$A^{2n}=I$ but $A^{n}neq I, -I$ [duplicate]
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This question already has an answer here:
If $A^{2n}=I$, then $A^n=pm I$ where $Ain M_n(mathbb{R})$
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Let $A$ be a $ntimes n$ real matrix. such that $A^{2n}=I$ but $A^{n}neq I, -I$, $ngeq 2?$ I have a example for $A^2=I$ but $Aneq I, -I$ but could not find a similar example for this question. I tried permuting basis vectors but could not an example. Any help is appreciated
linear-algebra matrices linear-transformations
asked Jan 16 at 11:38
user345777user345777
432312
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marked as duplicate by user1551 linear-algebra
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Jan 16 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
If $A^{2n}=I$, then $A^n=pm I$ where $Ain M_n(mathbb{R})$
1 answer
Let $A$ be a $ntimes n$ real matrix. such that $A^{2n}=I$ but $A^{n}neq I, -I$, $ngeq 2?$ I have a example for $A^2=I$ but $Aneq I, -I$ but could not find a similar example for this question. I tried permuting basis vectors but could not an example. Any help is appreciated
linear-algebra matrices linear-transformations
asked Jan 16 at 11:38
user345777user345777
432312
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marked as duplicate by user1551 linear-algebra
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Jan 16 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
If $A^{2n}=I$, then $A^n=pm I$ where $Ain M_n(mathbb{R})$
1 answer
Let $A$ be a $ntimes n$ real matrix. such that $A^{2n}=I$ but $A^{n}neq I, -I$, $ngeq 2?$ I have a example for $A^2=I$ but $Aneq I, -I$ but could not find a similar example for this question. I tried permuting basis vectors but could not an example. Any help is appreciated
linear-algebra matrices linear-transformations
asked Jan 16 at 11:38
user345777user345777
432312
$endgroup$
This question already has an answer here:
If $A^{2n}=I$, then $A^n=pm I$ where $Ain M_n(mathbb{R})$
1 answer
Let $A$ be a $ntimes n$ real matrix. such that $A^{2n}=I$ but $A^{n}neq I, -I$, $ngeq 2?$ I have a example for $A^2=I$ but $Aneq I, -I$ but could not find a similar example for this question. I tried permuting basis vectors but could not an example. Any help is appreciated
This question already has an answer here:
If $A^{2n}=I$, then $A^n=pm I$ where $Ain M_n(mathbb{R})$
1 answer
linear-algebra matrices linear-transformations
linear-algebra matrices linear-transformations
asked Jan 16 at 11:38
user345777user345777
432312
asked Jan 16 at 11:38
user345777user345777
432312
edited Jan 16 at 12:08
user345777
asked Jan 16 at 11:38
user345777user345777
432312
asked Jan 16 at 11:38
user345777user345777
432312
asked Jan 16 at 11:38
user345777user345777
432312
432312
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Jan 16 at 12:38
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Jan 16 at 12:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take $A$ to be diagonal, and its entries being $2n$-th roots of unity, and one of them not being a $n$-th of unity.
As far as real entries are concerned: let $r$ be a primitive $2n$-th root of unity. Take $A$ as a block-diagonal matrix, the second block being $I_{n-2}$ and the first block being the $2times 2$ real matrix associated with the complex number $r$.
answered Jan 16 at 11:42
MindlackMindlack
4,900211
$endgroup$
2
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
add a comment |
$begingroup$
Essentially this is just the answer by Mindlack, only pimped to work over $mathbb{R}$
Pick a $2$-dimensional subspace $V$ (which you can by assumption) and consider $mathbb{R}^n$ as $V oplus mathbb{R}^{n-2}$. then any automorphism $varphi'$ on $V$ induces an automorphism $varphi= varphioplus id$ on $mathbb{R}^n$, and in particular a matrix. Hence we only need to find a $varphi'$ that does the job. Now observe that by adding up with identities the requirement of $A^n neq -id$ collapses. hence it suffices to find a $varphi$ such that $varphi^{2n}= id$ but now you can just embed $mathbb{C}hookrightarrow mathbb{R}^{2times 2}cong mathrm{Aut}(mathbb{R}^2)cong mathrm{Aut}(V)$ and pick a $2n$-th root of unity in that image (this maps $1$ to $begin{pmatrix}1 & 0\0 & 1 end{pmatrix}$ and $i$ to $begin{pmatrix}0 & 1\-1 & 0 end{pmatrix}$, to prove that this indeed is an embedding of a subring is a beautiful exercise). This does the job.
Morally, it suffices to find a matrix that does the stuff on a subspace, and then lift it to the space you want to have.
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $A$ to be diagonal, and its entries being $2n$-th roots of unity, and one of them not being a $n$-th of unity.
As far as real entries are concerned: let $r$ be a primitive $2n$-th root of unity. Take $A$ as a block-diagonal matrix, the second block being $I_{n-2}$ and the first block being the $2times 2$ real matrix associated with the complex number $r$.
answered Jan 16 at 11:42
MindlackMindlack
4,900211
$endgroup$
2
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
add a comment |
$begingroup$
Take $A$ to be diagonal, and its entries being $2n$-th roots of unity, and one of them not being a $n$-th of unity.
As far as real entries are concerned: let $r$ be a primitive $2n$-th root of unity. Take $A$ as a block-diagonal matrix, the second block being $I_{n-2}$ and the first block being the $2times 2$ real matrix associated with the complex number $r$.
answered Jan 16 at 11:42
MindlackMindlack
4,900211
$endgroup$
2
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
add a comment |
$begingroup$
Take $A$ to be diagonal, and its entries being $2n$-th roots of unity, and one of them not being a $n$-th of unity.
As far as real entries are concerned: let $r$ be a primitive $2n$-th root of unity. Take $A$ as a block-diagonal matrix, the second block being $I_{n-2}$ and the first block being the $2times 2$ real matrix associated with the complex number $r$.
answered Jan 16 at 11:42
MindlackMindlack
4,900211
$endgroup$
Take $A$ to be diagonal, and its entries being $2n$-th roots of unity, and one of them not being a $n$-th of unity.
As far as real entries are concerned: let $r$ be a primitive $2n$-th root of unity. Take $A$ as a block-diagonal matrix, the second block being $I_{n-2}$ and the first block being the $2times 2$ real matrix associated with the complex number $r$.
answered Jan 16 at 11:42
MindlackMindlack
4,900211
edited Jan 16 at 12:19
answered Jan 16 at 11:42
MindlackMindlack
4,900211
answered Jan 16 at 11:42
MindlackMindlack
4,900211
answered Jan 16 at 11:42
MindlackMindlack
4,900211
4,900211
2
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
add a comment |
2
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
2
2
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
Slightly more simply: identity matrix, except replace one $1$ with a primitive $2n$-th root of unity.
$endgroup$
– Wojowu
Jan 16 at 11:45
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
$begingroup$
I am looking for a matrix with real entries. I will edit the question
$endgroup$
– user345777
Jan 16 at 12:08
add a comment |
$begingroup$
Essentially this is just the answer by Mindlack, only pimped to work over $mathbb{R}$
Pick a $2$-dimensional subspace $V$ (which you can by assumption) and consider $mathbb{R}^n$ as $V oplus mathbb{R}^{n-2}$. then any automorphism $varphi'$ on $V$ induces an automorphism $varphi= varphioplus id$ on $mathbb{R}^n$, and in particular a matrix. Hence we only need to find a $varphi'$ that does the job. Now observe that by adding up with identities the requirement of $A^n neq -id$ collapses. hence it suffices to find a $varphi$ such that $varphi^{2n}= id$ but now you can just embed $mathbb{C}hookrightarrow mathbb{R}^{2times 2}cong mathrm{Aut}(mathbb{R}^2)cong mathrm{Aut}(V)$ and pick a $2n$-th root of unity in that image (this maps $1$ to $begin{pmatrix}1 & 0\0 & 1 end{pmatrix}$ and $i$ to $begin{pmatrix}0 & 1\-1 & 0 end{pmatrix}$, to prove that this indeed is an embedding of a subring is a beautiful exercise). This does the job.
Morally, it suffices to find a matrix that does the stuff on a subspace, and then lift it to the space you want to have.
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
$endgroup$
add a comment |
$begingroup$
Essentially this is just the answer by Mindlack, only pimped to work over $mathbb{R}$
Pick a $2$-dimensional subspace $V$ (which you can by assumption) and consider $mathbb{R}^n$ as $V oplus mathbb{R}^{n-2}$. then any automorphism $varphi'$ on $V$ induces an automorphism $varphi= varphioplus id$ on $mathbb{R}^n$, and in particular a matrix. Hence we only need to find a $varphi'$ that does the job. Now observe that by adding up with identities the requirement of $A^n neq -id$ collapses. hence it suffices to find a $varphi$ such that $varphi^{2n}= id$ but now you can just embed $mathbb{C}hookrightarrow mathbb{R}^{2times 2}cong mathrm{Aut}(mathbb{R}^2)cong mathrm{Aut}(V)$ and pick a $2n$-th root of unity in that image (this maps $1$ to $begin{pmatrix}1 & 0\0 & 1 end{pmatrix}$ and $i$ to $begin{pmatrix}0 & 1\-1 & 0 end{pmatrix}$, to prove that this indeed is an embedding of a subring is a beautiful exercise). This does the job.
Morally, it suffices to find a matrix that does the stuff on a subspace, and then lift it to the space you want to have.
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
$endgroup$
add a comment |
$begingroup$
Essentially this is just the answer by Mindlack, only pimped to work over $mathbb{R}$
Pick a $2$-dimensional subspace $V$ (which you can by assumption) and consider $mathbb{R}^n$ as $V oplus mathbb{R}^{n-2}$. then any automorphism $varphi'$ on $V$ induces an automorphism $varphi= varphioplus id$ on $mathbb{R}^n$, and in particular a matrix. Hence we only need to find a $varphi'$ that does the job. Now observe that by adding up with identities the requirement of $A^n neq -id$ collapses. hence it suffices to find a $varphi$ such that $varphi^{2n}= id$ but now you can just embed $mathbb{C}hookrightarrow mathbb{R}^{2times 2}cong mathrm{Aut}(mathbb{R}^2)cong mathrm{Aut}(V)$ and pick a $2n$-th root of unity in that image (this maps $1$ to $begin{pmatrix}1 & 0\0 & 1 end{pmatrix}$ and $i$ to $begin{pmatrix}0 & 1\-1 & 0 end{pmatrix}$, to prove that this indeed is an embedding of a subring is a beautiful exercise). This does the job.
Morally, it suffices to find a matrix that does the stuff on a subspace, and then lift it to the space you want to have.
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
$endgroup$
Essentially this is just the answer by Mindlack, only pimped to work over $mathbb{R}$
Pick a $2$-dimensional subspace $V$ (which you can by assumption) and consider $mathbb{R}^n$ as $V oplus mathbb{R}^{n-2}$. then any automorphism $varphi'$ on $V$ induces an automorphism $varphi= varphioplus id$ on $mathbb{R}^n$, and in particular a matrix. Hence we only need to find a $varphi'$ that does the job. Now observe that by adding up with identities the requirement of $A^n neq -id$ collapses. hence it suffices to find a $varphi$ such that $varphi^{2n}= id$ but now you can just embed $mathbb{C}hookrightarrow mathbb{R}^{2times 2}cong mathrm{Aut}(mathbb{R}^2)cong mathrm{Aut}(V)$ and pick a $2n$-th root of unity in that image (this maps $1$ to $begin{pmatrix}1 & 0\0 & 1 end{pmatrix}$ and $i$ to $begin{pmatrix}0 & 1\-1 & 0 end{pmatrix}$, to prove that this indeed is an embedding of a subring is a beautiful exercise). This does the job.
Morally, it suffices to find a matrix that does the stuff on a subspace, and then lift it to the space you want to have.
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
edited Jan 16 at 12:27
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
answered Jan 16 at 12:22
EnkiduEnkidu
1,44429
1,44429
add a comment |
add a comment |
