I think I got it:
λ1 is a root of the characteristic polynomial if and only if A−λ1I is not invertible if and only if the eigenspace Eλ1 (A) is non-trivial if and only if A has an eigenvector with eigenvalue λ1.
Hi,
I've been thinking about something for the pasts days and since I'm on a trip I cannot site and write a whole proof.
Here is what I'm thinking about:
I'm given a matrix of a linear application in some bases. Then I want to find the eigenvectors of that application. So I use the method with the characteristic polynomial to find then eigenvalues ( find the root of det( A-I*t)=0 where A is my matrix in the give bases and I is the identity matrix and t is what I'm looking for ( scalar)). So what happens if I find a solution ( let say that s is a eigenvalue ). But the eigenvector of s is zero. ( which means that it is not an eigenvector by definition).
So
1. Is it possible to find such a root?
2. If not how can I prove it?
3. If it's possible what does it mean since the roots of the characteristic polynomial are eigenvalues.
Thanks a lot for clearing my mind!
I think I got it:
λ1 is a root of the characteristic polynomial if and only if A−λ1I is not invertible if and only if the eigenspace Eλ1 (A) is non-trivial if and only if A has an eigenvector with eigenvalue λ1.