# Cycle of generalized eigenvectors disjoint

• Dec 3rd 2009, 07:27 PM
Last_Singularity
Cycle of generalized eigenvectors disjoint
Question: Let $\gamma_1,...,\gamma_p$ be cycles of generalized eigenvectors of a linear operator $T$ with respect to eigenvalue $\lambda$. Prove that if the initial eigenvectors are distinct, then the cycles are disjoint.

Attempt: Suppose that $x,y$ are (distinct) generalized eigenvectors of $T$ and let $p,q$ be the smallest integer such that $(T- \lambda I)^p (x) = 0$ and $(T- \lambda I)^q (y) = 0$.

I tried to argue by contradiction. Say that $\gamma_1 \cup \gamma_2 \neq \emptyset$, then there exist $j < p$ and $k < q$ such that $(T-\lambda I)^j (x) = (T-\lambda I)^k (y)$. Then for all $j'>j$ and $k'>k$, we have $(T-\lambda I)^{j'} (x) = (T-\lambda I)^{k'} (y)$. And then, this needs to imply that $x,y$ are in fact not distinct anymore and thus creates a contradiction ...?
• Dec 3rd 2009, 09:12 PM
Jose27
Assume $p and apply $T-\lambda I \ \mbox{(p-j)}$ times on both sides of your second to last equation and use the minimality of p and q.
• Dec 4th 2009, 12:48 AM
Last_Singularity
$(T-\lambda I)^j (x) = (T-\lambda I)^k (y)$
$(T-\lambda I)^p (x) = (T-\lambda I)^{p+k-j} (y)$
$0 = (T-\lambda I)^{p+k-j} (y)$
But $-p < k-j < q-p$ so $0 < p+k-j < q$ contradicting that $q$ is the minimal number of times we apply $(T-\lambda I)$ to $y$ to get zero.

Thanks!