Classify all $Ain M_5(mathbbQ):A^8=I$ and $A^4not=I$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Classify all $Ain M_5(mathbbQ):A^8=I$ and $A^4not=I$
Attempt
Since $A^8=I$ it must be $m_A(x)|x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$. But $A^4not=I$ so $m_A(x)not=x-1,x+1,x^2-1,x^2+1,x^4-1,(x-1)(x^2+1),(x+1)(x^2+1)$
And $Ain M_5(mathbbQ)Rightarrow deg(chi_A(x))=5$
How should I continue? Hoe would you approach this question?
I want to find the similarity classes using rational normal form.
I know that $chi_A(x)=$product of the prime divisors of $A$ and $m_A(x)=$lcm of the prime divisors of $A$
linear-algebra abstract-algebra matrices
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
asked Sep 5 at 6:03
giannispapav
1,340223
add a comment |Â
up vote
2
down vote
favorite
Classify all $Ain M_5(mathbbQ):A^8=I$ and $A^4not=I$
Attempt
Since $A^8=I$ it must be $m_A(x)|x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$. But $A^4not=I$ so $m_A(x)not=x-1,x+1,x^2-1,x^2+1,x^4-1,(x-1)(x^2+1),(x+1)(x^2+1)$
And $Ain M_5(mathbbQ)Rightarrow deg(chi_A(x))=5$
How should I continue? Hoe would you approach this question?
I want to find the similarity classes using rational normal form.
I know that $chi_A(x)=$product of the prime divisors of $A$ and $m_A(x)=$lcm of the prime divisors of $A$
linear-algebra abstract-algebra matrices
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
asked Sep 5 at 6:03
giannispapav
1,340223
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Classify all $Ain M_5(mathbbQ):A^8=I$ and $A^4not=I$
Attempt
Since $A^8=I$ it must be $m_A(x)|x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$. But $A^4not=I$ so $m_A(x)not=x-1,x+1,x^2-1,x^2+1,x^4-1,(x-1)(x^2+1),(x+1)(x^2+1)$
And $Ain M_5(mathbbQ)Rightarrow deg(chi_A(x))=5$
How should I continue? Hoe would you approach this question?
I want to find the similarity classes using rational normal form.
I know that $chi_A(x)=$product of the prime divisors of $A$ and $m_A(x)=$lcm of the prime divisors of $A$
linear-algebra abstract-algebra matrices
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
asked Sep 5 at 6:03
giannispapav
1,340223
Classify all $Ain M_5(mathbbQ):A^8=I$ and $A^4not=I$
Attempt
Since $A^8=I$ it must be $m_A(x)|x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$. But $A^4not=I$ so $m_A(x)not=x-1,x+1,x^2-1,x^2+1,x^4-1,(x-1)(x^2+1),(x+1)(x^2+1)$
And $Ain M_5(mathbbQ)Rightarrow deg(chi_A(x))=5$
How should I continue? Hoe would you approach this question?
I want to find the similarity classes using rational normal form.
I know that $chi_A(x)=$product of the prime divisors of $A$ and $m_A(x)=$lcm of the prime divisors of $A$
linear-algebra abstract-algebra matrices
linear-algebra abstract-algebra matrices
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
asked Sep 5 at 6:03
giannispapav
1,340223
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
asked Sep 5 at 6:03
giannispapav
1,340223
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
edited Sep 5 at 15:17
Martin Sleziak
43.7k6113261
43.7k6113261
asked Sep 5 at 6:03
giannispapav
1,340223
asked Sep 5 at 6:03
giannispapav
1,340223
asked Sep 5 at 6:03
giannispapav
1,340223
1,340223
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Its minimal polynomial, and so its characteristic polynomial $p(x)$, must have $x^4+1$
as a factor. So $p(x)=(x-alpha)(x^4+1)$
for some $alpha$. But then $alphainBbb Q$ and there aren't many
possibilities for it...
In all cases $p(x)$ will be squarefree, so the RCF
for $A$ will be the companion matrix of $p(x)$.
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Its minimal polynomial, and so its characteristic polynomial $p(x)$, must have $x^4+1$
as a factor. So $p(x)=(x-alpha)(x^4+1)$
for some $alpha$. But then $alphainBbb Q$ and there aren't many
possibilities for it...
In all cases $p(x)$ will be squarefree, so the RCF
for $A$ will be the companion matrix of $p(x)$.
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
add a comment |Â
up vote
4
down vote
accepted
Its minimal polynomial, and so its characteristic polynomial $p(x)$, must have $x^4+1$
as a factor. So $p(x)=(x-alpha)(x^4+1)$
for some $alpha$. But then $alphainBbb Q$ and there aren't many
possibilities for it...
In all cases $p(x)$ will be squarefree, so the RCF
for $A$ will be the companion matrix of $p(x)$.
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Its minimal polynomial, and so its characteristic polynomial $p(x)$, must have $x^4+1$
as a factor. So $p(x)=(x-alpha)(x^4+1)$
for some $alpha$. But then $alphainBbb Q$ and there aren't many
possibilities for it...
In all cases $p(x)$ will be squarefree, so the RCF
for $A$ will be the companion matrix of $p(x)$.
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
Its minimal polynomial, and so its characteristic polynomial $p(x)$, must have $x^4+1$
as a factor. So $p(x)=(x-alpha)(x^4+1)$
for some $alpha$. But then $alphainBbb Q$ and there aren't many
possibilities for it...
In all cases $p(x)$ will be squarefree, so the RCF
for $A$ will be the companion matrix of $p(x)$.
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
answered Sep 5 at 6:11
Lord Shark the Unknown
89.6k955116
89.6k955116
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
add a comment |Â
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
So $alpha=pm 1$?? and characteristic polynomial $p(x)=(xpm1)(x^4+1)$ and minimal polynomial $m(x)=(x^4+1)(xpm1)$ ??
– giannispapav
Sep 5 at 6:16
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
That's right! @giannispapav
– Lord Shark the Unknown
Sep 5 at 6:23
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
ok thanks. Could you explain "must have $x^4+1$ as a factor"?? Is there a quick way to prove that? Because in order to exclude $m(x)=(x-1)(x^2+1)$ I calculated that $(A-I)(A^2+I)=0Rightarrow A^4=I$ by hand
– giannispapav
Sep 5 at 6:28
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
You did that in your answer! The point is that the other irreducible factors of $x^8-1$ are all factors of $x^4-1$.
– Lord Shark the Unknown
Sep 5 at 6:30
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
aaa ok I see, thanks!
– giannispapav
Sep 5 at 6:32
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2905923%2fclassify-all-a-in-m-5-mathbbqa8-i-and-a4-not-i%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
