Example of 2 matrices similar but not row equivalent












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$begingroup$


If two matrices are row equivalent, they may not be similar because all invertible matrices are row equivalent to $I$, yet not all invertible matrices have the same trace, eigenvalues etc.



Is it also true that if two matrices are similar, they may not be row equivalent? My instinct is that there is no reason that 2 similar matrices need to be row equivalent since having the same rank, eigenvalues, determinant etc does not necessarily make them row equivalent.



Any suggestions as to how to find a counter-example?



Thanks for help.






linear-algebra matrices







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    4












    $begingroup$


    If two matrices are row equivalent, they may not be similar because all invertible matrices are row equivalent to $I$, yet not all invertible matrices have the same trace, eigenvalues etc.



    Is it also true that if two matrices are similar, they may not be row equivalent? My instinct is that there is no reason that 2 similar matrices need to be row equivalent since having the same rank, eigenvalues, determinant etc does not necessarily make them row equivalent.



    Any suggestions as to how to find a counter-example?



    Thanks for help.






    linear-algebra matrices







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$


      If two matrices are row equivalent, they may not be similar because all invertible matrices are row equivalent to $I$, yet not all invertible matrices have the same trace, eigenvalues etc.



      Is it also true that if two matrices are similar, they may not be row equivalent? My instinct is that there is no reason that 2 similar matrices need to be row equivalent since having the same rank, eigenvalues, determinant etc does not necessarily make them row equivalent.



      Any suggestions as to how to find a counter-example?



      Thanks for help.






      linear-algebra matrices







      share|cite|improve this question









      $endgroup$




      If two matrices are row equivalent, they may not be similar because all invertible matrices are row equivalent to $I$, yet not all invertible matrices have the same trace, eigenvalues etc.



      Is it also true that if two matrices are similar, they may not be row equivalent? My instinct is that there is no reason that 2 similar matrices need to be row equivalent since having the same rank, eigenvalues, determinant etc does not necessarily make them row equivalent.



      Any suggestions as to how to find a counter-example?



      Thanks for help.






      linear-algebra matrices




      linear-algebra matrices






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 2:40









      AndrewAndrew

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      357213






















          2 Answers
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          $begingroup$

          Consider all $ntimes n$ diagonal matrices whose diagonal entries are $n$ distinct fixed numbers $d_1,d_2,ldots, d_n$. (there are $n!$ of them). Assume these numbers to be non-zero. Then it is easy to see that they are all similar to each other. (Use permutation matrices as done in the answer of Siong Thye Goh).



          But these matrices are not row equivalent: two row equivalent matrices will lead to systems of linear equations with identical solutions.
          Let RHS vector be $v=(1,2,ldots, n)^T$.



          The system $DX= v$ and $D'X=v$ will not not have same solution for $D,D'$ diagonal matrices satisfying the conditions above.






          share|cite|improve this answer









          $endgroup$





















            7












            $begingroup$

            $$begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}=begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$$



            $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$ and $begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$ are not row equivalent though they are similar.






            share|cite|improve this answer









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              2 Answers
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              2 Answers
              2






              active

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              active

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              3












              $begingroup$

              Consider all $ntimes n$ diagonal matrices whose diagonal entries are $n$ distinct fixed numbers $d_1,d_2,ldots, d_n$. (there are $n!$ of them). Assume these numbers to be non-zero. Then it is easy to see that they are all similar to each other. (Use permutation matrices as done in the answer of Siong Thye Goh).



              But these matrices are not row equivalent: two row equivalent matrices will lead to systems of linear equations with identical solutions.
              Let RHS vector be $v=(1,2,ldots, n)^T$.



              The system $DX= v$ and $D'X=v$ will not not have same solution for $D,D'$ diagonal matrices satisfying the conditions above.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Consider all $ntimes n$ diagonal matrices whose diagonal entries are $n$ distinct fixed numbers $d_1,d_2,ldots, d_n$. (there are $n!$ of them). Assume these numbers to be non-zero. Then it is easy to see that they are all similar to each other. (Use permutation matrices as done in the answer of Siong Thye Goh).



                But these matrices are not row equivalent: two row equivalent matrices will lead to systems of linear equations with identical solutions.
                Let RHS vector be $v=(1,2,ldots, n)^T$.



                The system $DX= v$ and $D'X=v$ will not not have same solution for $D,D'$ diagonal matrices satisfying the conditions above.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Consider all $ntimes n$ diagonal matrices whose diagonal entries are $n$ distinct fixed numbers $d_1,d_2,ldots, d_n$. (there are $n!$ of them). Assume these numbers to be non-zero. Then it is easy to see that they are all similar to each other. (Use permutation matrices as done in the answer of Siong Thye Goh).



                  But these matrices are not row equivalent: two row equivalent matrices will lead to systems of linear equations with identical solutions.
                  Let RHS vector be $v=(1,2,ldots, n)^T$.



                  The system $DX= v$ and $D'X=v$ will not not have same solution for $D,D'$ diagonal matrices satisfying the conditions above.






                  share|cite|improve this answer









                  $endgroup$



                  Consider all $ntimes n$ diagonal matrices whose diagonal entries are $n$ distinct fixed numbers $d_1,d_2,ldots, d_n$. (there are $n!$ of them). Assume these numbers to be non-zero. Then it is easy to see that they are all similar to each other. (Use permutation matrices as done in the answer of Siong Thye Goh).



                  But these matrices are not row equivalent: two row equivalent matrices will lead to systems of linear equations with identical solutions.
                  Let RHS vector be $v=(1,2,ldots, n)^T$.



                  The system $DX= v$ and $D'X=v$ will not not have same solution for $D,D'$ diagonal matrices satisfying the conditions above.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 3:27









                  P VanchinathanP Vanchinathan

                  15.6k12136




                  15.6k12136























                      7












                      $begingroup$

                      $$begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}=begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$$



                      $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$ and $begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$ are not row equivalent though they are similar.






                      share|cite|improve this answer









                      $endgroup$


















                        7












                        $begingroup$

                        $$begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}=begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$$



                        $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$ and $begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$ are not row equivalent though they are similar.






                        share|cite|improve this answer









                        $endgroup$
















                          7












                          7








                          7





                          $begingroup$

                          $$begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}=begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$$



                          $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$ and $begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$ are not row equivalent though they are similar.






                          share|cite|improve this answer









                          $endgroup$



                          $$begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}=begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$$



                          $begin{bmatrix} 0 & 1 \ 0 & 0end{bmatrix}$ and $begin{bmatrix} 0 & 0 \ 1 & 0end{bmatrix}$ are not row equivalent though they are similar.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 4 at 2:53









                          Siong Thye GohSiong Thye Goh

                          104k1468120




                          104k1468120






























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