Mixing
Let $Ain M_n(mathbb{Q})$ with $A^k=I_n$. If $j$ is a positive integer with $gcd(j,k)=1$, show that $...
$begingroup$
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
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 |
$begingroup$
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
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 |
$begingroup$
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
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
edited Jan 10 at 16:32
amWhy
1
asked Jan 10 at 10:59
w.sdkaw.sdka
36919
edited Jan 10 at 16:32
amWhy
1
asked Jan 10 at 10:59
w.sdkaw.sdka
36919
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
$begingroup$
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
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$
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
$begingroup$
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. $square$
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
$begingroup$
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
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$
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
$begingroup$
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
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$
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
$begingroup$
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
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$
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
$$
text{tr}(A) = sum_{i=1}^n lambda_i = sum_{i=1}^n lambda_i^j = text{tr}(A^j).
$$
answered Jan 10 at 11:40
SongSong
11.7k628
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
$begingroup$
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. $square$
answered Jan 10 at 12:38
loup blancloup blanc
23.1k21850
$endgroup$
add a comment |
$begingroup$
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. $square$
answered Jan 10 at 12:38
loup blancloup blanc
23.1k21850
$endgroup$
add a comment |
$begingroup$
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. $square$
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. $square$
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
answered Jan 10 at 12:38
loup blancloup blanc
23.1k21850
23.1k21850
add a comment |
add a comment |
