# Trace of power of a real matrix

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

• 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. Jul 8, 2019 at 11:46

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.

• How exactly does this follows from simultaneous Dirichlet approximation theorem (as stated in Wiki for instance) ? This not crystal clear to me ... Jul 8, 2019 at 10:59
• 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. Jul 8, 2019 at 11:04
• @JoseBrox I had to do some polishing, please see edit. Jul 8, 2019 at 11:11
• 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 $k\leq3$... 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.

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

• 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 ? Jul 10, 2019 at 10:35
• @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.
– user91684
Jul 11, 2019 at 16:16