TITLE: Trace of power of a real matrix
QUESTION [7 upvotes]: Suppose that $A \in M_n(\mathbb{R})$. Prove that there is a $k \in \mathbb{Z}_{\geq 1}$ such that $Tr (A^k) \geq 0$.

REPLY [5 votes]: Parcly Taxel gave a pretty solution; Yet, there is an elementary one.
We use the notations of the wikipedia note about the Newton formula; cf.
https://en.wikipedia.org/wiki/Newton%27s_identities
Let $(x_i)_i$ be the roots of a real polynomial, $(e_i)_i$ be the associated elementary symmetric polynomials and $(p_i)_i$ be the associated power sums. Note that $e_k=0$ when $k>n$.
$\textbf{Proposition 1}$. If $p_1<0,\cdots, p_n<0$, then $p_{n+1}\geq 0$.
$\textbf{Proof}$. Assume that $p_{n+1}<0$. According to the newton formulae, $e_1<0,e_2>0,\cdots,(-1)^ne_n>0,(-1)^{n+1}e_{n+1}>0$. In particular, $e_{n+1}\not= 0$, a contradiction.
EDIT. Proposition 1 shows that at least one element of the $(tr(A^k))_{k\leq n+1}$ is $\geq 0$. The following shows that we cannot do better than $n+1$.
$\textbf{Proposition 2}$. The polynomial $Q_n(x)=x^n+x^{n-1}+\cdots+1$ satisfies $p_1=\cdots=p_n=-1,p_{n+1}=n$.