Proof that if $\mathrm{tr}\,A^k=0$ for all $k=1,\ldots, n$, then $A^n = 0$

Question

Statement. Suppose we have a square matrix $A$ of order $n$ over a field $\mathbb{F}$ of characteristics $0$ or $p>n$. There is a theorem that if $\mathrm{tr}\,A^k=0$ for all $k=1,\ldots, n$, then $A$ is nilpotent and $A^n = 0$. The problem is to prove it.

What I did so far.

Let's consider characteristic polynomial $\chi(x)$ of our matrix $A$. The Newton identities relate $\mathrm{tr}\,A^k$ to the coefficients of $\chi(x)$: $$ s_k = \mathrm{tr}\,A^k $$
Since $\mathrm{tr}\,A^k=0$ for all $k=1,\ldots, n$, then $\chi(x) = x^n$ and (since we a over a field) all roots of $\chi$ are $0$. Therefor the only eigenvalue for $A$ is 0.

Now I have trouble proving first item and going from second to $A^n = 0$, How do I do those things?

score 1 · Accepted Answer · answered Sep 18 '15 at 07:56

1

Look up the Cayley-Hamilton Theorem.

answered Sep 18 '15 at 07:56

Christopher Carl Heckman

4,458

Then the OP shouldn't have any difficulty going from part (2) to the end result then. – Christopher Carl Heckman Sep 18 '15 at 08:02
The OP has trouble getting to part 2 altogether. Read the question. (But I'll admit I read too quickly too and thought the OP knew how to get from "$\chi_A = X^n$" to "$A$ is nilpotent") – Najib Idrissi Sep 18 '15 at 08:03
Am I OP? :) What's OP? – Glinka Sep 18 '15 at 11:36
Original Poster. In this case, you. – Christopher Carl Heckman Sep 19 '15 at 06:23

Proof that if $\mathrm{tr}\,A^k=0$ for all $k=1,\ldots, n$, then $A^n = 0$

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