There exists a linear operator with no proper invariant subspaces












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Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.



All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.






linear-algebra functional-analysis operator-algebras







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    5












    $begingroup$


    Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.



    All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.






    linear-algebra functional-analysis operator-algebras







    share|cite|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.



      All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.






      linear-algebra functional-analysis operator-algebras







      share|cite|improve this question









      $endgroup$




      Let $A$ be a bounded operator on a Hilbert space $H$ with two invariant subspaces $M$ and $N$ s.t. $N subset M$, dim$(M cap N^{perp})> 1$, and have no invariant subspaces between $N$ and $M$. Then, show that, there exists an operator $B$ on $H$ which has no proper invariant subspace.



      All I want a hint for constructing $B$ with the help of $A$ and given conditions, even a little hint will be appreciated. Thanks in advance.






      linear-algebra functional-analysis operator-algebras




      linear-algebra functional-analysis operator-algebras






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 8 at 11:38









      splifesplife

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          $begingroup$

          The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.






          share|cite|improve this answer









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            $begingroup$

            The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.






                share|cite|improve this answer









                $endgroup$



                The hint is to think about the invariant subspaces of $A$ the lie in $Mcap N^perp$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 13:53









                Martin ArgeramiMartin Argerami

                126k1182180




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