Prove that $tr(A)^p = tr(A^p)$ $mod$ $p$ where $A$ is a square integer matrix and $p$ is a prime number.












4












$begingroup$


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!






linear-algebra matrices modular-arithmetic trace







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    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
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 3:31








  • 1




    $begingroup$
    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 ;)
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:31
















4












$begingroup$


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!






linear-algebra matrices modular-arithmetic trace







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    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
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 3:31








  • 1




    $begingroup$
    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 ;)
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:31














4












4








4


2



$begingroup$


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!






linear-algebra matrices modular-arithmetic trace







share|cite|improve this question









$endgroup$




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!






linear-algebra matrices modular-arithmetic trace




linear-algebra matrices modular-arithmetic trace






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 2:29









Saad Saad

580211




580211








  • 1




    $begingroup$
    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
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 3:31








  • 1




    $begingroup$
    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 ;)
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:31














  • 1




    $begingroup$
    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
    $endgroup$
    – Omnomnomnom
    Dec 18 '18 at 3:31








  • 1




    $begingroup$
    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 ;)
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:31








1




1




$begingroup$
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
$endgroup$
– Omnomnomnom
Dec 18 '18 at 3:31






$begingroup$
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
$endgroup$
– Omnomnomnom
Dec 18 '18 at 3:31






1




1




$begingroup$
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 ;)
$endgroup$
– metamorphy
Dec 18 '18 at 6:31




$begingroup$
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 ;)
$endgroup$
– metamorphy
Dec 18 '18 at 6:31










1 Answer
1






active

oldest

votes


















4












$begingroup$

If $A,B$ are commuting square matrices over $mathbb{F}_p=mathbb{Z}/pmathbb{Z}$ of the same size, then
$$(A+B)^p=A^p+B^p$$
(the binomial formula is valid here, and $pmidbinom{p}{k}$ 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 $finmathbb{F}_p[x]$ (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 $ainmathbb{F}_p$).



Thus actually we have $chi_{A^p}(lambda^p)=chi_A(lambda^p)$, i.e. just $color{blue}{chi_{A^p}equivchi_A}$. 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 $ainmathbb{F}_p$).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain why the line about the characteristic polynomials being equal?
    $endgroup$
    – Saad
    Dec 18 '18 at 6:47










  • $begingroup$
    I meant the equality between characteristic polynomials right after "and therefore ..."
    $endgroup$
    – Saad
    Dec 18 '18 at 6:53






  • 1




    $begingroup$
    Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:55












  • $begingroup$
    Sorry, just saw that. It's clear now, thanks!
    $endgroup$
    – Saad
    Dec 18 '18 at 6:56










  • $begingroup$
    @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 8:01













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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

votes









4












$begingroup$

If $A,B$ are commuting square matrices over $mathbb{F}_p=mathbb{Z}/pmathbb{Z}$ of the same size, then
$$(A+B)^p=A^p+B^p$$
(the binomial formula is valid here, and $pmidbinom{p}{k}$ 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 $finmathbb{F}_p[x]$ (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 $ainmathbb{F}_p$).



Thus actually we have $chi_{A^p}(lambda^p)=chi_A(lambda^p)$, i.e. just $color{blue}{chi_{A^p}equivchi_A}$. 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 $ainmathbb{F}_p$).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain why the line about the characteristic polynomials being equal?
    $endgroup$
    – Saad
    Dec 18 '18 at 6:47










  • $begingroup$
    I meant the equality between characteristic polynomials right after "and therefore ..."
    $endgroup$
    – Saad
    Dec 18 '18 at 6:53






  • 1




    $begingroup$
    Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:55












  • $begingroup$
    Sorry, just saw that. It's clear now, thanks!
    $endgroup$
    – Saad
    Dec 18 '18 at 6:56










  • $begingroup$
    @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 8:01


















4












$begingroup$

If $A,B$ are commuting square matrices over $mathbb{F}_p=mathbb{Z}/pmathbb{Z}$ of the same size, then
$$(A+B)^p=A^p+B^p$$
(the binomial formula is valid here, and $pmidbinom{p}{k}$ 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 $finmathbb{F}_p[x]$ (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 $ainmathbb{F}_p$).



Thus actually we have $chi_{A^p}(lambda^p)=chi_A(lambda^p)$, i.e. just $color{blue}{chi_{A^p}equivchi_A}$. 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 $ainmathbb{F}_p$).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you explain why the line about the characteristic polynomials being equal?
    $endgroup$
    – Saad
    Dec 18 '18 at 6:47










  • $begingroup$
    I meant the equality between characteristic polynomials right after "and therefore ..."
    $endgroup$
    – Saad
    Dec 18 '18 at 6:53






  • 1




    $begingroup$
    Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:55












  • $begingroup$
    Sorry, just saw that. It's clear now, thanks!
    $endgroup$
    – Saad
    Dec 18 '18 at 6:56










  • $begingroup$
    @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 8:01
















4












4








4





$begingroup$

If $A,B$ are commuting square matrices over $mathbb{F}_p=mathbb{Z}/pmathbb{Z}$ of the same size, then
$$(A+B)^p=A^p+B^p$$
(the binomial formula is valid here, and $pmidbinom{p}{k}$ 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 $finmathbb{F}_p[x]$ (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 $ainmathbb{F}_p$).



Thus actually we have $chi_{A^p}(lambda^p)=chi_A(lambda^p)$, i.e. just $color{blue}{chi_{A^p}equivchi_A}$. 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 $ainmathbb{F}_p$).






share|cite|improve this answer











$endgroup$



If $A,B$ are commuting square matrices over $mathbb{F}_p=mathbb{Z}/pmathbb{Z}$ of the same size, then
$$(A+B)^p=A^p+B^p$$
(the binomial formula is valid here, and $pmidbinom{p}{k}$ 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 $finmathbb{F}_p[x]$ (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 $ainmathbb{F}_p$).



Thus actually we have $chi_{A^p}(lambda^p)=chi_A(lambda^p)$, i.e. just $color{blue}{chi_{A^p}equivchi_A}$. 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 $ainmathbb{F}_p$).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 18:23

























answered Dec 18 '18 at 6:20









metamorphymetamorphy

3,6821621




3,6821621












  • $begingroup$
    Could you explain why the line about the characteristic polynomials being equal?
    $endgroup$
    – Saad
    Dec 18 '18 at 6:47










  • $begingroup$
    I meant the equality between characteristic polynomials right after "and therefore ..."
    $endgroup$
    – Saad
    Dec 18 '18 at 6:53






  • 1




    $begingroup$
    Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:55












  • $begingroup$
    Sorry, just saw that. It's clear now, thanks!
    $endgroup$
    – Saad
    Dec 18 '18 at 6:56










  • $begingroup$
    @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 8:01




















  • $begingroup$
    Could you explain why the line about the characteristic polynomials being equal?
    $endgroup$
    – Saad
    Dec 18 '18 at 6:47










  • $begingroup$
    I meant the equality between characteristic polynomials right after "and therefore ..."
    $endgroup$
    – Saad
    Dec 18 '18 at 6:53






  • 1




    $begingroup$
    Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 6:55












  • $begingroup$
    Sorry, just saw that. It's clear now, thanks!
    $endgroup$
    – Saad
    Dec 18 '18 at 6:56










  • $begingroup$
    @reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 8:01


















$begingroup$
Could you explain why the line about the characteristic polynomials being equal?
$endgroup$
– Saad
Dec 18 '18 at 6:47




$begingroup$
Could you explain why the line about the characteristic polynomials being equal?
$endgroup$
– Saad
Dec 18 '18 at 6:47












$begingroup$
I meant the equality between characteristic polynomials right after "and therefore ..."
$endgroup$
– Saad
Dec 18 '18 at 6:53




$begingroup$
I meant the equality between characteristic polynomials right after "and therefore ..."
$endgroup$
– Saad
Dec 18 '18 at 6:53




1




1




$begingroup$
Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
$endgroup$
– metamorphy
Dec 18 '18 at 6:55






$begingroup$
Yes, I've edited this right before. We take $det$ of the both sides of $(lambda I-A)^p=lambda^p I-A^p$.
$endgroup$
– metamorphy
Dec 18 '18 at 6:55














$begingroup$
Sorry, just saw that. It's clear now, thanks!
$endgroup$
– Saad
Dec 18 '18 at 6:56




$begingroup$
Sorry, just saw that. It's clear now, thanks!
$endgroup$
– Saad
Dec 18 '18 at 6:56












$begingroup$
@reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
$endgroup$
– metamorphy
Dec 18 '18 at 8:01






$begingroup$
@reuns: for non-commuting $X,Y$, $(X+Y)^p=X^p+Y^p$ may fail to hold.
$endgroup$
– metamorphy
Dec 18 '18 at 8:01




















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