48 AM
Also sprach Zarathustra
Quote:

Originally Posted by Glitch
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}$
• Jan 16th 2011, 03:49 AM
FernandoRevilla
$\displaystyle \textrm{r}(A)=1\Rightarrow \dim(\ker A)=2-1=1$

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