# $A \in M_{n\times n}(\mathbb{Q})$ with $A\neq I_n$, $A^p=I_n$ then $p\leq n+1$.

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.)

• 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 $n+1$). Feb 4, 2022 at 12:12

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 $$m_A(t)\neq t-1$$.