How do I find out that the following two matrices are similar?












2














How do I find out that the following two matrices are similar?
$N =
begin{pmatrix}
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{pmatrix}$



and $M=
begin{pmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 1 \
0 & 0 & 0 & 0
end{pmatrix}$



I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried



$P=
begin{pmatrix}
0 & 0 & 1 & 0 \
0 & 1 & 0 & 0 \
1 & 0 & 0 & 0 \
0 & 0 & 0 & 1
end{pmatrix}$



such that $PN = begin{pmatrix}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 0 & 0
end{pmatrix}$ but then $PNP^{-1} neq M$.



My linear algebra is a bit rusty. Is there a more elaborate way to do this?






linear-algebra matrices







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    2














    How do I find out that the following two matrices are similar?
    $N =
    begin{pmatrix}
    0 & 1 & 0 & 0 \
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0
    end{pmatrix}$



    and $M=
    begin{pmatrix}
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 1 \
    0 & 0 & 0 & 0
    end{pmatrix}$



    I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried



    $P=
    begin{pmatrix}
    0 & 0 & 1 & 0 \
    0 & 1 & 0 & 0 \
    1 & 0 & 0 & 0 \
    0 & 0 & 0 & 1
    end{pmatrix}$



    such that $PN = begin{pmatrix}
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 \
    0 & 1 & 0 & 0 \
    0 & 0 & 0 & 0
    end{pmatrix}$     but then $PNP^{-1} neq M$.



    My linear algebra is a bit rusty. Is there a more elaborate way to do this?






    linear-algebra matrices







    share|cite|improve this question



























      2












      2








      2


      1





      How do I find out that the following two matrices are similar?
      $N =
      begin{pmatrix}
      0 & 1 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0
      end{pmatrix}$



      and $M=
      begin{pmatrix}
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 1 \
      0 & 0 & 0 & 0
      end{pmatrix}$



      I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried



      $P=
      begin{pmatrix}
      0 & 0 & 1 & 0 \
      0 & 1 & 0 & 0 \
      1 & 0 & 0 & 0 \
      0 & 0 & 0 & 1
      end{pmatrix}$



      such that $PN = begin{pmatrix}
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 1 & 0 & 0 \
      0 & 0 & 0 & 0
      end{pmatrix}$       but then $PNP^{-1} neq M$.



      My linear algebra is a bit rusty. Is there a more elaborate way to do this?






      linear-algebra matrices







      share|cite|improve this question















      How do I find out that the following two matrices are similar?
      $N =
      begin{pmatrix}
      0 & 1 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0
      end{pmatrix}$



      and $M=
      begin{pmatrix}
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 1 \
      0 & 0 & 0 & 0
      end{pmatrix}$



      I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried



      $P=
      begin{pmatrix}
      0 & 0 & 1 & 0 \
      0 & 1 & 0 & 0 \
      1 & 0 & 0 & 0 \
      0 & 0 & 0 & 1
      end{pmatrix}$



      such that $PN = begin{pmatrix}
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 1 & 0 & 0 \
      0 & 0 & 0 & 0
      end{pmatrix}$       but then $PNP^{-1} neq M$.



      My linear algebra is a bit rusty. Is there a more elaborate way to do this?






      linear-algebra matrices




      linear-algebra matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 1 hour ago









      José Carlos Santos

      151k22122223




      151k22122223










      asked 1 hour ago









      MPB94

      25016




      25016






















          2 Answers
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          8














          Let $(e_1,e_2,e_3,e_3)$ be the standard basis of $mathbb{R}^4$. You have





          • $N.e_1=0$;


          • $N.e_2=e_1$;


          • $N.e_3=0$;


          • $N.e_4=0$.


          You also have





          • $M.e_3=0$;


          • $M.e_4=e_3$;


          • $M.e_1=0$;


          • $M.e_2=0$.


          So, if you see $M$ as a linear map from $mathbb{R}^4$ into itself, the matrix of $M$ with respect to the basis $(e_3,e_4,e_1,e_2)$ is the matrix $N$. Therefore, $N$ and $M$ are similar.



          Or you can take$$P=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix},$$which is basically the same thing.






          share|cite|improve this answer





























            3














            The two matrices are made of Jordan blocks; in $2times2$ block format, they are
            $$
            N=begin{bmatrix} J & 0 \ 0 & 0 end{bmatrix}
            qquad
            M=begin{bmatrix} 0 & 0 \ 0 & J end{bmatrix}
            $$
            You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
            $$
            M=begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix} N
            begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix}
            $$






            share|cite|improve this answer





















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              2 Answers
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              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              8














              Let $(e_1,e_2,e_3,e_3)$ be the standard basis of $mathbb{R}^4$. You have





              • $N.e_1=0$;


              • $N.e_2=e_1$;


              • $N.e_3=0$;


              • $N.e_4=0$.


              You also have





              • $M.e_3=0$;


              • $M.e_4=e_3$;


              • $M.e_1=0$;


              • $M.e_2=0$.


              So, if you see $M$ as a linear map from $mathbb{R}^4$ into itself, the matrix of $M$ with respect to the basis $(e_3,e_4,e_1,e_2)$ is the matrix $N$. Therefore, $N$ and $M$ are similar.



              Or you can take$$P=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix},$$which is basically the same thing.






              share|cite|improve this answer


























                8














                Let $(e_1,e_2,e_3,e_3)$ be the standard basis of $mathbb{R}^4$. You have





                • $N.e_1=0$;


                • $N.e_2=e_1$;


                • $N.e_3=0$;


                • $N.e_4=0$.


                You also have





                • $M.e_3=0$;


                • $M.e_4=e_3$;


                • $M.e_1=0$;


                • $M.e_2=0$.


                So, if you see $M$ as a linear map from $mathbb{R}^4$ into itself, the matrix of $M$ with respect to the basis $(e_3,e_4,e_1,e_2)$ is the matrix $N$. Therefore, $N$ and $M$ are similar.



                Or you can take$$P=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix},$$which is basically the same thing.






                share|cite|improve this answer
























                  8












                  8








                  8






                  Let $(e_1,e_2,e_3,e_3)$ be the standard basis of $mathbb{R}^4$. You have





                  • $N.e_1=0$;


                  • $N.e_2=e_1$;


                  • $N.e_3=0$;


                  • $N.e_4=0$.


                  You also have





                  • $M.e_3=0$;


                  • $M.e_4=e_3$;


                  • $M.e_1=0$;


                  • $M.e_2=0$.


                  So, if you see $M$ as a linear map from $mathbb{R}^4$ into itself, the matrix of $M$ with respect to the basis $(e_3,e_4,e_1,e_2)$ is the matrix $N$. Therefore, $N$ and $M$ are similar.



                  Or you can take$$P=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix},$$which is basically the same thing.






                  share|cite|improve this answer












                  Let $(e_1,e_2,e_3,e_3)$ be the standard basis of $mathbb{R}^4$. You have





                  • $N.e_1=0$;


                  • $N.e_2=e_1$;


                  • $N.e_3=0$;


                  • $N.e_4=0$.


                  You also have





                  • $M.e_3=0$;


                  • $M.e_4=e_3$;


                  • $M.e_1=0$;


                  • $M.e_2=0$.


                  So, if you see $M$ as a linear map from $mathbb{R}^4$ into itself, the matrix of $M$ with respect to the basis $(e_3,e_4,e_1,e_2)$ is the matrix $N$. Therefore, $N$ and $M$ are similar.



                  Or you can take$$P=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix},$$which is basically the same thing.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  José Carlos Santos

                  151k22122223




                  151k22122223























                      3














                      The two matrices are made of Jordan blocks; in $2times2$ block format, they are
                      $$
                      N=begin{bmatrix} J & 0 \ 0 & 0 end{bmatrix}
                      qquad
                      M=begin{bmatrix} 0 & 0 \ 0 & J end{bmatrix}
                      $$
                      You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
                      $$
                      M=begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix} N
                      begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix}
                      $$






                      share|cite|improve this answer


























                        3














                        The two matrices are made of Jordan blocks; in $2times2$ block format, they are
                        $$
                        N=begin{bmatrix} J & 0 \ 0 & 0 end{bmatrix}
                        qquad
                        M=begin{bmatrix} 0 & 0 \ 0 & J end{bmatrix}
                        $$
                        You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
                        $$
                        M=begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix} N
                        begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix}
                        $$






                        share|cite|improve this answer
























                          3












                          3








                          3






                          The two matrices are made of Jordan blocks; in $2times2$ block format, they are
                          $$
                          N=begin{bmatrix} J & 0 \ 0 & 0 end{bmatrix}
                          qquad
                          M=begin{bmatrix} 0 & 0 \ 0 & J end{bmatrix}
                          $$
                          You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
                          $$
                          M=begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix} N
                          begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix}
                          $$






                          share|cite|improve this answer












                          The two matrices are made of Jordan blocks; in $2times2$ block format, they are
                          $$
                          N=begin{bmatrix} J & 0 \ 0 & 0 end{bmatrix}
                          qquad
                          M=begin{bmatrix} 0 & 0 \ 0 & J end{bmatrix}
                          $$
                          You get similar matrices if you perform a row switch together with the corresponding column switch; in this case there is only one possible switch:
                          $$
                          M=begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix} N
                          begin{bmatrix} 0 & I_2 \ I_2 & 0 end{bmatrix}
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          egreg

                          178k1484201




                          178k1484201






























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