Prove that $T$ is the orthonal projection on the range of the right shift












-2












$begingroup$


Consider the following operator on $ell_{mathbb{N}^*}^2(mathbb{C})$ defined as:



$$ T(x_1,x_2,cdots)=(0,x_2,x_3,cdots).$$




I want prove that
$$sigma(T)={0,1}.$$




I have the following idea: let us consider the right shift
$$S_r(x_1,x_2,cdots)=(0,x_1,x_2,x_3,cdots).$$
If I show that $T$ is the orthonal projection on the range of $S_r$, then clearly
$$sigma(T)={0,1}.$$



Because, it is well known that the spectrum of an Orthogonal Projection Operator is equal to ${0,1}$.



Note that
$$T=S_rT_ell,$$
where
$$S_ell(x_1,x_2,cdots)=(x_2,x_3,cdots).$$






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand why all this downvotes? Is the question is obvious?
    $endgroup$
    – Schüler
    Dec 9 '18 at 16:35
















-2












$begingroup$


Consider the following operator on $ell_{mathbb{N}^*}^2(mathbb{C})$ defined as:



$$ T(x_1,x_2,cdots)=(0,x_2,x_3,cdots).$$




I want prove that
$$sigma(T)={0,1}.$$




I have the following idea: let us consider the right shift
$$S_r(x_1,x_2,cdots)=(0,x_1,x_2,x_3,cdots).$$
If I show that $T$ is the orthonal projection on the range of $S_r$, then clearly
$$sigma(T)={0,1}.$$



Because, it is well known that the spectrum of an Orthogonal Projection Operator is equal to ${0,1}$.



Note that
$$T=S_rT_ell,$$
where
$$S_ell(x_1,x_2,cdots)=(x_2,x_3,cdots).$$






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand why all this downvotes? Is the question is obvious?
    $endgroup$
    – Schüler
    Dec 9 '18 at 16:35














-2












-2








-2





$begingroup$


Consider the following operator on $ell_{mathbb{N}^*}^2(mathbb{C})$ defined as:



$$ T(x_1,x_2,cdots)=(0,x_2,x_3,cdots).$$




I want prove that
$$sigma(T)={0,1}.$$




I have the following idea: let us consider the right shift
$$S_r(x_1,x_2,cdots)=(0,x_1,x_2,x_3,cdots).$$
If I show that $T$ is the orthonal projection on the range of $S_r$, then clearly
$$sigma(T)={0,1}.$$



Because, it is well known that the spectrum of an Orthogonal Projection Operator is equal to ${0,1}$.



Note that
$$T=S_rT_ell,$$
where
$$S_ell(x_1,x_2,cdots)=(x_2,x_3,cdots).$$






functional-analysis operator-theory hilbert-spaces







share|cite|improve this question











$endgroup$




Consider the following operator on $ell_{mathbb{N}^*}^2(mathbb{C})$ defined as:



$$ T(x_1,x_2,cdots)=(0,x_2,x_3,cdots).$$




I want prove that
$$sigma(T)={0,1}.$$




I have the following idea: let us consider the right shift
$$S_r(x_1,x_2,cdots)=(0,x_1,x_2,x_3,cdots).$$
If I show that $T$ is the orthonal projection on the range of $S_r$, then clearly
$$sigma(T)={0,1}.$$



Because, it is well known that the spectrum of an Orthogonal Projection Operator is equal to ${0,1}$.



Note that
$$T=S_rT_ell,$$
where
$$S_ell(x_1,x_2,cdots)=(x_2,x_3,cdots).$$






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








edited Dec 9 '18 at 19:43







Schüler

















asked Dec 9 '18 at 13:58









SchülerSchüler

1,4841421




1,4841421












  • $begingroup$
    I don't understand why all this downvotes? Is the question is obvious?
    $endgroup$
    – Schüler
    Dec 9 '18 at 16:35


















  • $begingroup$
    I don't understand why all this downvotes? Is the question is obvious?
    $endgroup$
    – Schüler
    Dec 9 '18 at 16:35
















$begingroup$
I don't understand why all this downvotes? Is the question is obvious?
$endgroup$
– Schüler
Dec 9 '18 at 16:35




$begingroup$
I don't understand why all this downvotes? Is the question is obvious?
$endgroup$
– Schüler
Dec 9 '18 at 16:35










2 Answers
2






active

oldest

votes


















2












$begingroup$

I cannot see why you want to involve the shift at all.



You have $T^2-T=0$, so $${0}={lambda^2-lambda: lambdainsigma(T)}.$$So $sigma(T)={0,1}$.



For a way to prove this without any theory, look at this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
    $endgroup$
    – Schüler
    Dec 9 '18 at 20:35










  • $begingroup$
    Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
    $endgroup$
    – Martin Argerami
    Dec 9 '18 at 20:47












  • $begingroup$
    It is idempotent and self adjoint.
    $endgroup$
    – Schüler
    Dec 9 '18 at 21:00










  • $begingroup$
    So i think we should show that T is self adjoint
    $endgroup$
    – Schüler
    Dec 9 '18 at 21:01










  • $begingroup$
    That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
    $endgroup$
    – Martin Argerami
    Dec 9 '18 at 21:13



















0












$begingroup$

First of all, let $e_1 = left( 1, 0, dots right) $ be the first element of the canonical basis for $ell^2(mathbb{C})$ and $W= e_1^{bot}=langle e_irangle_{igeq 2})$ (here $langle bullet rangle$ identifies the linear span).



Both $langle e_1rangle$ and $W$ are $T$-invariant, so $T-lambda Id$ is invertible iff $T-lambda Id |_{langle e_1 rangle} $ and $T-lambda Id|_W$ are both invertible.



$T-lambda Id |_{langle e_1 rangle} equiv -lambda Id $ so it is invertible iff $lambda neq 0$.



$T-lambda Id|_W equiv left(1-lambdaright)Id $ so it is invertible iff $lambda neq 1$.






share|cite|improve this answer









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    2 Answers
    2






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    2 Answers
    2






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    2












    $begingroup$

    I cannot see why you want to involve the shift at all.



    You have $T^2-T=0$, so $${0}={lambda^2-lambda: lambdainsigma(T)}.$$So $sigma(T)={0,1}$.



    For a way to prove this without any theory, look at this.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
      $endgroup$
      – Schüler
      Dec 9 '18 at 20:35










    • $begingroup$
      Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 20:47












    • $begingroup$
      It is idempotent and self adjoint.
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:00










    • $begingroup$
      So i think we should show that T is self adjoint
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:01










    • $begingroup$
      That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 21:13
















    2












    $begingroup$

    I cannot see why you want to involve the shift at all.



    You have $T^2-T=0$, so $${0}={lambda^2-lambda: lambdainsigma(T)}.$$So $sigma(T)={0,1}$.



    For a way to prove this without any theory, look at this.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
      $endgroup$
      – Schüler
      Dec 9 '18 at 20:35










    • $begingroup$
      Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 20:47












    • $begingroup$
      It is idempotent and self adjoint.
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:00










    • $begingroup$
      So i think we should show that T is self adjoint
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:01










    • $begingroup$
      That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 21:13














    2












    2








    2





    $begingroup$

    I cannot see why you want to involve the shift at all.



    You have $T^2-T=0$, so $${0}={lambda^2-lambda: lambdainsigma(T)}.$$So $sigma(T)={0,1}$.



    For a way to prove this without any theory, look at this.






    share|cite|improve this answer











    $endgroup$



    I cannot see why you want to involve the shift at all.



    You have $T^2-T=0$, so $${0}={lambda^2-lambda: lambdainsigma(T)}.$$So $sigma(T)={0,1}$.



    For a way to prove this without any theory, look at this.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 10 '18 at 3:11

























    answered Dec 9 '18 at 20:21









    Martin ArgeramiMartin Argerami

    126k1182180




    126k1182180












    • $begingroup$
      Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
      $endgroup$
      – Schüler
      Dec 9 '18 at 20:35










    • $begingroup$
      Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 20:47












    • $begingroup$
      It is idempotent and self adjoint.
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:00










    • $begingroup$
      So i think we should show that T is self adjoint
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:01










    • $begingroup$
      That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 21:13


















    • $begingroup$
      Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
      $endgroup$
      – Schüler
      Dec 9 '18 at 20:35










    • $begingroup$
      Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 20:47












    • $begingroup$
      It is idempotent and self adjoint.
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:00










    • $begingroup$
      So i think we should show that T is self adjoint
      $endgroup$
      – Schüler
      Dec 9 '18 at 21:01










    • $begingroup$
      That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
      $endgroup$
      – Martin Argerami
      Dec 9 '18 at 21:13
















    $begingroup$
    Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
    $endgroup$
    – Schüler
    Dec 9 '18 at 20:35




    $begingroup$
    Thank you Professor. I don't understant why $T^2-T=0$ implies that $${0}={lambda^2-lambda: lambdainsigma(T)}?$$ My goal to introduce the shift is to show that T is an orthogonal projection.
    $endgroup$
    – Schüler
    Dec 9 '18 at 20:35












    $begingroup$
    Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
    $endgroup$
    – Martin Argerami
    Dec 9 '18 at 20:47






    $begingroup$
    Look up Spectral Mapping Theorem. As for the shift, it is completely irrelevant here. What's your definition of "orthogonal projection"?
    $endgroup$
    – Martin Argerami
    Dec 9 '18 at 20:47














    $begingroup$
    It is idempotent and self adjoint.
    $endgroup$
    – Schüler
    Dec 9 '18 at 21:00




    $begingroup$
    It is idempotent and self adjoint.
    $endgroup$
    – Schüler
    Dec 9 '18 at 21:00












    $begingroup$
    So i think we should show that T is self adjoint
    $endgroup$
    – Schüler
    Dec 9 '18 at 21:01




    $begingroup$
    So i think we should show that T is self adjoint
    $endgroup$
    – Schüler
    Dec 9 '18 at 21:01












    $begingroup$
    That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
    $endgroup$
    – Martin Argerami
    Dec 9 '18 at 21:13




    $begingroup$
    That was not your question (you asked to prove that the spectrum of $T$ was ${0,1}$). But that $T$ is orthogonal is trivial, by definition: you just calculate $langle Tx,yrangle$ and $langle x,Tyrangle$.
    $endgroup$
    – Martin Argerami
    Dec 9 '18 at 21:13











    0












    $begingroup$

    First of all, let $e_1 = left( 1, 0, dots right) $ be the first element of the canonical basis for $ell^2(mathbb{C})$ and $W= e_1^{bot}=langle e_irangle_{igeq 2})$ (here $langle bullet rangle$ identifies the linear span).



    Both $langle e_1rangle$ and $W$ are $T$-invariant, so $T-lambda Id$ is invertible iff $T-lambda Id |_{langle e_1 rangle} $ and $T-lambda Id|_W$ are both invertible.



    $T-lambda Id |_{langle e_1 rangle} equiv -lambda Id $ so it is invertible iff $lambda neq 0$.



    $T-lambda Id|_W equiv left(1-lambdaright)Id $ so it is invertible iff $lambda neq 1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      First of all, let $e_1 = left( 1, 0, dots right) $ be the first element of the canonical basis for $ell^2(mathbb{C})$ and $W= e_1^{bot}=langle e_irangle_{igeq 2})$ (here $langle bullet rangle$ identifies the linear span).



      Both $langle e_1rangle$ and $W$ are $T$-invariant, so $T-lambda Id$ is invertible iff $T-lambda Id |_{langle e_1 rangle} $ and $T-lambda Id|_W$ are both invertible.



      $T-lambda Id |_{langle e_1 rangle} equiv -lambda Id $ so it is invertible iff $lambda neq 0$.



      $T-lambda Id|_W equiv left(1-lambdaright)Id $ so it is invertible iff $lambda neq 1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        First of all, let $e_1 = left( 1, 0, dots right) $ be the first element of the canonical basis for $ell^2(mathbb{C})$ and $W= e_1^{bot}=langle e_irangle_{igeq 2})$ (here $langle bullet rangle$ identifies the linear span).



        Both $langle e_1rangle$ and $W$ are $T$-invariant, so $T-lambda Id$ is invertible iff $T-lambda Id |_{langle e_1 rangle} $ and $T-lambda Id|_W$ are both invertible.



        $T-lambda Id |_{langle e_1 rangle} equiv -lambda Id $ so it is invertible iff $lambda neq 0$.



        $T-lambda Id|_W equiv left(1-lambdaright)Id $ so it is invertible iff $lambda neq 1$.






        share|cite|improve this answer









        $endgroup$



        First of all, let $e_1 = left( 1, 0, dots right) $ be the first element of the canonical basis for $ell^2(mathbb{C})$ and $W= e_1^{bot}=langle e_irangle_{igeq 2})$ (here $langle bullet rangle$ identifies the linear span).



        Both $langle e_1rangle$ and $W$ are $T$-invariant, so $T-lambda Id$ is invertible iff $T-lambda Id |_{langle e_1 rangle} $ and $T-lambda Id|_W$ are both invertible.



        $T-lambda Id |_{langle e_1 rangle} equiv -lambda Id $ so it is invertible iff $lambda neq 0$.



        $T-lambda Id|_W equiv left(1-lambdaright)Id $ so it is invertible iff $lambda neq 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 14:21









        Lord_BanquoLord_Banquo

        417




        417






























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