Generalized Eigenvectors

I remember reading a theorem that said that for an n x n matrix A, there exists a basis of Cn consisting of generalized eigenvectors of A.

For A = [1 1 1; 0 1 0; 0 0 1] (the semicolons indicate a new row so that A should be 3 x 3 with a first row consisting of all 1's and a diagonal of 1's). The eigenvalue of A is 1 with an algebraic multiplicity of 3 and geometric multiplicity of 2. So I can pull out a basis of 2 vectors from Ker(A-uI). For the third generalized eigenvector, it should be in Ker((A-uI)3), but that produces 0v = 0, in which v could be any vector in Cn...did I do something wrong here?
 Blog Entries: 5 Recognitions: Homework Help Science Advisor It's been really long since I did linear algebra, so forgive me if I say anything stupid. You get two "true" eigenvectors (1, 0, 0) and (0, 1, -1) (or scalar multiples thereof). Then shouldn't the third generalised eigenvector satisfy $$\left. (A - \lambda I) \right|_{\lambda = 1} \vec v = (1, 0, 0)^\mathrm{T}$$, with the first eigenvector on the right hand side rather than 0?
 Recognitions: Homework Help Science Advisor this is explained in great detail in the notes for math 4050 in my webpage. briefly, you factor the characteristic polynomial, find the bases of the kernels of the corresponding powers (T-c)^r. this is called a jordan basis. it is tedious to do in practice.

Generalized Eigenvectors

 Quote by mathwonk this is explained in great detail in the notes for math 4050 in my webpage.
 Quote by CompuChip You get two "true" eigenvectors (1, 0, 0) and (0, 1, -1) (or scalar multiples thereof). Then shouldn't the third generalised eigenvector satisfy $$\left. (A - \lambda I) \right|_{\lambda = 1} \vec v = (1, 0, 0)^\mathrm{T}$$, with the first eigenvector on the right hand side rather than 0?
You're right, though I haven't fully read how so yet.
 Blog Entries: 5 Recognitions: Homework Help Science Advisor I was just thinking that this actually makes sense. The whole point that you need generalised eigenvectors here, is that the algebraic multiplicity of the eigenvalue is larger, so if you calculate them in the normal way, you will get a null eigenvector somewhere, which is not allowed. And if you solve the equation (A - I) v2 = v1, where v1 satisfies (A - I)v1 = 0, then you will indeed get an eigenvector in ker((A - I)²) (which in this case is equal to the whole space since (A - I)² is the zero matrix). So you are right that the generalised eigenvector you are looking for is inside ker((A - I)²), it is just not any vector in it. But as I said, it's been a long time, so we're both better of reading mathwonk's notes, I think :-)
 Recognitions: Homework Help Science Advisor in reference to post #1, you want an exponent of 2 instead of 3, on (A-uI).