Properties of integer matrices $A$ such that $A^{p}=I$ for $p$ a prime integer.

Question

Problem Statement: Let $p$ be an integer prime, and let $A$ be an $n\times n$ integer matrix such that $A^{p} = I$ but $A \neq I$. Prove that $n \geq p − 1$.

We have been learning factoring of polynomials in $\mathbb{Z}[x], \mathbb{Q}[x], \mathbb{F}_{p}[x]$, etc. and my professor gave us this problem.

I have been trying to understand the structure of such a matrix $A$, but we have never discussed how the dimension of a matrix relates to powers of a matrix.

I have tried looking it in many different ways, and I am still not sure which direction to go into.

First, it is clear that $A$ is invertible, as $$1=\det(I)=\det(A^{p})=\det(A)^{p}$$ so either $\det(A)=1$ or $p=2$. But I don't think this is much help.

Also, since $A^{P}=I$, $A$ must be diagonalizable. I tried to break up the structure to find out more information, as follows: $$A^{\frac{p-1}{2}}AA^{\frac{p-1}{2}}=I$$ but this causes an issue, as $(A^{\frac{p-1}{2}})^{-1}\neq A^{\frac{p-1}{2}}$.

I also observed that we have $A^{p}-I=\mathbf{0}$, and $A^{p}-I$ factors as $$A^{p}-I=(A-I)(A^{p-1}+A^{p-2}+\cdots+A^{2}+A+I)=0.$$

We know $A-I\neq\mathbf{0}$, but that does not necessarily imply that $A^{p-1}+A^{p-2}+\cdots+A^{2}+A+I=\mathbf{0}$, correct?, because it is possible to have non-trivial products equal to $\mathbf{0}$. Either way, I still am not seeing a connection to the power of a matrix and its dimension.

I also observed that in that factorization $A^{p-1}+A^{p-2}+\cdots+A^{2}+A+I=\mathbf{0}$ takes the same form as the $p$-th cyclotonic polynomial, which we did learn about in class, but my book also does not cover much about it.

This chapter doesn't even mention matrices, which is why I am unsure how to approach this problem, as we have never discussed anything similar to this.

Any suggestions for how I should approach this?

Answer 1

The minimal polynomial of $A$, say $m_A$, divides $X^p-1=(X-1)(X^{p-1}+\cdots+X+1)$. Since $X-1$ and $X^{p-1}+\cdots+X+1$ are irreducible over $\mathbb Q$, we have three cases:
$m_A=X-1$, and then $A=I$, impossible.
$m_A=X^{p-1}+\cdots+X+1$. But $\deg m_A\le n$, so $p-1\le n$.
$m_A=X^p-1$. Then $n\ge\deg m_A=p>p-1$.

Answer 2

The minimal polynomial of such a matrix divides $X^p-1=(X-1)(X^{p-1}+\cdots+1)$. Since your matrix is not the identity, we see that its minimal polynomial must be divisible $X^{p-1}+\cdots+1$ because this polynomial is irreducible. In particular, its degree is at least $p-1$.

Now the minimal polynomial has degree at most the size $n$ of the matrix, because of the theorem of Cayley-Hamilton, so $p-1\leq n$, as you wanted.