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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.

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    $\begingroup$ Notably, a matrix can be expressed as $M = AB - BA$ if (and only if) it has trace $0$, as is proven here for example. $\endgroup$ Mar 28, 2018 at 22:31
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    $\begingroup$ 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 $p$) $\endgroup$ Mar 28, 2018 at 22:34
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    $\begingroup$ Put it simply, when $p$ is a prime, primitive $p$-th roots of unity are $\mathbb Q$-linearly independent. $\endgroup$
    – user1551
    Mar 29, 2018 at 4:06

1 Answer 1

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Let $C=AB-BA$, and $P_C$ be the minimal polynomial of $C$.
Let $P(x)=x^{71}-1$. Since $P(C)=C^{71}-I=0$ we get $P_C(x)\mid P(x)$. But $P$ splits into linear factors and it's a square-free polynomial, and so is $P_C$. 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 rational numbers of $Z$, i.e., the Cyclotomic polynomial is $\Phi_n(x)=\sum_{i=0}^{70} x^n$. which contradict our trace equation.

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    $\begingroup$ Now that I think about it, this should also work when the dimension of the matrix is $70$, right? $\endgroup$
    – AndrewC
    Mar 29, 2018 at 15:55
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    $\begingroup$ yes in fact it will be true for any dimension under $70$. $\endgroup$
    – Hamza
    Mar 29, 2018 at 17:55

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