It may be a quite silly question but I am having trouble with this.
My question is that if a nonzero vector $v\in V$ is a generalized eigenvector for a linear operator $T: V\to V$ such that $(T-\lambda_1)^{d_1}v=0$ and $(T-\lambda_2)^{d_2}v=0$ where $d_1$ and $d_2$ are positive integers, is it necessarily true that $\lambda_1 = \lambda_2$?
For example, suppose T is a nilpotent linear operator. Then every vector is a generalized eigenvector with eigenvalue 0. Isn't there any generalized eigenvector with nonzero eigenvalue?
Yes, it is necessarily true that $\lambda_1 = \lambda_2$. In particular: suppose that $(T - \lambda_1)^{d_1} v = 0$ and $(T - \lambda_1)^{d_1 - 1}v \neq 0$. Then $w = (T - \lambda_1)^{d_1 - 1} v$ is non-zero and satisfies $Tw = \lambda_1 w$.
It follows that if $\lambda_2 \neq \lambda_1$, we have $$ \begin{align} (T - \lambda_1)^{d_1 - 1}[(T - \lambda_2)^{d_2}v] &= (T - \lambda_2)^{d_2}[(T - \lambda_1)^{d_1 - 1}v] \\ & = (T - \lambda_2)^{d_2} w = (\lambda_1 - \lambda_2)^{d_2}w \neq 0. \end{align} $$ It follows that $(T - \lambda_2)^{d_2}v \neq 0$.