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I would like to have a result similar to "Traces of all positive powers of a matrix are zero implies it is nilpotent". Namely:

Let $R$ be a commutative $\mathbb{C}$-algebra, $A \in \mathcal{M}_n(R)$ such that $\mathrm{tr}(A)=\cdots=\mathrm{tr}(A^n)=0$. Does it follow that $A^n=0$?

I have no idea weather or not the analogy of eigenvalue exists for matrices over algebra so I can't simply generalize considerations from cited post. Any help will be appreciated.

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J.E.M.S
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    An analogue of eigenvalues does exist: The characteristic polynomial of $A$ is a monic polynomial over $R$ and thus can be factored into linear factors over its universal splitting algebra (which is a ring that contains $R$ as a subring and is finite free as an $R$-module). See Dan Laksov's Splitting algebras and Gysin homomorphisms ( https://people.kth.se/~laksov/art/splittingfroeberg.pdf ) for an introduction to splitting algebras, and Dan Laksov's Diagonalization of matrices over rings ( http://www.sciencedirect.com/science/article/pii/S0021869312005601 ) for an application ... – darij grinberg Jun 02 '16 at 04:04
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    ... to the use of eigenvalues over rings (i.e., pretty much what you are looking for). Other papers by Laksov (sometimes joint with Anders Thorup) are probably also relevant, although I don't know them well. Anyway, your claim has simpler proofs as well, but I'd be just as happy as you if anyone actually writes them up as opposed to handwaving them into existence. – darij grinberg Jun 02 '16 at 04:05
  • Oh, and $R$ does not have to be a $\mathbb{C}$-algebra. Being a $\mathbb{Q}$-algebra is perfectly enough. – darij grinberg Jun 02 '16 at 04:06
  • Thanks a lot, I think you can make an answear fromm this since it partially answears my question. – J.E.M.S Jun 02 '16 at 10:07

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The answer to your question is positive. For a self-contained proof, see Corollary 4.1 (b) in my note The trace Cayley-Hamilton theorem.

darij grinberg
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