Let $A \in M_{n\times n}(\mathbb{Q})$ with $A\neq I_n$. I want to prove the following: if $A^p =I_n$ for a prime $p$, then $p\leq n+1$.

Hmm, At this moment, I have no good idea for this problem. I just list some of my ideas (which seem not useful)....

If $A^p=I_n$, then $\det(A) \neq 0$ so $A$ is invertible.

For matrix over $\mathbb{C}$, $A^p-I=0$ implies $m_A(t)|t^p-1$, and for this case since $\mathbb{C}$ is algebraically closed, minimal polynomial factors out to monic polynomial and hence $A$ is diagonalizable. So $\lambda^p=1$... (This approach seems not good.)

  • $\begingroup$ Just a cultural point : if $p$ is not prime and is the smallest number such that $A^p = I_n$, then you still have a bound for $p$ that only depends on $n$ (but this bound is larger that $n+1$). $\endgroup$ Feb 4, 2022 at 12:12

1 Answer 1


Hint: the minimal polynomial $m_{A}(t)$ of $A$ has degree at most $n$ (why?) and has rational coefficients. On the other hand, $m_A(t)$ divides $t^p-1=(t-1)(t^{p-1}+\dots+t+1)$ where $t^{p-1}+\dots+t+1$ is irreducible over $\mathbb{Q}$ (why?). Thus, $\deg m_A(t)\geq p-1$ due to $m_A(t)\neq t-1$.


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