Did I make a mistake in finding the third generalized eigenvector for A?

[Gear300]

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?

 

[CompuChip]

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?

 

[mathwonk]

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.

 

[Gear300]

this is explained in great detail in the notes for math 4050 in my webpage.

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.

 

[CompuChip]

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 :-)

 

[mathwonk]

in reference to post #1, you want an exponent of 2 instead of 3, on (A-uI).