System of homogen linear equations in a division ring












2












$begingroup$


Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
$$left{begin{array}{rcl}
a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
vdotsqquadqquadqquad&vdots&vdots\
a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
end{array}right.$$
where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.



Is it also true in the non commutative case?
Thanks in advance.






linear-algebra noncommutative-algebra







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
    $$left{begin{array}{rcl}
    a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
    a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
    vdotsqquadqquadqquad&vdots&vdots\
    a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
    end{array}right.$$
    where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.



    Is it also true in the non commutative case?
    Thanks in advance.






    linear-algebra noncommutative-algebra







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
      $$left{begin{array}{rcl}
      a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
      a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
      vdotsqquadqquadqquad&vdots&vdots\
      a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
      end{array}right.$$
      where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.



      Is it also true in the non commutative case?
      Thanks in advance.






      linear-algebra noncommutative-algebra







      share|cite|improve this question









      $endgroup$




      Let $K$ be a division ring (one does not suppose that $K$ is commutative) and $m,n$ two positive integers such that $m<n$. Consider the system of homogen linear equations
      $$left{begin{array}{rcl}
      a_{11}x_1+a_{12}x_2+cdots+a_{1n}x_n&=&0\
      a_{21}x_1+a_{22}x_2+cdots+a_{2n}x_n&=&0\
      vdotsqquadqquadqquad&vdots&vdots\
      a_{m1}x_1+a_{m2}x_2+cdots+a_{mn}x_n&=&0\
      end{array}right.$$
      where the $a_{ij}$'s are in $K$. Can we assert that this system admits an infinity of solutions. Obviously, it is true when $K$ is commutative. But when $K$ is not, I do not know if this results still holds.



      Is it also true in the non commutative case?
      Thanks in advance.






      linear-algebra noncommutative-algebra




      linear-algebra noncommutative-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 27 '18 at 2:37









      joaopajoaopa

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          $begingroup$

          The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.




          Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.






          share|cite|improve this answer









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            $begingroup$

            The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.




            Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.




              Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.




                Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.






                share|cite|improve this answer









                $endgroup$



                The system has infinitely many solutions, even when $K$ is non-commutative. The reason is that row reduction is still possible over a non-commutative division ring.




                Side remark: if $K$ is commutative, the system has infinitely many solutions only if $K$ is infinite. If $K$ is not commutative, then it is automatically infinite by Wedderburn's little theorem.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 27 '18 at 13:20









                Pierre-Guy PlamondonPierre-Guy Plamondon

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