Eigenvalues and powers 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...
The statement is true. In general you have 
$$Ker(L-aI)\subseteq Ker(L-aI)^2\subseteq\cdots $$
This sequence stabilises, i.e. there is an $m$ such that
$$Ker(L-aI)^i=Ker(L-aI)^{i+1}$$
for all $i\geq m$. This $m$ is precisely the $m$ which is the exponent of $x-a$ in the minimal polynomial. In particular you have
$$Ker(L-aI)^m=Ker(L-aI)^{m+1}.$$

You write that the kernel of $L-aI$ restricted to $Ker(L-aI)^m$ is zero. Indeed it is, but for trivial reasons, since the whole source is the zero space.

Edit: The sequence stabilises since the dimension is bounded from above by the dimension of $V$. To see that the $m$ for which this happens is precisely the power in the minimal polynomial, one has to carefully analise the proof of the Caley Hamilton Theorem. This is done in any basic book on linear algebra. The key idea is that there is a squence of spaces with decreasing dimension which finally will be the image of the characteristic polynomial evaluated at $L$. This shows that the image is trivial. One observes carefully that we may reduce the power of the linear coefficients precisely to $m$ to obtain the similar result for the minimal polynomial.
But $\,L-aI\,$ cannot have nullspace equal to zero as it is a singular map: $$\exists\,0\neq v\in V\,\,s.t.\,\,Lv=av\Longleftrightarrow (L-aI)(v)=0\,$$ and this seems to disprove your idea.

Now, if $\,Tv=0\,$ for some lin. transf., then $\,T^n(v)=0\,,\,\forall n\in\mathbb{N}\,$ , so both $\,(L-aI)^m\,\,,\,\,(L-aI)^{m+1}\,$ vanish at $\,v\,$, but not that from one we can deduce the other.