Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive integer with $gcd(j,k)=1$, show that $...

3

$begingroup$

Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive
integer with $gcd(j,k)=1$, show that $ operatorname{tr}(A)=
operatorname{tr}(A^j)$.

I don't know how to start to prove that. I tried to find the matrix $B$ similar with $A$, but I am stuck...

Thank you.

linear-algebra matrices trace

share|cite|improve this question

edited Jan 10 at 16:32

amWhy

1

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

$endgroup$

closed as off-topic by Saad, amWhy, Holo, Arnaud D., RRL Jan 10 at 17:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Holo, Arnaud D., RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

3

$begingroup$

Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive
integer with $gcd(j,k)=1$, show that $ operatorname{tr}(A)=
operatorname{tr}(A^j)$.

I don't know how to start to prove that. I tried to find the matrix $B$ similar with $A$, but I am stuck...

Thank you.

linear-algebra matrices trace

share|cite|improve this question

edited Jan 10 at 16:32

amWhy

1

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

$endgroup$

closed as off-topic by Saad, amWhy, Holo, Arnaud D., RRL Jan 10 at 17:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Holo, Arnaud D., RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

3

3

3

0

$begingroup$

Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive
integer with $gcd(j,k)=1$, show that $ operatorname{tr}(A)=
operatorname{tr}(A^j)$.

I don't know how to start to prove that. I tried to find the matrix $B$ similar with $A$, but I am stuck...

Thank you.

linear-algebra matrices trace

share|cite|improve this question

edited Jan 10 at 16:32

amWhy

1

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

$endgroup$

Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive
integer with $gcd(j,k)=1$, show that $ operatorname{tr}(A)=
operatorname{tr}(A^j)$.

I don't know how to start to prove that. I tried to find the matrix $B$ similar with $A$, but I am stuck...

Thank you.

linear-algebra matrices trace

linear-algebra matrices trace

share|cite|improve this question

edited Jan 10 at 16:32

amWhy

1

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

share|cite|improve this question

edited Jan 10 at 16:32

amWhy

1

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

share|cite|improve this question

share|cite|improve this question

edited Jan 10 at 16:32

amWhy

1

edited Jan 10 at 16:32

amWhy

1

edited Jan 10 at 16:32

amWhy

1

1

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

asked Jan 10 at 10:59

w.sdkaw.sdka

36919

36919

closed as off-topic by Saad, amWhy, Holo, Arnaud D., RRL Jan 10 at 17:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Holo, Arnaud D., RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Saad, amWhy, Holo, Arnaud D., RRL Jan 10 at 17:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, amWhy, Holo, Arnaud D., RRL

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

10

$begingroup$

Note that eigenvalues $lambda_i, i=1,ldots, n$ of $A$ are roots of the rational polynomial
$$
p(t) = det(tI-A)=prod_{i=1}^n (t-lambda_i).
$$ Since $Ax=lambda_i x$ implies $A^kx = x = lambda_i^k x$, we have ${lambda_i}subset mu_k ={zetainmathbb{C};|;zeta^k=1}$. Let $omega$ be the $k$-th primitive root of unity. Since $(j,k)=1$, we can define $sigmain text{Aut}(mathbb{Q}(mu_k)/mathbb{Q})$ by letting
$$
sigma(omega)= omega^j.
$$ Then, it holds
$$
p(t) = sigma(p(t)) = sigmaleft[prod_{i=1}^n (t-lambda_i)right]=prod_{i=1}^n (t-sigma(lambda_i))=prod_{i=1}^n (t-lambda_i^j).
$$ Now, it follows
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$

share|cite|improve this answer

edited Jan 10 at 17:34

answered Jan 10 at 11:40

SongSong

11.7k628

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
  $endgroup$
  – Tobias Kildetoft
  Jan 10 at 11:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this answer is wonderful..... I sincerely appreciate you!
  $endgroup$
  – w.sdka
  Jan 10 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
  $endgroup$
  – M. Van
  Jan 10 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
  $endgroup$
  – Song
  Jan 10 at 13:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
  $endgroup$
  – w.sdka
  Jan 10 at 13:43
  
  
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

We must use the fact that $Ain M_n(mathbb{Q})$ because the result is false over $M_n(mathbb{C})$. Indeed, consider $A=diag(e^{2ipi/3},i)$ where $k=12$; then $trace(A)not= trace(A^5)$.

$textbf{Proposition.}$ When $Ain M_n(mathbb{Q})$ and $(j,k)=1$, $spectrum(A)=spectrum(A^j)$.

$textbf{Proof}.$
We may assume that $order(A)=k$ ($k=min{l;A^l=I_2}$) and that $j<k$. Clearly, $order(A^j)=l$. $A$ is diagonalizable over $mathbb{C}$ and its minimal polynomial divides $x^k-1$.

Therefore, $chi_A$, the characteristic polynomial of $A$, is a product of (irreducible) cyclotomic polynomials $phi_a(x)=Pi_{(u,a)=1}(x-e^{2ipi u/a})inmathbb{Q}[x]$ where $a|k$. Since $(j,a)=1$, it is easy to see that $phi_a(x)=Pi_{(u,a)=1}(x-(e^{2ipi u/a})^j)$ and we are done. $square$

share|cite|improve this answer

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

10

$begingroup$

Note that eigenvalues $lambda_i, i=1,ldots, n$ of $A$ are roots of the rational polynomial
$$
p(t) = det(tI-A)=prod_{i=1}^n (t-lambda_i).
$$ Since $Ax=lambda_i x$ implies $A^kx = x = lambda_i^k x$, we have ${lambda_i}subset mu_k ={zetainmathbb{C};|;zeta^k=1}$. Let $omega$ be the $k$-th primitive root of unity. Since $(j,k)=1$, we can define $sigmain text{Aut}(mathbb{Q}(mu_k)/mathbb{Q})$ by letting
$$
sigma(omega)= omega^j.
$$ Then, it holds
$$
p(t) = sigma(p(t)) = sigmaleft[prod_{i=1}^n (t-lambda_i)right]=prod_{i=1}^n (t-sigma(lambda_i))=prod_{i=1}^n (t-lambda_i^j).
$$ Now, it follows
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$

share|cite|improve this answer

edited Jan 10 at 17:34

answered Jan 10 at 11:40

SongSong

11.7k628

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
  $endgroup$
  – Tobias Kildetoft
  Jan 10 at 11:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this answer is wonderful..... I sincerely appreciate you!
  $endgroup$
  – w.sdka
  Jan 10 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
  $endgroup$
  – M. Van
  Jan 10 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
  $endgroup$
  – Song
  Jan 10 at 13:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
  $endgroup$
  – w.sdka
  Jan 10 at 13:43
  
  
  

  
  
  
  

 | 
show 2 more comments

10

$begingroup$

Note that eigenvalues $lambda_i, i=1,ldots, n$ of $A$ are roots of the rational polynomial
$$
p(t) = det(tI-A)=prod_{i=1}^n (t-lambda_i).
$$ Since $Ax=lambda_i x$ implies $A^kx = x = lambda_i^k x$, we have ${lambda_i}subset mu_k ={zetainmathbb{C};|;zeta^k=1}$. Let $omega$ be the $k$-th primitive root of unity. Since $(j,k)=1$, we can define $sigmain text{Aut}(mathbb{Q}(mu_k)/mathbb{Q})$ by letting
$$
sigma(omega)= omega^j.
$$ Then, it holds
$$
p(t) = sigma(p(t)) = sigmaleft[prod_{i=1}^n (t-lambda_i)right]=prod_{i=1}^n (t-sigma(lambda_i))=prod_{i=1}^n (t-lambda_i^j).
$$ Now, it follows
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$

share|cite|improve this answer

edited Jan 10 at 17:34

answered Jan 10 at 11:40

SongSong

11.7k628

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
  $endgroup$
  – Tobias Kildetoft
  Jan 10 at 11:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this answer is wonderful..... I sincerely appreciate you!
  $endgroup$
  – w.sdka
  Jan 10 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
  $endgroup$
  – M. Van
  Jan 10 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
  $endgroup$
  – Song
  Jan 10 at 13:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
  $endgroup$
  – w.sdka
  Jan 10 at 13:43
  
  
  

  
  
  
  

 | 
show 2 more comments

10

10

10

$begingroup$

Note that eigenvalues $lambda_i, i=1,ldots, n$ of $A$ are roots of the rational polynomial
$$
p(t) = det(tI-A)=prod_{i=1}^n (t-lambda_i).
$$ Since $Ax=lambda_i x$ implies $A^kx = x = lambda_i^k x$, we have ${lambda_i}subset mu_k ={zetainmathbb{C};|;zeta^k=1}$. Let $omega$ be the $k$-th primitive root of unity. Since $(j,k)=1$, we can define $sigmain text{Aut}(mathbb{Q}(mu_k)/mathbb{Q})$ by letting
$$
sigma(omega)= omega^j.
$$ Then, it holds
$$
p(t) = sigma(p(t)) = sigmaleft[prod_{i=1}^n (t-lambda_i)right]=prod_{i=1}^n (t-sigma(lambda_i))=prod_{i=1}^n (t-lambda_i^j).
$$ Now, it follows
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$

share|cite|improve this answer

edited Jan 10 at 17:34

answered Jan 10 at 11:40

SongSong

11.7k628

$endgroup$

Note that eigenvalues $lambda_i, i=1,ldots, n$ of $A$ are roots of the rational polynomial
$$
p(t) = det(tI-A)=prod_{i=1}^n (t-lambda_i).
$$ Since $Ax=lambda_i x$ implies $A^kx = x = lambda_i^k x$, we have ${lambda_i}subset mu_k ={zetainmathbb{C};|;zeta^k=1}$. Let $omega$ be the $k$-th primitive root of unity. Since $(j,k)=1$, we can define $sigmain text{Aut}(mathbb{Q}(mu_k)/mathbb{Q})$ by letting
$$
sigma(omega)= omega^j.
$$ Then, it holds
$$
p(t) = sigma(p(t)) = sigmaleft[prod_{i=1}^n (t-lambda_i)right]=prod_{i=1}^n (t-sigma(lambda_i))=prod_{i=1}^n (t-lambda_i^j).
$$ Now, it follows
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$

share|cite|improve this answer

edited Jan 10 at 17:34

answered Jan 10 at 11:40

SongSong

11.7k628

share|cite|improve this answer

share|cite|improve this answer

edited Jan 10 at 17:34

edited Jan 10 at 17:34

edited Jan 10 at 17:34

answered Jan 10 at 11:40

SongSong

11.7k628

answered Jan 10 at 11:40

SongSong

11.7k628

answered Jan 10 at 11:40

SongSong

11.7k628

11.7k628

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
  $endgroup$
  – Tobias Kildetoft
  Jan 10 at 11:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this answer is wonderful..... I sincerely appreciate you!
  $endgroup$
  – w.sdka
  Jan 10 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
  $endgroup$
  – M. Van
  Jan 10 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
  $endgroup$
  – Song
  Jan 10 at 13:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
  $endgroup$
  – w.sdka
  Jan 10 at 13:43
  
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
  $endgroup$
  – Tobias Kildetoft
  Jan 10 at 11:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this answer is wonderful..... I sincerely appreciate you!
  $endgroup$
  – w.sdka
  Jan 10 at 12:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
  $endgroup$
  – M. Van
  Jan 10 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
  $endgroup$
  – Song
  Jan 10 at 13:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
  $endgroup$
  – w.sdka
  Jan 10 at 13:43
  
  
  

  
  
  
  

1

1

$begingroup$
This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
$endgroup$
– Tobias Kildetoft
Jan 10 at 11:46

$begingroup$
This is a nice answer that also nicely shows why we do need to do this over $mathbb{Q}$, since it will fail in suitable extensions.
$endgroup$
– Tobias Kildetoft
Jan 10 at 11:46

$begingroup$
I think this answer is wonderful..... I sincerely appreciate you!
$endgroup$
– w.sdka
Jan 10 at 12:00

$begingroup$
I think this answer is wonderful..... I sincerely appreciate you!
$endgroup$
– w.sdka
Jan 10 at 12:00

$begingroup$
why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
$endgroup$
– M. Van
Jan 10 at 13:19

$begingroup$
why does an eigenvalue $lambda_i$ occur with the same algebraic multiplicity as an eigenvalue $lambda_i^j$ of $A^j$?
$endgroup$
– M. Van
Jan 10 at 13:19

$begingroup$
@M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
$endgroup$
– Song
Jan 10 at 13:32

$begingroup$
@M.Van Because ${lambda_i} = {lambda_i^j}$ ..? Equality not as a set, but counted with multiplicity.
$endgroup$
– Song
Jan 10 at 13:32

$begingroup$
Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
$endgroup$
– w.sdka
Jan 10 at 13:43

$begingroup$
Umm... Could you explain why $lambda_i in mu_k$?.... why $prod_{i=1}^n lambda_i^k =1$ implies $lambda_i in mu_k$??
$endgroup$
– w.sdka
Jan 10 at 13:43

 | 
show 2 more comments

1

$begingroup$

We must use the fact that $Ain M_n(mathbb{Q})$ because the result is false over $M_n(mathbb{C})$. Indeed, consider $A=diag(e^{2ipi/3},i)$ where $k=12$; then $trace(A)not= trace(A^5)$.

$textbf{Proposition.}$ When $Ain M_n(mathbb{Q})$ and $(j,k)=1$, $spectrum(A)=spectrum(A^j)$.

$textbf{Proof}.$
We may assume that $order(A)=k$ ($k=min{l;A^l=I_2}$) and that $j<k$. Clearly, $order(A^j)=l$. $A$ is diagonalizable over $mathbb{C}$ and its minimal polynomial divides $x^k-1$.

Therefore, $chi_A$, the characteristic polynomial of $A$, is a product of (irreducible) cyclotomic polynomials $phi_a(x)=Pi_{(u,a)=1}(x-e^{2ipi u/a})inmathbb{Q}[x]$ where $a|k$. Since $(j,a)=1$, it is easy to see that $phi_a(x)=Pi_{(u,a)=1}(x-(e^{2ipi u/a})^j)$ and we are done. $square$

share|cite|improve this answer

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

$endgroup$

add a comment | 

1

$begingroup$

We must use the fact that $Ain M_n(mathbb{Q})$ because the result is false over $M_n(mathbb{C})$. Indeed, consider $A=diag(e^{2ipi/3},i)$ where $k=12$; then $trace(A)not= trace(A^5)$.

$textbf{Proposition.}$ When $Ain M_n(mathbb{Q})$ and $(j,k)=1$, $spectrum(A)=spectrum(A^j)$.

$textbf{Proof}.$
We may assume that $order(A)=k$ ($k=min{l;A^l=I_2}$) and that $j<k$. Clearly, $order(A^j)=l$. $A$ is diagonalizable over $mathbb{C}$ and its minimal polynomial divides $x^k-1$.

Therefore, $chi_A$, the characteristic polynomial of $A$, is a product of (irreducible) cyclotomic polynomials $phi_a(x)=Pi_{(u,a)=1}(x-e^{2ipi u/a})inmathbb{Q}[x]$ where $a|k$. Since $(j,a)=1$, it is easy to see that $phi_a(x)=Pi_{(u,a)=1}(x-(e^{2ipi u/a})^j)$ and we are done. $square$

share|cite|improve this answer

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

$endgroup$

add a comment | 

1

1

1

$begingroup$

We must use the fact that $Ain M_n(mathbb{Q})$ because the result is false over $M_n(mathbb{C})$. Indeed, consider $A=diag(e^{2ipi/3},i)$ where $k=12$; then $trace(A)not= trace(A^5)$.

$textbf{Proposition.}$ When $Ain M_n(mathbb{Q})$ and $(j,k)=1$, $spectrum(A)=spectrum(A^j)$.

$textbf{Proof}.$
We may assume that $order(A)=k$ ($k=min{l;A^l=I_2}$) and that $j<k$. Clearly, $order(A^j)=l$. $A$ is diagonalizable over $mathbb{C}$ and its minimal polynomial divides $x^k-1$.

Therefore, $chi_A$, the characteristic polynomial of $A$, is a product of (irreducible) cyclotomic polynomials $phi_a(x)=Pi_{(u,a)=1}(x-e^{2ipi u/a})inmathbb{Q}[x]$ where $a|k$. Since $(j,a)=1$, it is easy to see that $phi_a(x)=Pi_{(u,a)=1}(x-(e^{2ipi u/a})^j)$ and we are done. $square$

share|cite|improve this answer

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

$endgroup$

We must use the fact that $Ain M_n(mathbb{Q})$ because the result is false over $M_n(mathbb{C})$. Indeed, consider $A=diag(e^{2ipi/3},i)$ where $k=12$; then $trace(A)not= trace(A^5)$.

$textbf{Proposition.}$ When $Ain M_n(mathbb{Q})$ and $(j,k)=1$, $spectrum(A)=spectrum(A^j)$.

$textbf{Proof}.$
We may assume that $order(A)=k$ ($k=min{l;A^l=I_2}$) and that $j<k$. Clearly, $order(A^j)=l$. $A$ is diagonalizable over $mathbb{C}$ and its minimal polynomial divides $x^k-1$.

Therefore, $chi_A$, the characteristic polynomial of $A$, is a product of (irreducible) cyclotomic polynomials $phi_a(x)=Pi_{(u,a)=1}(x-e^{2ipi u/a})inmathbb{Q}[x]$ where $a|k$. Since $(j,a)=1$, it is easy to see that $phi_a(x)=Pi_{(u,a)=1}(x-(e^{2ipi u/a})^j)$ and we are done. $square$

share|cite|improve this answer

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

answered Jan 10 at 12:38

loup blancloup blanc

23.1k21850

23.1k21850

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here