# Are there matrices such that $(AB-BA)^{71}=I_{69}$?

Are there matrices $A,B \in \mathcal{M}_{69}(\mathbb{C})$ such that $$(AB-BA)^{71}=I_{69}?$$ Here $I_{69}$ denotes the $69 \times 69$ matrix with $1$ on its main diagonal and $0$ everywhere else.

My strong guess is that there are not. Let $C=AB-BA$. Then $\text{tr }C=0$ and $\text{tr }C^{71}=69$. If $\lambda$ is an eigenvalue of $C$, then $\lambda^{71}=1$. So \begin{align*} \lambda_1+\lambda_2+\dots+\lambda_{69}=0 \\ \lambda_1^{71}+\lambda_2^{71}+\dots+\lambda_{69}^{71}=69 \end{align*} I think some contradiction may come from here, but I don't see it yet.
I also tried working with polynomials. If $p=X^{71}-1$, then $p(C)=0$. If we worked in $\mathcal{M}_{69}(\mathbb{Q})$ it would have been easier because the minimal polynomial of $C$ divides $p$ and since $(X-1)$ is the only irreductible factor of $p$ with its degree less or equal than $69$, we would have that the minimal polynomial of $C$ is $(X-1)$ and so $C=I_{69}$, which leads to a contradiction after applying trace.

• Notably, a matrix can be expressed as $M = AB - BA$ if (and only if) it has trace $0$, as is proven here for example. – Omnomnomnom Mar 28 '18 at 22:31
• My gut feeling here is that the fact that $71$ is a prime number is the key point (otherwise, we would have other small irreducible factors for $p$) – Omnomnomnom Mar 28 '18 at 22:34
• Put it simply, when $p$ is a prime, primitive $p$-th roots of unity are $\mathbb Q$-linearly independent. – user1551 Mar 29 '18 at 4:06

## 1 Answer

Let $C=AB-BA$, and $P_C$ be the minimal polynomial of $C$.

Since $P(C)=C^{71}-I=0$ we get $P(x)=Q(x).P_C(x)$, but $P$ split in linear factors and square-free polynomial, on so on $P_c$, in particular we get that the spectrum of $C$ is a subset of $$\{1, Z,Z^2,\dots ,Z^{70} \}$$ where $Z=e^{\frac{2i\pi}{71}}$. Using that $\operatorname{Tr}(C)=0$ we get, $$0=\operatorname{Tr}(C)=\sum_{n\leq 70}\epsilon_nZ^n$$ Where $\epsilon_n\in \mathbb{N}$, such that $\sum \epsilon_n=69$. but since $71$ is prime, the minimal polynomial over the field of the rational numbers of $Z$ (i.e) the Cyclotomic polynomial is $\Phi_n(x)=\sum_{i=0}^{70} x^n$. which contradict our trace equation.

• Now that I think about it, this should also work when the dimension of the matrix is $70$, right? – AndrewC Mar 29 '18 at 15:55
• yes in fact it will be true for any dimension under $70$. – Hamza Mar 29 '18 at 17:55