# How to prove that this linear operator is nilpotent?

Let $A\in M_n(\mathbb{C})$ be an arbitrary matrix , $\mathbb{C}$ is complex fields, and $L$ a mapping that is defined by

$L:M_n(\mathbb{C})\to M_n(\mathbb{C})$, $L(X):=AX+XA$. How can we show that $A$ is nilpotent iff $L$ be nilpotent?

Thanks for any hint.

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Is nilpotent here of order $k$ or do you define just for $k=2$? – Git Gud Jan 19 '13 at 18:45
it sounds k=2 Icompute $L^i$ for i=1,2,3,4 but why for i greater than 2 is not nilpotent? – Maisam Hedyelloo Jan 19 '13 at 18:53
Not sure what you mean. What is your definition of nilpotent? – Git Gud Jan 19 '13 at 19:00
i know definition of nilpotent($\exists k \in N$ {$A^k=0$} but my qustion is why nilpotently order is not K≥3 – Maisam Hedyelloo Jan 19 '13 at 19:06
It can happen that $k\ge 3$, it doesn't have to be $k=2$. – Git Gud Jan 19 '13 at 19:07

First, show that for $i=1,2,\ldots$, there exist some positive integers $c_{i0},\ldots,c_{ii}$ such that $L^i(X) = \sum_{j=0}^i c_{ij} A^j X A^{i-j}$.

Now, if $A^k=0$, show that $L^{2k-1}=0$.

Conversely, if $L^i=0$ for some natural number $i$, by considering $L^i(A)$, show that $A^{i+1}=0$.

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+1 you did it nice as well. – Babak S. Jan 19 '13 at 19:31
• Look at L(A)
• Look at LL(X), LLL(X), and maybe LLLL(X).
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Is it useful here that the $M_n(\mathbb C)$ is an PI ring? – Babak S. Jan 19 '13 at 18:43
you mean let $A^k=0$ and compute $L^i(x)$ to find nilpotently order of L and vice versa – Maisam Hedyelloo Jan 19 '13 at 18:45
+1 What a succinct answer. – user1551 Jan 19 '13 at 19:29
@Maisam: Yep; once you've looked at them, that's the next idea to have. I tried to keep the hint minimal both for the sake of being a hint to give you the chance to make the connections, and to emphasize the phase of problem solving where you are still finding leads to explore without actually knowing where they're going to lead yet. – Hurkyl Jan 19 '13 at 19:40