# Math Help - How to find the corrosponding eigenvector?

1. ## How to find the corrosponding eigenvector?

Im am nearly there to find the corrosponding eigenvector.

I solved the equation (A - lambda * E)*x = 0, where x and 0 are vectors

and the result is
(0,1,0)
(0,0,1)
(0,0,0)

how do I find the corrosponding vector? Maple says (1,0,0), but how do I reach that conclusion?

2. ## Re: How to find the corrosponding eigenvector?

You have reduced the homogenous system to:
$\left( \begin{array}{ccc} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{array}\right)\left( \begin{array}{ccc} x\\y\\z \end{array} \right)= \left( \begin{array}{ccc} 0\\0\\0 \end{array} \right)$
That means the solution of the system is:
$\left \{ \begin{array}{ll} y=0 \\ z=0 \end{array} \right.$

But because we have a zero row we can choose (a parameter) $x=t$ thus the eigenspace of the given eigenvalue is:
$\left \{ \left( \begin{array}{ccc} t\\0\\0 \end{array}\right) | t \in \mathbb{R} \right \}$

If you choose $t=1$ that will give you the eigenvector:
$\left(\begin{array}{ccc} 1\\0\\0 \end{array} \right)$

3. ## Re: How to find the corrosponding eigenvector?

Thanks for a very good explanation, I wasnt totally clueless on that one. However, thats the case for

lambda2 = i*sqrt(3) taken off the characteristic polynomial

The matrix A is then given as;
(6-i*sqrt3,0,1)
(0,-i*sqrt3,-1)
(0,3,-i*sqrt3)

and the eigenvector, says maple, is (1/39(-6-i*sqrt(3) , i/sqrt(3) , 1)

How do I reach that eigenvector?

I rowreduce A to (1,0,0)(0,1,0)(0,0,1)

4. ## Re: How to find the corrosponding eigenvector?

I guess you have know:
$\left \{ \begin{array}{lll} (6-i\sqrt{3})x+z=0 \\ -i\sqrt{3}y-z=0 \\ 3y-i\sqrt{3}z=0 \end{array} \right.$
$\Leftrightarrow \left \{ \begin{array}{lll} (6-i\sqrt{3})x+z=0 \\ -i\sqrt{3}y-z=0 \\ y=\frac{i\sqrt{3}z}{3} \end{array} \right.$

If we substitute $y$ into the second equation we obtain:
$\left \{ \begin{array}{lll} (6-i\sqrt{3})x+z=0 \\ z=1 \\ y=\frac{i\sqrt{3}z}{3} \end{array} \right.$

Go further to find the $x$ component of the eigenvector.