Suppose that $A \in M_n(\mathbb{R})$. Prove that there is a $k \in \mathbb{Z}_{\geq 1}$ such that $Tr (A^k) \geq 0$.

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    $\begingroup$ Explaining my downvote: This post is phrased as an isolated "homework"-like problem without context and missing the part where you show us what you have done. $\endgroup$ Jul 8, 2019 at 11:46

2 Answers 2


For each eigenvalue $\lambda_j$ of $A$, write it as $r_je^{2\pi i\alpha_j}$ with $\alpha\in[0,1)$. Recalling that the eigenvalues of $A^k$ are $\lambda_j^k$, and that the trace is the sum of eigenvalues, we see that the real part of $\lambda_j^k$ contributes $r_j^k\cos2k\pi\alpha_j$ to $A^k$'s trace, the complex parts cancelling out.

By the simultaneous Dirichlet's approximation theorem, we can always find integers $n_j$ and a positive integer $k$ such that for all $j$ $$|k\alpha_j-n_j|\le\frac14$$ Rewriting this using $\bmod$, we can always find a single positive integer $k$ such that $$k\alpha_j\bmod1\in[0,1/4]\cup[3/4,1)$$ Once this $k$ is found, $\cos2k\pi\alpha_j\ge0$ for all $j$, whereby all the trace contributions, and thus $A^k$'s trace, become non-negative.

  • $\begingroup$ How exactly does this follows from simultaneous Dirichlet approximation theorem (as stated in Wiki for instance) ? This not crystal clear to me ... $\endgroup$
    – Olivier
    Jul 8, 2019 at 10:59
  • $\begingroup$ I do not exactly see $k\alpha_j \mod 1\leq1/2$, but that $k\alpha_j$ is centered around some $p_j$ with radius less or equal than $1/2$ (@Olivier pick $q:=k>0$ and mutliply the inequation by $q$, with $N=2^d$). Nevertheless I don't see that this condition gives nonnegativity: if $k\alpha_j=1$ then $\cos(k\alpha_j\pi)=-1$ ! We'd need some way to assure that all the $p_j$ can be even at the same time. $\endgroup$
    – Jose Brox
    Jul 8, 2019 at 11:04
  • $\begingroup$ @JoseBrox I had to do some polishing, please see edit. $\endgroup$ Jul 8, 2019 at 11:11
  • $\begingroup$ Yes, you nailed it now! I'm curious about the sharpness of the bound in the exponent $k$ such that tr$(A^k)\geq0$. With your proof we get $k\leq 2^{2n}$ where $n$ is the order of the matrix, by Dirichlet's approximation theorem. But we can easily show that for $n=2$ we have $k\leq3$... $\endgroup$
    – Jose Brox
    Jul 8, 2019 at 12:54

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.


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 $p_1=\cdots=p_n=-1,p_{n+1}=n$.

  • $\begingroup$ Nice !!! (although the use of word "elementary" when compared to the pigeonhole pple that underlies Dirichlet approximation seems brave to me) - but your bound is much better indeed. Is the use of such inequalities classical when dealing with the symmetric polynomials ? $\endgroup$
    – Olivier
    Jul 10, 2019 at 10:35
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    $\begingroup$ @Olivier , usually, we use the Newton's formulae (due also to Girard) to write relations as $P((x_i)_i)=0$ or $P((x_i)_i)\not= 0$ where $P$ is a symmetric polynomial. On the other hand, if the discriminant of a polynomial $\sum_i a_ix^i$ is a square in $\mathbb{Q}[(a_i)_i]$, then its Galois group cannot be $S_n$. Finally, the signum of the discriminant gives information about the roots; or, also, cf. Descartes’ Rule of Signs. But it is true that we do not often encounter questions using signs of symm. pol. In particular, I did not know the above result required by the OP. $\endgroup$
    – user91684
    Jul 11, 2019 at 16:16

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