show that $ker(Id-T) = ker(Id-T)^{star}$












2












$begingroup$


$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$



by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$



it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$



but I fail to see how one must proceed.






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
    $endgroup$
    – mechanodroid
    Jan 7 at 21:59
















2












$begingroup$


$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$



by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$



it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$



but I fail to see how one must proceed.






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
    $endgroup$
    – mechanodroid
    Jan 7 at 21:59














2












2








2





$begingroup$


$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$



by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$



it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$



but I fail to see how one must proceed.






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question









$endgroup$




$H$ is a hilbert space and $T$ is a bounded linear operator on $H$, also $|T| leq 1$



by calculating $|Tx-x|^2$ I have shown the following string of equivalences $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, Txrangle = |x|^2 $$



it's supposed to help me prove $$ker(Id-T) = ker(Id-T)^{star}$$



but I fail to see how one must proceed.






functional-analysis operator-theory hilbert-spaces




functional-analysis operator-theory hilbert-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 21:24









rapidracimrapidracim

1,5331319




1,5331319












  • $begingroup$
    Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
    $endgroup$
    – mechanodroid
    Jan 7 at 21:59


















  • $begingroup$
    Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
    $endgroup$
    – mechanodroid
    Jan 7 at 21:59
















$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59




$begingroup$
Note that if $|T| < 1$ then $I-T$ and $I-T^*$ are both invertible so in particular they both have trivial kernels.
$endgroup$
– mechanodroid
Jan 7 at 21:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

You made an error in your equivalence chain.



In fact, you have:



$$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$



You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
, this leads to
$$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$



and concludes the proof.






share|cite|improve this answer









$endgroup$













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    1 Answer
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    $begingroup$

    You made an error in your equivalence chain.



    In fact, you have:



    $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$



    You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
    , this leads to
    $$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$



    and concludes the proof.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You made an error in your equivalence chain.



      In fact, you have:



      $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$



      You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
      , this leads to
      $$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$



      and concludes the proof.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You made an error in your equivalence chain.



        In fact, you have:



        $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$



        You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
        , this leads to
        $$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$



        and concludes the proof.






        share|cite|improve this answer









        $endgroup$



        You made an error in your equivalence chain.



        In fact, you have:



        $$Tx = x iff langle,Tx, xrangle = |x|^2 iff langle,x, T^* xrangle = |x|^2.$$



        You also have $Vert T^* Vert le Vert T Vert le 1$. In conjonction with $langle,x, T^* xrangle = |x|^2$
        , this leads to
        $$langle,x, T^* xrangle = |x|^2 iff T^*x=x$$



        and concludes the proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 21:39









        mathcounterexamples.netmathcounterexamples.net

        26k21955




        26k21955






























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