<s>[INST] Interesting and fun eigenvalue problem for a matrix $B \in M_{n}(\mathbb{Q})$ I've been thinking about ways on how to tackle this particular linear algebra problem related to eigenvalues from an old algebra prelim. The problem is that suppose we are given a matrix $B \in M_{n}(\mathbb{Q})$ such that $B^5 =I$, i.e. the $n \times n$ identity matrix, and no eigenvalue of $B$ is equal to $1$. We have to show that $n$ is divisible by $4$.  
My attempt: I take it that $B^5 =I \implies B^5 -I=0$ (the zero matrix). Now by definition, the eigenvalues of a matrix $B$ are the roots of the characteristic polynomial $c_B(x)$ of $B$ and we also know that $c_B(x)$ annihilates the matrix $B$, i.e. $c_B(B)=0$. Now in this problem, the key condition is that $1$ cannot be an eigenvalue of $B$, so does this mean, via the factorization $$x^5 -1= (x-1)(x^4 +x^3 +x^2 +x+1)$$ over $\mathbb{Q}$, that $x^4 + x^3 + x^2 +x+1$ divides $c_B(x)$, which has degree $n$? Is this enough to say that $n$ is divisible by $4$, or are there any other necessary things to be considered carefully before coming to that conclusion? 
 [/INST] Since $B^5 = I$, the minimal polynomial of $B$ over the rationals divides 
$x^5 - 1$.  Now $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$, where 
$x^4 + x^3 + x^2 + x + 1$ is irreducible (it's the fifth cyclotomic polynomial), and since $1$ is not an eigenvalue the minimal polynomial can only be $x^4 + x^3 + x^2 + x + 1$.  All irreducible factors of the characteristic polynomial divide the minimal polynomial, so the characteristic polynomial is 
a power of $x^4 + x^3 + x^2 + x + 1$, and this implies that $n$ is a multiple of $4$.
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