Question about matrices and eigenvectors












3












$begingroup$


For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29
















3












$begingroup$


For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29














3












3








3


1



$begingroup$


For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra







share|cite|improve this question











$endgroup$




For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 14:44









user376343

3,6783827




3,6783827










asked Dec 13 '18 at 12:12









Real_GaloisReal_Galois

255




255












  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29


















  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29
















$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14




$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14




4




4




$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19






$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19






2




2




$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23




$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23




1




1




$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27




$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27












$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29




$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29










2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



You should be able to continue from here since now $AB= BA$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Show $B=(A-I)^{-1}A=(B-I) A$.
    $endgroup$
    – user9077
    Dec 13 '18 at 13:49








  • 2




    $begingroup$
    If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
    $endgroup$
    – Widawensen
    Dec 13 '18 at 16:31






  • 1




    $begingroup$
    Nice observations, thanks!
    $endgroup$
    – user9077
    Dec 13 '18 at 16:48



















1












$begingroup$

With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

    votes









    4












    $begingroup$

    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48
















    4












    $begingroup$

    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48














    4












    4








    4





    $begingroup$

    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.






    share|cite|improve this answer











    $endgroup$



    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 13 '18 at 13:58

























    answered Dec 13 '18 at 13:25









    user9077user9077

    1,239612




    1,239612








    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48














    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48








    1




    1




    $begingroup$
    Show $B=(A-I)^{-1}A=(B-I) A$.
    $endgroup$
    – user9077
    Dec 13 '18 at 13:49






    $begingroup$
    Show $B=(A-I)^{-1}A=(B-I) A$.
    $endgroup$
    – user9077
    Dec 13 '18 at 13:49






    2




    2




    $begingroup$
    If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
    $endgroup$
    – Widawensen
    Dec 13 '18 at 16:31




    $begingroup$
    If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
    $endgroup$
    – Widawensen
    Dec 13 '18 at 16:31




    1




    1




    $begingroup$
    Nice observations, thanks!
    $endgroup$
    – user9077
    Dec 13 '18 at 16:48




    $begingroup$
    Nice observations, thanks!
    $endgroup$
    – user9077
    Dec 13 '18 at 16:48











    1












    $begingroup$

    With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
    Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
      Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
        Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






        share|cite|improve this answer









        $endgroup$



        With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
        Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 14:41









        user376343user376343

        3,6783827




        3,6783827






























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