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Math Help - Cycle of generalized eigenvectors disjoint

  1. #1
    Member Last_Singularity's Avatar
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    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 ...?
    Last edited by Last_Singularity; December 3rd 2009 at 07:03 PM.
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  2. #2
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    Assume p<q 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.
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  3. #3
    Member Last_Singularity's Avatar
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    (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!
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