Prove there exists a square matrix $B$ such that $A^m=A^m+1B$

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$A$ is a square matrix. Prove



  1. there exists $m$ such that $operatornamerank(A^m)=operatornamerank(A^m+1)$

  2. there exists a square matrix $B$ such that $A^m=A^m+1B$

I already knew the first one , since $A^m+1=A^mcdot A$ , then $operatornamerank(A^m)ge operatornamerank(A^m+1) $. If there is no $m$ to satisfy $operatornamerank(A^m)=operatornamerank(A^m+1)$ , then $operatornamerank(A^m)gt operatornamerank(A^m+1) $. And there will be $operatornamerank(A^k)=0$, it follows $operatornamerank(A^k+1)=0$. Now I really don't know how to deal with the second one. Is there any connection?




linear-algebra matrices




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    $A$ is a square matrix. Prove



    1. there exists $m$ such that $operatornamerank(A^m)=operatornamerank(A^m+1)$

    2. there exists a square matrix $B$ such that $A^m=A^m+1B$

    I already knew the first one , since $A^m+1=A^mcdot A$ , then $operatornamerank(A^m)ge operatornamerank(A^m+1) $. If there is no $m$ to satisfy $operatornamerank(A^m)=operatornamerank(A^m+1)$ , then $operatornamerank(A^m)gt operatornamerank(A^m+1) $. And there will be $operatornamerank(A^k)=0$, it follows $operatornamerank(A^k+1)=0$. Now I really don't know how to deal with the second one. Is there any connection?




    linear-algebra matrices




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      $A$ is a square matrix. Prove



      1. there exists $m$ such that $operatornamerank(A^m)=operatornamerank(A^m+1)$

      2. there exists a square matrix $B$ such that $A^m=A^m+1B$

      I already knew the first one , since $A^m+1=A^mcdot A$ , then $operatornamerank(A^m)ge operatornamerank(A^m+1) $. If there is no $m$ to satisfy $operatornamerank(A^m)=operatornamerank(A^m+1)$ , then $operatornamerank(A^m)gt operatornamerank(A^m+1) $. And there will be $operatornamerank(A^k)=0$, it follows $operatornamerank(A^k+1)=0$. Now I really don't know how to deal with the second one. Is there any connection?




      linear-algebra matrices




      share|cite|improve this question














      $A$ is a square matrix. Prove



      1. there exists $m$ such that $operatornamerank(A^m)=operatornamerank(A^m+1)$

      2. there exists a square matrix $B$ such that $A^m=A^m+1B$

      I already knew the first one , since $A^m+1=A^mcdot A$ , then $operatornamerank(A^m)ge operatornamerank(A^m+1) $. If there is no $m$ to satisfy $operatornamerank(A^m)=operatornamerank(A^m+1)$ , then $operatornamerank(A^m)gt operatornamerank(A^m+1) $. And there will be $operatornamerank(A^k)=0$, it follows $operatornamerank(A^k+1)=0$. Now I really don't know how to deal with the second one. Is there any connection?





      linear-algebra matrices





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      edited Aug 24 at 8:02









      Bernard

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      111k635103










      asked Aug 24 at 5:21









      Jaqen Chou

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          Hint:



          • Show that $A^m$ and $A^m+1$ have the same column space, using one inclusion and equality of dimensions.


          • Use multiplication on the right by elementary matrices (which amounts to invertible column operations).






          share|cite|improve this answer




















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            up vote
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            down vote













            Hint:



            • Show that $A^m$ and $A^m+1$ have the same column space, using one inclusion and equality of dimensions.


            • Use multiplication on the right by elementary matrices (which amounts to invertible column operations).






            share|cite|improve this answer
























              up vote
              2
              down vote













              Hint:



              • Show that $A^m$ and $A^m+1$ have the same column space, using one inclusion and equality of dimensions.


              • Use multiplication on the right by elementary matrices (which amounts to invertible column operations).






              share|cite|improve this answer






















                up vote
                2
                down vote










                up vote
                2
                down vote









                Hint:



                • Show that $A^m$ and $A^m+1$ have the same column space, using one inclusion and equality of dimensions.


                • Use multiplication on the right by elementary matrices (which amounts to invertible column operations).






                share|cite|improve this answer












                Hint:



                • Show that $A^m$ and $A^m+1$ have the same column space, using one inclusion and equality of dimensions.


                • Use multiplication on the right by elementary matrices (which amounts to invertible column operations).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 24 at 5:28









                Arnaud Mortier

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