Thread: 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?

You have reduced the homogenous system to:

That means the solution of the system is:


But because we have a zero row we can choose (a parameter) thus the eigenspace of the given eigenvalue is:


If you choose that will give you the 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)

I guess you have know:



If we substitute into the second equation we obtain:


Go further to find the component of the eigenvector.