$A^{2n}=I$ but $A^{n}neq I, -I$ [duplicate]












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  • 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










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Jan 16 at 12:38


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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










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    Jan 16 at 12:38


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      1












      1








      1





      $begingroup$



      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










      share|cite|improve this question











      $endgroup$





      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






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      edited Jan 16 at 12:08







      user345777

















      asked Jan 16 at 11:38









      user345777user345777

      432312




      432312




      marked as duplicate by user1551 linear-algebra
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      Jan 16 at 12:38


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      marked as duplicate by user1551 linear-algebra
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      Jan 16 at 12:38


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          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          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 $r$.






          share|cite|improve this answer











          $endgroup$









          • 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



















          1












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            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 $r$.






            share|cite|improve this answer











            $endgroup$









            • 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
















            5












            $begingroup$

            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 $r$.






            share|cite|improve this answer











            $endgroup$









            • 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














            5












            5








            5





            $begingroup$

            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 $r$.






            share|cite|improve this answer











            $endgroup$



            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 $r$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 16 at 12:19

























            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














            • 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











            1












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$



                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 16 at 12:27

























                answered Jan 16 at 12:22









                EnkiduEnkidu

                1,44429




                1,44429















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