# 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:

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

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:

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

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

$$\left( {\begin{array}{cc} -1-\sqrt{3}i & 2i \\ 2i & 1-\sqrt{3}i \\ \end{array} } \right)$$
$\begin{pmatrix}
x_1+iy_1
\\
x_2+iy_2
\end{pmatrix}=\begin{pmatrix}
0
\\
0
\end{pmatrix}$

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

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