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I am learning generalized eigenvectors at the moment, and I am not totally sure I get them yet.


$v\in V$ is a generalized eigenvector if $v\ne 0$ and $v\in null(T-\lambda I)^j$ for some $j\gt 0$. Where $T\in L(V)$ I.e. it's a linear operator on $V$.

Apparently we can take $j=n=\dim V$


So essentially if I take any transformation, let's say the identity, the generalised eigenvectors are the $(v_1,v_2)$ that satisfy the following: $$\begin{bmatrix}(1-\lambda)^2&0\\0&(1-\lambda)^2\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=0$$

Is that correct?

Here they are $v_1(1-\lambda)^2=0$ and $v_2(1-\lambda)^2=0$

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  • $\begingroup$ $\lambda $ is an eigenvalue so if your matrix is $I$ then $\lambda =1$ and any vector is a generalized eigenvector. Clearly, in this case any vector is also an eigenvetor. $\endgroup$ – Ofir Schnabel Jun 22 '15 at 9:46
  • $\begingroup$ @OfirSchnabel is there a notion of generalized eigenvalues? $\endgroup$ – I cut trees Jun 22 '15 at 10:02
  • $\begingroup$ No, there are eigenvalues. Eigenvectors and generalized eigenvectors are with respect to eigenvalues. $\endgroup$ – Ofir Schnabel Jun 22 '15 at 10:05

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