Mixing
Minimal polynomial?
0
$begingroup$
Let $K$ a field, $p,nin mathbb{N}$, $Bin mathcal{M}_p({K})$ and let denote $S_B={Xin mathcal{M}_p(K) mid X^n=B}$.
If $Xin S_B$, I have to prove that $mu_X$ (the minimal polynomial of $X$) divides $mu_B(xi^n)$.
If $Xin S_B$ then we can deduce that $X^n-B=0$ but after I do not know how to link this with $mu_X$ or $mu_B$...
Thanks in advance !
linear-algebra matrices polynomials minimal-polynomials
edited Jan 11 at 19:09
Omnomnomnom
128k791182
asked Jan 11 at 19:07
MamanMaman
1,189722
$endgroup$
add a comment |
0
$begingroup$
Let $K$ a field, $p,nin mathbb{N}$, $Bin mathcal{M}_p({K})$ and let denote $S_B={Xin mathcal{M}_p(K) mid X^n=B}$.
If $Xin S_B$, I have to prove that $mu_X$ (the minimal polynomial of $X$) divides $mu_B(xi^n)$.
If $Xin S_B$ then we can deduce that $X^n-B=0$ but after I do not know how to link this with $mu_X$ or $mu_B$...
Thanks in advance !
linear-algebra matrices polynomials minimal-polynomials
edited Jan 11 at 19:09
Omnomnomnom
128k791182
asked Jan 11 at 19:07
MamanMaman
1,189722
$endgroup$
add a comment |
0
0
0
$begingroup$
Let $K$ a field, $p,nin mathbb{N}$, $Bin mathcal{M}_p({K})$ and let denote $S_B={Xin mathcal{M}_p(K) mid X^n=B}$.
If $Xin S_B$, I have to prove that $mu_X$ (the minimal polynomial of $X$) divides $mu_B(xi^n)$.
If $Xin S_B$ then we can deduce that $X^n-B=0$ but after I do not know how to link this with $mu_X$ or $mu_B$...
Thanks in advance !
linear-algebra matrices polynomials minimal-polynomials
edited Jan 11 at 19:09
Omnomnomnom
128k791182
asked Jan 11 at 19:07
MamanMaman
1,189722
$endgroup$
Let $K$ a field, $p,nin mathbb{N}$, $Bin mathcal{M}_p({K})$ and let denote $S_B={Xin mathcal{M}_p(K) mid X^n=B}$.
If $Xin S_B$, I have to prove that $mu_X$ (the minimal polynomial of $X$) divides $mu_B(xi^n)$.
If $Xin S_B$ then we can deduce that $X^n-B=0$ but after I do not know how to link this with $mu_X$ or $mu_B$...
Thanks in advance !
linear-algebra matrices polynomials minimal-polynomials
linear-algebra matrices polynomials minimal-polynomials
edited Jan 11 at 19:09
Omnomnomnom
128k791182
asked Jan 11 at 19:07
MamanMaman
1,189722
edited Jan 11 at 19:09
Omnomnomnom
128k791182
asked Jan 11 at 19:07
MamanMaman
1,189722
edited Jan 11 at 19:09
Omnomnomnom
128k791182
edited Jan 11 at 19:09
Omnomnomnom
128k791182
edited Jan 11 at 19:09
Omnomnomnom
128k791182
128k791182
asked Jan 11 at 19:07
MamanMaman
1,189722
asked Jan 11 at 19:07
MamanMaman
1,189722
asked Jan 11 at 19:07
MamanMaman
1,189722
1,189722
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Hint: It suffices to note that if $X in S_B$, then $mu_B(X^n) = 0$
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
$endgroup$
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Hint: It suffices to note that if $X in S_B$, then $mu_B(X^n) = 0$
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
$endgroup$
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
add a comment |
2
$begingroup$
Hint: It suffices to note that if $X in S_B$, then $mu_B(X^n) = 0$
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
$endgroup$
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
add a comment |
2
2
2
$begingroup$
Hint: It suffices to note that if $X in S_B$, then $mu_B(X^n) = 0$
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
$endgroup$
Hint: It suffices to note that if $X in S_B$, then $mu_B(X^n) = 0$
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
answered Jan 11 at 19:11
OmnomnomnomOmnomnomnom
128k791182
128k791182
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
add a comment |
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
Does it come from the fact that $mu_B(B)=0$ by definition ?
$endgroup$
– Maman
Jan 11 at 19:15
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
@Maman exactly.
$endgroup$
– Omnomnomnom
Jan 11 at 19:16
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
Then I think we have : $mu_{X^n}(X^N)=0$ ? What is the link with $xi^n$ ?
$endgroup$
– Maman
Jan 12 at 21:27
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
If $p(A)=0$, then the minimal polynomial of $A$ divides $p$.
$endgroup$
– Omnomnomnom
Jan 13 at 1:49
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
$begingroup$
@Maman $xi$ is used to emphasize the polynomial itself. That is, $p(xi) = mu_B(xi^n)$ is a polynomial satisfying $p(X) = 0$
$endgroup$
– Omnomnomnom
Jan 13 at 16:54
add a comment |
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