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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...

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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.

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    Thank you, Simon, may I ask why it has to stabilize for $m$? So $(L-aI)^m$ kills off all the Jordan blocks with eigenvalue $a$, but then how does the conclusion follow?2012-05-21
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    Can the person who downvoted my answer please elaborate?2012-05-21
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    @Alagna I tried to explain it in the edit. This is pretty hard to do without repeating the whole proof. I hope this helps.2012-05-21
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    Thank you, Simon! Sorry for the delay in replying, I have been looking up the Cayley Hamilton theorem you mentioned :)2012-05-21
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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.

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    May I ask why the downvote? Thanks2012-05-21
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    You basically say that you can not deduce that $(L-aI)^m$ vanishes when $(L-aI)^{m+1}$ vanishes and vice versa. Well, one direction is _always_ true, the other one is true under the additional hypthesis you didn't use.2012-05-21
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    The direction the OP wanted implied, as he wrote, that $\, ker(L-aI)=0\,$, which is impossible as it is given that $\,a\,$ is an eigenvalue of $\,L\,$, and *this* is what I remarked.2012-05-21
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    The reason the OP mentioned is indeed not correct but she suspected it herself. Your answer doesn't clarify where her mistake was. The last line is just wrong and you provide no answer to the question.2012-05-21