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












6












$begingroup$


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











$endgroup$

















    6












    $begingroup$


    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











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      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











      $endgroup$




      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




      share|cite|improve this question








      edited Jan 3 at 10:41









      José Carlos Santos

      159k22126229




      159k22126229










      asked Jan 3 at 10:22









      MPB94MPB94

      27017




      27017






















          3 Answers
          3






          active

          oldest

          votes


















          14












          $begingroup$

          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











          $endgroup$





















            8












            $begingroup$

            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









            $endgroup$





















              0












              $begingroup$

              Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.






              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









                14












                $begingroup$

                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











                $endgroup$


















                  14












                  $begingroup$

                  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











                  $endgroup$
















                    14












                    14








                    14





                    $begingroup$

                    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











                    $endgroup$



                    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








                    edited Jan 3 at 23:20

























                    answered Jan 3 at 10:30









                    José Carlos SantosJosé Carlos Santos

                    159k22126229




                    159k22126229























                        8












                        $begingroup$

                        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









                        $endgroup$


















                          8












                          $begingroup$

                          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









                          $endgroup$
















                            8












                            8








                            8





                            $begingroup$

                            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









                            $endgroup$



                            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 Jan 3 at 10:40









                            egregegreg

                            181k1485203




                            181k1485203























                                0












                                $begingroup$

                                Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 3 at 16:16









                                    AcccumulationAcccumulation

                                    6,8942618




                                    6,8942618






























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