# Thread: Part of an eigenvector problem...

1. ## 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?

2. 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}$

3. $\displaystyle \textrm{r}(A)=1\Rightarrow \dim(\ker A)=2-1=1$

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