# $\mathrm{tr}(A^4)$ is a rational number [closed]

Let $$A \in \mathcal{M}_3(\mathbb{C})$$ such that $$\mathrm{tr}(A^2) = \mathrm{tr}(A^3) \in \mathbb{Q},$$ where $$\mathrm{tr}(A)$$ is the trace of $$A.$$ It is possible to prove that $$\mathrm{tr}(A^4) \in \mathbb{Q}?$$

## closed as off-topic by Travis, Especially Lime, John B, Mars Plastic, user21820Sep 15 at 14:23

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• Let $x,y,z$ be the eigenvalues of $A$, may be repeated. Can you write down an expression for $tr(A^n)$ in terms of $x,y,z$? – астон вілла олоф мэллбэрг Sep 11 at 4:33
• $tr(A^{n})= x^n+y^n+z^n$ – gaurav saini Sep 11 at 5:00

We fix some $$t\in \Bbb C$$. Let $$a,b,c$$ be the complex roots of the equation $$X^3-6tX^2+18t^2 X -36 t^3\ .$$ Let $$e_n$$ be the elementary symmetric polynomial of degree $$n$$ in $$a,b,c$$.
Let $$p_n$$ be the Newton symmetric polynomial of degree $$n$$ in $$a,b,c$$.
By Vieta, \left\{ \begin{aligned} e_1 &= a+b+c = 6t\ ,\\ e_2 &= ab+bc+ca = 18t^2\ ,\\ e_3 &= abc = 36t^3\ . \end{aligned} \right. Let $$A$$ be the matrix with diagonal entries $$a,b,c$$. Then, using the Newton identities, \begin{aligned} \operatorname{Trace}(A^2) &=a^2+b^2+c^2\\ &=p_2\\ &=e_1p_1-2e_2\\ &=6t\cdot 6t-2\cdot 18t^2=0\ , \\ \operatorname{Trace}(A^3) &=a^3+b^3+c^3\\ &=p_3\\ &=e_1p_2-e_2p_1+3e_3\\ &=0-18t^2\cdot 6t+3\cdot 36t^3=0\ , \\ \operatorname{Trace}(A^4) &=a^4+b^4+c^4\\ &=p_4\\ &=e_1p_3-e_2p_2+e_3p_1-4e_4\\ &=0-0+36t^3\cdot 6t-0\\ &=216t^4\ . \end{aligned} It is clear that we can chose $$t$$, such that the last expression is not rational. This gives and explicit counterexample.
Later edit: The general family of cubics with roots $$a,b,c$$, so that $$p_1=p_1(a,b,c)=e_1=e_1(a,b,c)=s$$ and $$p_2=p_3=p_2(a,b,c) =p_3(a,b,c)=q$$, thus parametrized by $$s,q$$, is obtained by solving \begin{aligned} e_1&=p_1=s\ ,\\ 2e_2 &=e_1p_1-p_2=s^2-q\ ,\\ 3e_3 &=e_2p_1-e_1p_2+p_3=\frac 12(s^2-q)s-sq+q\ ,\\ \end{aligned} and $$p_4$$ is obtained algebraically as $$p_4=e_1p_3-e_2p_2+e_3p_1-4e_4$$. (Here, $$e_4=0$$.) The question "It is still possible to find a counterexample so that $$s$$ (trace of $$A$$) is a natural (or even rational) number?" has a negative answer. If both $$s,q$$ are rational, than $$p_4$$ is rational, as an algebraic expression in $$s,q$$.
• Thanks! Nice solution! It is still possible to find a counterexample even that $\mathrm{tr}(A) \in \mathbb{N}^{*}?$ – 674123173797 - 4 Sep 11 at 10:25
• @674123173797-4 No - by those same Newton identities, $\mathrm{tr}(A^4)$ is a polynomial in $\mathrm{tr}(A^3)$, $\mathrm{tr}(A^2)$ and $\mathrm{tr}(A)$. – Steven Stadnicki Sep 11 at 17:09
• @Steven Stadnicki. Yes, I realized this later, and the same argument show that $\mathrm{tr}(A^k) \in \mathbb{Q}$ for all $k \in \mathbb{N}.$ – 674123173797 - 4 Sep 12 at 4:09