There is a theorem in Freidberg that goes like this. This is exposition leading to Jordan Form discussion.
Theorem 7.6. Let T be a linear operator on a vector space $\mathrm{V},$ and let
$\lambda$ be an eigenvalue of T. Suppose that $\gamma_{1}, \gamma_{2}, \ldots, \gamma_{q}$ are cycles of generalized eigenvectors of T corresponding to $\lambda$ such that the initial vectors of the $\gamma_{i}$ 's are distinct and form a linearly independent set. Then the $\gamma_{i}$ 's are disjoint, and their union $\gamma=\bigcup_{i=1}^{q} \gamma_{i}$ is linearly independent.
Is it true then that the only way a particular eigenvalue $\lambda_{particular}$ can have multiple disjoint cycles $\gamma_{1}, \gamma_{2}, .., \gamma_{i} $ associated with it if it had at least that many ($i$) lin.indep (non-generalized) eigenvectors associated with it to begin with?
In other words, if i got a defective 3×3 matrix $A$ with only one eigenvalue $\lambda_{particular} = 1$ then it is a given that it has a cycle of three lin. indep. generalized eigenvectors but also it is a given that it cannot have multiple cycles associated with it and this holds for arbitrary dimensions of $A$.
I would appreciate if somebody could confirm this or provide a counter example. Thank you tons.
Best Answer
Yes. In particular, there are exactly as many generalized eigenvector cycles as there are linearly independent eigenvectors. To see this, the last element in any chain is an eigenvector; conversely, for any eigenvector we can build a cycle that has length at least $1$.