Suppose BA is nilpotent, is it always true that AB is nilpotent?












0












$begingroup$


Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?



My thoughts so far:



$BA$ nilpotent $Rightarrow (BA)^{k} = 0$ for some positive integer k.



$(BA)^{k} = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^{k-1}A = 0$



But I have no idea how to proceed from here. If its not true, is there any counterexample?






linear-algebra







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$endgroup$












  • $begingroup$
    You can also explain the other way. AB and BA share eigenvalues. Eigenvalues of BA are zeros consequently also eigenvalues of AB. Hence AB is nilpotent.
    $endgroup$
    – Widawensen
    1 hour ago
















0












$begingroup$


Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?



My thoughts so far:



$BA$ nilpotent $Rightarrow (BA)^{k} = 0$ for some positive integer k.



$(BA)^{k} = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^{k-1}A = 0$



But I have no idea how to proceed from here. If its not true, is there any counterexample?






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can also explain the other way. AB and BA share eigenvalues. Eigenvalues of BA are zeros consequently also eigenvalues of AB. Hence AB is nilpotent.
    $endgroup$
    – Widawensen
    1 hour ago














0












0








0





$begingroup$


Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?



My thoughts so far:



$BA$ nilpotent $Rightarrow (BA)^{k} = 0$ for some positive integer k.



$(BA)^{k} = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^{k-1}A = 0$



But I have no idea how to proceed from here. If its not true, is there any counterexample?






linear-algebra







share|cite|improve this question









$endgroup$




Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?



My thoughts so far:



$BA$ nilpotent $Rightarrow (BA)^{k} = 0$ for some positive integer k.



$(BA)^{k} = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^{k-1}A = 0$



But I have no idea how to proceed from here. If its not true, is there any counterexample?






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









aaaaaa

274




274












  • $begingroup$
    You can also explain the other way. AB and BA share eigenvalues. Eigenvalues of BA are zeros consequently also eigenvalues of AB. Hence AB is nilpotent.
    $endgroup$
    – Widawensen
    1 hour ago


















  • $begingroup$
    You can also explain the other way. AB and BA share eigenvalues. Eigenvalues of BA are zeros consequently also eigenvalues of AB. Hence AB is nilpotent.
    $endgroup$
    – Widawensen
    1 hour ago
















$begingroup$
You can also explain the other way. AB and BA share eigenvalues. Eigenvalues of BA are zeros consequently also eigenvalues of AB. Hence AB is nilpotent.
$endgroup$
– Widawensen
1 hour ago




$begingroup$
You can also explain the other way. AB and BA share eigenvalues. Eigenvalues of BA are zeros consequently also eigenvalues of AB. Hence AB is nilpotent.
$endgroup$
– Widawensen
1 hour ago










3 Answers
3






active

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3












$begingroup$

$(BA)^{k} = (BA)(BA)....(BA) = BABA...BA =0$



Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^{k+1}=0$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    You just have to replace $A$ by $B$ in your computation. Then $(AB)^{k +1}= A(BA)^k B$. I assume you can finish from here.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      You can multiply by A on the left side and B on the right side, and you get $(AB)^{k+1} = 0$






      share|cite|improve this answer









      $endgroup$














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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

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        3












        $begingroup$

        $(BA)^{k} = (BA)(BA)....(BA) = BABA...BA =0$



        Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^{k+1}=0$$






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          $(BA)^{k} = (BA)(BA)....(BA) = BABA...BA =0$



          Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^{k+1}=0$$






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            $(BA)^{k} = (BA)(BA)....(BA) = BABA...BA =0$



            Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^{k+1}=0$$






            share|cite|improve this answer









            $endgroup$



            $(BA)^{k} = (BA)(BA)....(BA) = BABA...BA =0$



            Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^{k+1}=0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Fareed AFFareed AF

            677112




            677112























                4












                $begingroup$

                You just have to replace $A$ by $B$ in your computation. Then $(AB)^{k +1}= A(BA)^k B$. I assume you can finish from here.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  You just have to replace $A$ by $B$ in your computation. Then $(AB)^{k +1}= A(BA)^k B$. I assume you can finish from here.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    You just have to replace $A$ by $B$ in your computation. Then $(AB)^{k +1}= A(BA)^k B$. I assume you can finish from here.






                    share|cite|improve this answer









                    $endgroup$



                    You just have to replace $A$ by $B$ in your computation. Then $(AB)^{k +1}= A(BA)^k B$. I assume you can finish from here.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Severin SchravenSeverin Schraven

                    6,7602936




                    6,7602936























                        2












                        $begingroup$

                        You can multiply by A on the left side and B on the right side, and you get $(AB)^{k+1} = 0$






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          You can multiply by A on the left side and B on the right side, and you get $(AB)^{k+1} = 0$






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            You can multiply by A on the left side and B on the right side, and you get $(AB)^{k+1} = 0$






                            share|cite|improve this answer









                            $endgroup$



                            You can multiply by A on the left side and B on the right side, and you get $(AB)^{k+1} = 0$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            GatgatGatgat

                            1675




                            1675






























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