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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?

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    $\begingroup$ A important result in Jordan-normal-form theory is $$ V=R_{\lambda_1}\left(\mathscr{A}\right)\oplus R_{\lambda_2}\left(\mathscr{A}\right)\oplus\cdots\oplus R_{\lambda_l}\left(\mathscr{A}\right) $$ where $R_{\lambda_i}\left(\mathscr{A}\right)=\mathrm{Ker}\left(\mathscr{A}-\lambda_i\mathrm{id}\right)^{k_i}$. If you have got here, the answer of your problem is obvious. Maybe refer to the textbook again is more helpful. $\endgroup$ – user823011 Sep 17 at 10:03
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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 $(T - \lambda_2)^{d_2}v \neq 0$.

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  • $\begingroup$ Thank you for your clear proof $\endgroup$ – Henry Choi Sep 17 at 10:10

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