# Is it possible for a generalized eigenvector to have two different eigenvalues?

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?

• 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. – user823011 Sep 17 at 10:03

## 1 Answer

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 $$(T - \lambda_2)^{d_2}v \neq 0$$.

• Thank you for your clear proof – Henry Choi Sep 17 at 10:10