# generalized eigenvectors - questions

We introduced generalized eigenvectors and then we were given this example:

Given an endomorphism $\varphi: V \to V$ with $\chi_\varphi(t) = (-1)^n (t-\lambda)^n$, then $U(\lambda) = V$ since $(\varphi-\lambda · \mathrm{id})^n = 0$.

I don't really get this. Why is it that $(\varphi-\lambda · \mathrm{id})^n = 0$? If this is the case then clearly $U(\lambda) = V$ but I can't justify this.

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## 1 Answer

This is by the Cayley–Hamilton theorem: $0=\chi_\varphi(\varphi)=(-1)^n(\varphi-\lambda I)^n$

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