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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?

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    $\begingroup$ Interesting but tricky? I don‘t get that part. $\endgroup$ – Keba Jan 8 '15 at 0:34
  • $\begingroup$ Probably just missing something really simple, perhaps not that tricky. Interesting might be the better way to put it. $\endgroup$ – Libertron Jan 8 '15 at 0:36
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    $\begingroup$ "Interesting but tricky" somewhat implies that normally "interesting" and "tricky" don‘t go well together. I don‘t agree with that. ;) $\endgroup$ – Keba Jan 8 '15 at 0:37
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    $\begingroup$ I changed it to "interesting and fun." $\endgroup$ – Libertron Jan 8 '15 at 0:39
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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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All eigenvalues of $B$ are (complex) fifth roots of unity, and since $1$ is excluded they must be $\omega,\omega^{-1},\omega^2,\omega^{-2}$, where $$\omega=e^{2\pi i/5}\ .$$ The characteristic polynomial of $B$ has the form $$(z-\omega)^a(z-\omega^{-1})^b(z-\omega^2)^c(z-\omega^{-2})^d\ ,$$ where $a,b,c,d$ are non-negative integers and $a+b+c+d=n$. However this polynomial must have rational coefficients, which forces $a=b=c=d$; so $n=4a$ is a multiple of $4$.

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  • $\begingroup$ Oh that's right. $x^4 + x^3 + x^2 + x+1$ is the fifth cyclotomic polynomial and the minimal polynomial of primitive fifth root of unity. $\endgroup$ – Libertron Jan 8 '15 at 0:42
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    $\begingroup$ Not sure that's exactly the point. If $B$ were allowed to have complex entries then $a,b,c,d$ could be anything. And if it were allowed to have real entries, then $a=b$ and $c=d$ but there would be no reason for all four to be equal. So it is essential that we are talking about a matrix with rational entries. $\endgroup$ – David Jan 8 '15 at 0:44
  • $\begingroup$ To be clear, your argument only works for $M_{n}(\mathbb{C})$, am I correct? $\endgroup$ – Libertron Jan 8 '15 at 1:02
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    $\begingroup$ No, it only works for $M_n(\Bbb Q)$. In $M_n(\Bbb C)$ we could have any value for $n$, in $M_n(\Bbb R)$ any even value. $\endgroup$ – David Jan 8 '15 at 1:07

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