Find all square matrices $A$ satisfying ${}^t!A = -A$.












0












$begingroup$


I am given an $n times n$ matrix $A$.



If $A = A^2$ and ${}^t!A = -A$, I need to find matrix of $A$.



$$A^2=A⟹A^2⋅{}^t!A=A⋅{}^t!A⟹A^2⋅(-A)=A⋅(-A)$$



How do I show the matrix of A from here ?






linear-algebra matrices matrix-equations transpose







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to transpose $A^2$ and see what kind of a result you get.
    $endgroup$
    – orion
    Sep 17 '18 at 7:49
















0












$begingroup$


I am given an $n times n$ matrix $A$.



If $A = A^2$ and ${}^t!A = -A$, I need to find matrix of $A$.



$$A^2=A⟹A^2⋅{}^t!A=A⋅{}^t!A⟹A^2⋅(-A)=A⋅(-A)$$



How do I show the matrix of A from here ?






linear-algebra matrices matrix-equations transpose







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to transpose $A^2$ and see what kind of a result you get.
    $endgroup$
    – orion
    Sep 17 '18 at 7:49














0












0








0





$begingroup$


I am given an $n times n$ matrix $A$.



If $A = A^2$ and ${}^t!A = -A$, I need to find matrix of $A$.



$$A^2=A⟹A^2⋅{}^t!A=A⋅{}^t!A⟹A^2⋅(-A)=A⋅(-A)$$



How do I show the matrix of A from here ?






linear-algebra matrices matrix-equations transpose







share|cite|improve this question











$endgroup$




I am given an $n times n$ matrix $A$.



If $A = A^2$ and ${}^t!A = -A$, I need to find matrix of $A$.



$$A^2=A⟹A^2⋅{}^t!A=A⋅{}^t!A⟹A^2⋅(-A)=A⋅(-A)$$



How do I show the matrix of A from here ?






linear-algebra matrices matrix-equations transpose




linear-algebra matrices matrix-equations transpose






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 16:19







user593746

















asked Sep 17 '18 at 7:45









RecklessSerenadeRecklessSerenade

295




295












  • $begingroup$
    Try to transpose $A^2$ and see what kind of a result you get.
    $endgroup$
    – orion
    Sep 17 '18 at 7:49


















  • $begingroup$
    Try to transpose $A^2$ and see what kind of a result you get.
    $endgroup$
    – orion
    Sep 17 '18 at 7:49
















$begingroup$
Try to transpose $A^2$ and see what kind of a result you get.
$endgroup$
– orion
Sep 17 '18 at 7:49




$begingroup$
Try to transpose $A^2$ and see what kind of a result you get.
$endgroup$
– orion
Sep 17 '18 at 7:49










3 Answers
3






active

oldest

votes


















1












$begingroup$

$A^t=-A$, hence $A=A^2=(-A^t)^2=(A^t)^2=(A^2)^t=A^t$.



Can you proceed ?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
    $endgroup$
    – RecklessSerenade
    Sep 17 '18 at 9:05










  • $begingroup$
    We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
    $endgroup$
    – Fred
    Sep 17 '18 at 9:12










  • $begingroup$
    Thank you very much. I understand it.
    $endgroup$
    – RecklessSerenade
    Sep 17 '18 at 9:47



















1












$begingroup$

The transpose of $A^{2}$ is the square of the transpose of $A$. From the given hypothesis you get $A=-A$ so $A=0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Another way to solve this is to look at the (monic) minimal polynomial $mu(x)$ of $A$. Note that $A^2=A$ implies that $mu(x)$ divides $x^2-x$. Thus, any eigenvalue of $A$ can only be $0$ or $1$. Now, if $lambda$ is an eigenvalue of $A$, then $A^t=-A$ implies that $-lambda$ is also an eigenvalue of $A$ (unless your matrix $A$ is defined over a field of characteristic $2$, where $-lambda=lambda$). This proves that $1$ cannot be an eigenvalue of $A$. That is, $0$ is the only eigenvalue of $A$, whence $mu(x)=x$, and so $A=0$.



    In characteristic $2$, $A$ can be any diagonalizable symmetric matrix whose spectrum is a subset of ${0,1}$. There are many such matrices $A$ in this case.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      $A^t=-A$, hence $A=A^2=(-A^t)^2=(A^t)^2=(A^2)^t=A^t$.



      Can you proceed ?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:05










      • $begingroup$
        We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
        $endgroup$
        – Fred
        Sep 17 '18 at 9:12










      • $begingroup$
        Thank you very much. I understand it.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:47
















      1












      $begingroup$

      $A^t=-A$, hence $A=A^2=(-A^t)^2=(A^t)^2=(A^2)^t=A^t$.



      Can you proceed ?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:05










      • $begingroup$
        We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
        $endgroup$
        – Fred
        Sep 17 '18 at 9:12










      • $begingroup$
        Thank you very much. I understand it.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:47














      1












      1








      1





      $begingroup$

      $A^t=-A$, hence $A=A^2=(-A^t)^2=(A^t)^2=(A^2)^t=A^t$.



      Can you proceed ?






      share|cite|improve this answer









      $endgroup$



      $A^t=-A$, hence $A=A^2=(-A^t)^2=(A^t)^2=(A^2)^t=A^t$.



      Can you proceed ?







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Sep 17 '18 at 7:51









      FredFred

      46.9k1848




      46.9k1848












      • $begingroup$
        I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:05










      • $begingroup$
        We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
        $endgroup$
        – Fred
        Sep 17 '18 at 9:12










      • $begingroup$
        Thank you very much. I understand it.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:47


















      • $begingroup$
        I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:05










      • $begingroup$
        We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
        $endgroup$
        – Fred
        Sep 17 '18 at 9:12










      • $begingroup$
        Thank you very much. I understand it.
        $endgroup$
        – RecklessSerenade
        Sep 17 '18 at 9:47
















      $begingroup$
      I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
      $endgroup$
      – RecklessSerenade
      Sep 17 '18 at 9:05




      $begingroup$
      I was able to understand its transformation, but I still don't understand why it will be A = 0 from its transformation as other respondents say.
      $endgroup$
      – RecklessSerenade
      Sep 17 '18 at 9:05












      $begingroup$
      We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
      $endgroup$
      – Fred
      Sep 17 '18 at 9:12




      $begingroup$
      We have $A=A^t=-A$, hence $2A=0$. This gives $A=0$.
      $endgroup$
      – Fred
      Sep 17 '18 at 9:12












      $begingroup$
      Thank you very much. I understand it.
      $endgroup$
      – RecklessSerenade
      Sep 17 '18 at 9:47




      $begingroup$
      Thank you very much. I understand it.
      $endgroup$
      – RecklessSerenade
      Sep 17 '18 at 9:47











      1












      $begingroup$

      The transpose of $A^{2}$ is the square of the transpose of $A$. From the given hypothesis you get $A=-A$ so $A=0$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The transpose of $A^{2}$ is the square of the transpose of $A$. From the given hypothesis you get $A=-A$ so $A=0$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The transpose of $A^{2}$ is the square of the transpose of $A$. From the given hypothesis you get $A=-A$ so $A=0$.






          share|cite|improve this answer









          $endgroup$



          The transpose of $A^{2}$ is the square of the transpose of $A$. From the given hypothesis you get $A=-A$ so $A=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 17 '18 at 7:51









          Kavi Rama MurthyKavi Rama Murthy

          61.2k42262




          61.2k42262























              1












              $begingroup$

              Another way to solve this is to look at the (monic) minimal polynomial $mu(x)$ of $A$. Note that $A^2=A$ implies that $mu(x)$ divides $x^2-x$. Thus, any eigenvalue of $A$ can only be $0$ or $1$. Now, if $lambda$ is an eigenvalue of $A$, then $A^t=-A$ implies that $-lambda$ is also an eigenvalue of $A$ (unless your matrix $A$ is defined over a field of characteristic $2$, where $-lambda=lambda$). This proves that $1$ cannot be an eigenvalue of $A$. That is, $0$ is the only eigenvalue of $A$, whence $mu(x)=x$, and so $A=0$.



              In characteristic $2$, $A$ can be any diagonalizable symmetric matrix whose spectrum is a subset of ${0,1}$. There are many such matrices $A$ in this case.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Another way to solve this is to look at the (monic) minimal polynomial $mu(x)$ of $A$. Note that $A^2=A$ implies that $mu(x)$ divides $x^2-x$. Thus, any eigenvalue of $A$ can only be $0$ or $1$. Now, if $lambda$ is an eigenvalue of $A$, then $A^t=-A$ implies that $-lambda$ is also an eigenvalue of $A$ (unless your matrix $A$ is defined over a field of characteristic $2$, where $-lambda=lambda$). This proves that $1$ cannot be an eigenvalue of $A$. That is, $0$ is the only eigenvalue of $A$, whence $mu(x)=x$, and so $A=0$.



                In characteristic $2$, $A$ can be any diagonalizable symmetric matrix whose spectrum is a subset of ${0,1}$. There are many such matrices $A$ in this case.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Another way to solve this is to look at the (monic) minimal polynomial $mu(x)$ of $A$. Note that $A^2=A$ implies that $mu(x)$ divides $x^2-x$. Thus, any eigenvalue of $A$ can only be $0$ or $1$. Now, if $lambda$ is an eigenvalue of $A$, then $A^t=-A$ implies that $-lambda$ is also an eigenvalue of $A$ (unless your matrix $A$ is defined over a field of characteristic $2$, where $-lambda=lambda$). This proves that $1$ cannot be an eigenvalue of $A$. That is, $0$ is the only eigenvalue of $A$, whence $mu(x)=x$, and so $A=0$.



                  In characteristic $2$, $A$ can be any diagonalizable symmetric matrix whose spectrum is a subset of ${0,1}$. There are many such matrices $A$ in this case.






                  share|cite|improve this answer











                  $endgroup$



                  Another way to solve this is to look at the (monic) minimal polynomial $mu(x)$ of $A$. Note that $A^2=A$ implies that $mu(x)$ divides $x^2-x$. Thus, any eigenvalue of $A$ can only be $0$ or $1$. Now, if $lambda$ is an eigenvalue of $A$, then $A^t=-A$ implies that $-lambda$ is also an eigenvalue of $A$ (unless your matrix $A$ is defined over a field of characteristic $2$, where $-lambda=lambda$). This proves that $1$ cannot be an eigenvalue of $A$. That is, $0$ is the only eigenvalue of $A$, whence $mu(x)=x$, and so $A=0$.



                  In characteristic $2$, $A$ can be any diagonalizable symmetric matrix whose spectrum is a subset of ${0,1}$. There are many such matrices $A$ in this case.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 10 '18 at 13:43

























                  answered Sep 17 '18 at 8:55







                  user593746





































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