ad (A^n) is a polynomial in ad A ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:58:55Z http://mathoverflow.net/feeds/question/101023 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101023/ad-an-is-a-polynomial-in-ad-a ad (A^n) is a polynomial in ad A ? darij grinberg 2012-06-30T22:41:27Z 2012-07-01T10:14:33Z <p>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 $\mathrm{End}_k\left(\mathrm{M}_n\left(k\right)\right)$.</p> <p>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 $P\left(0\right)=0$ ?</p> <p>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).</p> http://mathoverflow.net/questions/101023/ad-an-is-a-polynomial-in-ad-a/101055#101055 Answer by Konstantin Ardakov for ad (A^n) is a polynomial in ad A ? Konstantin Ardakov 2012-07-01T10:14:33Z 2012-07-01T10:14:33Z <p>I don't think it's true even when $A$ is diagonalisable.</p> <p>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)$, so </p> <p>$P(\rm{ad}$$(A)) = \rm{diag}$$ (P (\lambda_i - \lambda_j) )$</p> <p>for any polynomial $P(t)$.</p> <p>On the other hand, $A^m = \rm{diag}$$(\lambda_i^m) so</p> <p>\rm{ad}$$(A^m) = \rm{diag}$$(\lambda_i^m - \lambda_j^m).</p> <p>So if P(\rm{ad}$$(A)) = $$\rm{ad}$$(A^m)$ then $P(\lambda_i - \lambda_j) = \lambda_i^m -\lambda_j^m$ for all $i,j$ such that $1\leq i,j\leq n$. </p> <p>But it can happen that $\lambda_i - \lambda_j = \lambda_a - \lambda_b$ for two pairs of indices $(i,j)$ and $(a,b)$ with $\lambda_i^m - \lambda_j^m \neq \lambda_a^m - \lambda_b^m$.</p> <p>For a concrete example we can take $A = \rm{diag}$$(2,1,4,3)$.</p>