# Cycle of generalized eigenvectors disjoint

• Dec 3rd 2009, 06:27 PM
Last_Singularity
Cycle of generalized eigenvectors disjoint
Question: Let $\displaystyle \gamma_1,...,\gamma_p$ be cycles of generalized eigenvectors of a linear operator $\displaystyle T$ with respect to eigenvalue $\displaystyle \lambda$. Prove that if the initial eigenvectors are distinct, then the cycles are disjoint.

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

I tried to argue by contradiction. Say that $\displaystyle \gamma_1 \cup \gamma_2 \neq \emptyset$, then there exist $\displaystyle j < p$ and $\displaystyle k < q$ such that $\displaystyle (T-\lambda I)^j (x) = (T-\lambda I)^k (y)$. Then for all $\displaystyle j'>j$ and $\displaystyle k'>k$, we have $\displaystyle (T-\lambda I)^{j'} (x) = (T-\lambda I)^{k'} (y)$. And then, this needs to imply that $\displaystyle x,y$ are in fact not distinct anymore and thus creates a contradiction ...?
• Dec 3rd 2009, 08:12 PM
Jose27
Assume $\displaystyle p<q$ and apply $\displaystyle T-\lambda I \ \mbox{(p-j)}$ times on both sides of your second to last equation and use the minimality of p and q.
• Dec 3rd 2009, 11:48 PM
Last_Singularity
$\displaystyle (T-\lambda I)^j (x) = (T-\lambda I)^k (y)$
$\displaystyle (T-\lambda I)^p (x) = (T-\lambda I)^{p+k-j} (y)$
$\displaystyle 0 = (T-\lambda I)^{p+k-j} (y)$
But $\displaystyle -p < k-j < q-p$ so $\displaystyle 0 < p+k-j < q$ contradicting that $\displaystyle q$ is the minimal number of times we apply $\displaystyle (T-\lambda I)$ to $\displaystyle y$ to get zero.

Thanks!