Find the characteristic & minimal polynomials, eigenvectors, and dimension of the eigenspace for this 6x6 jordan matrix?

Question

3 1 0 0 0 0 
0 3 1 0 0 0 
0 0 3 0 0 0 
0 0 0 1 1 0 
0 0 0 0 1 0 
0 0 0 0 0 1 

I think the characteristic polynomial is: ((x-3)^3)((x-1)^3) Found by taking the number of each eigenvalue along the diagonal, 3 3's, 3 1's I think the minimal polynomial is: ((x-3)^3)((x-1)^2) Found by taking each eigenvalue, start with an exponent 1 and add 1 for the maximum number of 1's in a row along that eigenvalue's subdiagonal. Since there's 2 1's in a row for eigenvalue 3 we have 2+1=3 for an exponent, since there's only 1 1's in a row for the subdiagonal of eigenvalue 1, (x-1) has 2 for an exponent. Am I thinking about this correctly?How do I quickly find the eigenvectors and dimension? Thanks.

Answer 1

You've got the polynomials right. The eigenvectors are obvious --- no, really --- $(1,0,0,0,0,0)$ is an eigenvector for the eigenvalue 3, and both $(0,0,0,1,0,0)$ and $(0,0,0,0,0,1)$ are eigenvectors for the eigenvalue 1.

But, perhaps you also want the generalized eigenvectors?