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:
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)