Eigenvalue kind of nilpotent problem

1. Jun 29, 2008

jostpuur

If a vector $$v\in V$$ and a linear mapping $$T:V\to V$$ are fixed, and there exists numbers $$\lambda_1\in\mathbb{C}$$, $$n_1\in\mathbb{N}$$ so that

$$(T - \lambda_1)^{n_1}v = 0,$$

is it possible that there exists some $$\lambda_2\neq\lambda_1$$, and $$n_2\in\mathbb{N}$$ so that

$$(T - \lambda_2)^{n_2}v = 0?$$

(Here complex numbers are interpreted as multiplication operators $$V\to V$$, $$v\mapsto \lambda v$$, as usual.)

2. Jun 29, 2008

morphism

That happens iff v=0. This is the statement that the intersection of different generalized eigenspaces is {0}.

3. Jun 30, 2008

gel

Furthermore, if f,g are any two polynomials with no common factor and f(T)v=g(T)v=0 then you can conclude that v=0.