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Let $k$ be a field and $n$ a nonnegative integer. For any matrix $U\in\mathrm{M}_n\left(k\right)$, let $\mathrm{ad} U$ denote the map $\mathrm{M}_n\left(k\right)\to \mathrm{M}_n\left(k\right),\ V\mapsto UV-VU$. Thus, $\mathrm{ad} U$ is an element of the $k$-algebra $\mathrm{End}_k\left(\mathrm{M}_n\left(k\right)\right)$.

Is it true that for every $n\times n$-matrix $A$ over $k$, and for every $m\in\mathbb N$, the endomorphism $\mathrm{ad}\left(A^m\right)$ can be written in the form $P\left(\mathrm{ad}A\right)$ for some polynomial $P\in k\left[X\right]$ satisfying $P\left(0\right)=0$ ?

I know that this holds when $A$ is diagonalizable, and in that case it is used in the proof of Cartan's Lemma from Lie algebra theory. If it holds generally and can be proven neatly, it could be used to tidy up the proof of Cartan's Lemma (which, in the form I know it, is rather ugly, requiring an algebraic extension of the ground field and the use of Jordan's normal form).

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To check whether it is true, I would try to use the Jordan normal form of a matrix –  user24527 Jul 1 '12 at 5:51
    
NN: Unfortunately this isn't a fact which clearly holds for a matrix if it holds for each Jordan block. –  darij grinberg Jul 1 '12 at 8:42
    
is it on purpose that the power ($n$ in $ad(A^n)$) coincides with the size $n$ of the matrix? –  Yves Cornulier Jul 1 '12 at 10:07
    
Oops! No! Thanks for poinitng out the typo. –  darij grinberg Jul 1 '12 at 10:08
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up vote 7 down vote accepted

I don't think it's true even when $A$ is diagonalisable.

Suppose that $A$ is a diagonal matrix, with $(i,i)$th entry $\lambda_i$; thus $A = \rm{diag}$$(\lambda_i)$. Then we can write $\rm{ad}(A) = \rm{diag}$$(\lambda_i - \lambda_j)$, so

$P(\rm{ad}$$(A)) = \rm{diag}$$ (P (\lambda_i - \lambda_j) )$

for any polynomial $P(t)$.

On the other hand, $A^m = \rm{diag}$$(\lambda_i^m)$ so

$\rm{ad}$$(A^m) = \rm{diag}$$(\lambda_i^m - \lambda_j^m)$.

So if $P(\rm{ad}$$(A)) = $$\rm{ad}$$(A^m)$ then $P(\lambda_i - \lambda_j) = \lambda_i^m -\lambda_j^m$ for all $i,j$ such that $1\leq i,j\leq n$.

But it can happen that $\lambda_i - \lambda_j = \lambda_a - \lambda_b$ for two pairs of indices $(i,j)$ and $(a,b)$ with $\lambda_i^m - \lambda_j^m \neq \lambda_a^m - \lambda_b^m$.

For a concrete example we can take $A = \rm{diag}$$(2,1,4,3)$.

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Sorry! My argument was completely wrong. –  darij grinberg Jul 1 '12 at 10:38
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