Is this statement true?
$L:V\to V$ is a linear map with eigenvalue (not necessarily the only one) $a$. Suppose $(L-aI)^{m+1}(v)=0$ where $m$ is the power of the $(x-aI)$ term in the minimal polynomial of $L$. Then $(L-aI)^m(v)=0$ also.
Some thoughts:
So this is essentially saying that $(L-aI)[(L-aI)^m(v)]=0\implies (L-aI)^m(v)=0$. In other words the nullspace of $ (L-aI)$ is $\{0\}$. Therefore I don't think the statement is necessarily true. Is there a way of constructing an explicit example to disprove this?
Ah wait, I haven't used the fact that $m$ is the power of the term in the minimal polynomial. So this may yet be true...