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I'm looking for an elementary proof (one which does not use Galois theory). For the case $p = 3$, we have that $tr(A^3) = tr(A)^3 - 3e_1e_2 + 3e_3$ where the $e_i$ are coefficients of the characterstic polynomial of $A$, and are thus integers, so the result follows. I cannot see a way to generalize this for arbitrary $p$. Any help would be great!

Arturo Magidin
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John
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    It may help to consider the equivalent formulation $$ \operatorname{tr}(n ,A^p - \operatorname{tr}(A)^p I) = 0 $$ where $n$ is the size of the matrix – Ben Grossmann Dec 18 '18 at 03:31
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    I'm not sure whether my answer below is "elementary enough" (I'm still using the language of finite fields, but not any "deep theory" of). And Galois theory seems to be an overkill to apply here ;) – metamorphy Dec 18 '18 at 06:31
  • Note: The result is true for any matrix $A$ over a commutative ring, not just for integer matrices. For integer matrices, you can additionally simplify it by replacing $\left(\operatorname{Tr}A\right)^p$ as $\operatorname{Tr}A$ (since each integer $m$ satisfies $m^p \equiv m \mod p$). – darij grinberg Sep 27 '25 at 18:32

1 Answers1

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If $A,B$ are commuting square matrices over $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$ of the same size, then $$(A+B)^p=A^p+B^p$$ (the binomial formula is valid here, and $p\mid\binom{p}{k}$ for $0<k<p$). (The same holds for $A,B$ over $\mathbb{F}_p[\lambda]$, for the same reason.) It follows, for a matrix $A$ over $\mathbb{F}_p$, denoting by $\chi_A(\lambda)=\det(\lambda I-A)$ its characteristic polynomial, that $(\lambda I-A)^p=\lambda^p I-A^p$ and therefore $\chi_{A^p}(\lambda^p)=\big(\chi_A(\lambda)\big)^p$. But $\big(f(x)\big)^p=f(x^p)$ for any polynomial $f\in\mathbb{F}_p[x]$ (this is usually proven by induction on the degree of $f$ using the same "binomial argument" as above, and the fact that $a^p=a$ for any $a\in\mathbb{F}_p$).

Thus actually we have $\chi_{A^p}(\lambda^p)=\chi_A(\lambda^p)$, i.e. just $\color{blue}{\chi_{A^p}\equiv\chi_A}$. As a corollary (comparing the coefficients), we get $\operatorname{tr}(A^p)=\operatorname{tr}(A)$, and we're done (again by $a^p=a$ for any $a\in\mathbb{F}_p$).

metamorphy
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  • Could you explain why the line about the characteristic polynomials being equal? – John Dec 18 '18 at 06:47
  • I meant the equality between characteristic polynomials right after "and therefore ..." – John Dec 18 '18 at 06:53
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    Yes, I've edited this right before. We take $\det$ of the both sides of $(\lambda I-A)^p=\lambda^p I-A^p$. – metamorphy Dec 18 '18 at 06:55
  • Sorry, just saw that. It's clear now, thanks! – John Dec 18 '18 at 06:56
  • @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold. – metamorphy Dec 18 '18 at 08:01
  • Oops you are right. So I agree that looking at $(A+\lambda I)^p$ is the only way then – reuns Dec 18 '18 at 08:04
  • I think this argument needs some additional explanation because we end up with an argument that when two polynomial expressions take identical values, the underlying polynomials must have identical coefficients.

    This is true on $\mathbb R$ or $\mathbb C$, but in $\mathbb F_p$ not. E.g. $X^p$ and $X$ take identical values on $\mathbb F_p$ but do not have identical coefficients.

     – Chris Jan 14 '24 at 11:16
  • @Chris: I don't think so. The equalities above (including the last paragraph) that involve polynomials in $\lambda$ are in $\mathbb{F}_p[\lambda]$. That is, these are equalities of polynomials, not just of the values taken at any fixed $\lambda$. – metamorphy Jan 15 '24 at 05:23
  • "$\color{blue}{\chi_{A^p}\equiv\chi_A}$": This is not literally correct. You mean something slightly different. – darij grinberg Sep 27 '25 at 18:32