I found out that the equation $(AB-BA)^m=I_n$ does have solution when $n=km$, where $k$ is an arbitrary integer number.

To prove, we just need to consider $C=$diag($r_1,..., r_m$) where $r_1,..., r_m$ are the roots of $x^m-1=0$, i.e., the eigenvalues of matrix $C$.

In this case we know trace($C$)=0, and so there is two matrices $A$ and $B$ such that $C=AB-BA$, and also it is vivid that $C^m=I_m$. For each $n=km$, we can duplicate the matrix $C$, $k$ times to get an $mk\times mk$ matrix to consider as a new matrix $C$.

Now, my question is this:

Does the equation $(AB-BA)^m=I_n$ have answer if and only if $n=mk$?

  • $\begingroup$ If $p$ is a number which divides $m$, then any solution to $(AB - BA)^{p} = I_n$ also solves $(AB - BA)^{m} = I_n$. So for instance, a solution to $(AB - BA)^{4}$ exists for all even integers. $\endgroup$ – Alex Zorn Oct 10 '15 at 16:48
  • $\begingroup$ However, I would guess that this statement is true when $m$ is prime. $\endgroup$ – Alex Zorn Oct 10 '15 at 16:48
  • $\begingroup$ The answers to your previous question (with $m=2$) have already shown that your hypothesis is true for some fields and false for some others. $\endgroup$ – user1551 Oct 10 '15 at 22:35
  • $\begingroup$ Thanks Alex. I had brought up my question there inside of an answer to more discuss, but according Mice Elf's comment, I asked the question here. Actually, the problem is discussing on Complex numbers as a fields. $\endgroup$ – Majid Oct 11 '15 at 19:56

As Alex wrote, let $m$ be a prime and assume that $char(K)=0$; $x^m-1=(x-1)f_m(x)$ where $f_m$ is irreducible over $\mathbb{Q}$. Let $1,s_2,\cdots,s_{m}$ be the roots of $x^m-1$. Note that $spectrum(C)=\{1,\alpha_1;s_2,\alpha_2; \cdots;s_{m},\alpha_{m}\}$ where the $(\alpha_i)_i$ are integers , $\alpha_2s_2+\cdots +\alpha_ms_m=-\alpha_1$ and $\sum_{i=2}^m\alpha_i=n-\alpha_1$. Since $s_2,\cdots,s_m$ is a free system over $\mathbb{Q}$, the relations $-1=s_2+\cdots+s_m=\alpha_2/\alpha_1s_2+\cdots+\alpha_m/\alpha_1s_m$ imply $\alpha_1=\cdots=\alpha_m$.

Then the characteristic polynomial of $C=AB-BA$ is $\chi_C(x)=(x^m-1)^{\alpha_1}$; therefore $n=\alpha_1m$.

  • $\begingroup$ Thank you so much. Actually, I am not that much interested in the answers of the equation. I just wanted to know that if the abovementioned claim correct. Now, I think it is correct when $m$ is a prime number. $\endgroup$ – Majid Oct 11 '15 at 19:59

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