Can we deduce Surjectivity from a property of Unitary linear map?

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Many books define the unitary linear map $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ as a bijective linear map such that
$left< x,y right>_1=left<Ux,Uy right>_2$.



And I have been trying to disprove that if $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ is a linear map such that
$left< x,y right>_1=left<Ux,Uy right>_2$ then $U$ is surjective.



I am pretty sure that the statement above is false since definition specify the bijectivity of unitary map and otherwise they don't need to do that. I hope someone can help me to come up with the counter example.




functional-analysis hilbert-spaces




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    Many books define the unitary linear map $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ as a bijective linear map such that
    $left< x,y right>_1=left<Ux,Uy right>_2$.



    And I have been trying to disprove that if $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ is a linear map such that
    $left< x,y right>_1=left<Ux,Uy right>_2$ then $U$ is surjective.



    I am pretty sure that the statement above is false since definition specify the bijectivity of unitary map and otherwise they don't need to do that. I hope someone can help me to come up with the counter example.




    functional-analysis hilbert-spaces




    share|cite|improve this question





















      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      Many books define the unitary linear map $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ as a bijective linear map such that
      $left< x,y right>_1=left<Ux,Uy right>_2$.



      And I have been trying to disprove that if $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ is a linear map such that
      $left< x,y right>_1=left<Ux,Uy right>_2$ then $U$ is surjective.



      I am pretty sure that the statement above is false since definition specify the bijectivity of unitary map and otherwise they don't need to do that. I hope someone can help me to come up with the counter example.




      functional-analysis hilbert-spaces




      share|cite|improve this question











      Many books define the unitary linear map $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ as a bijective linear map such that
      $left< x,y right>_1=left<Ux,Uy right>_2$.



      And I have been trying to disprove that if $U: (mathbbH_1,left<right>_1) rightarrow (mathbbH_2,left<right>_2)$ is a linear map such that
      $left< x,y right>_1=left<Ux,Uy right>_2$ then $U$ is surjective.



      I am pretty sure that the statement above is false since definition specify the bijectivity of unitary map and otherwise they don't need to do that. I hope someone can help me to come up with the counter example.





      functional-analysis hilbert-spaces





      share|cite|improve this question










      share|cite|improve this question




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      asked Jul 26 at 2:35









      Lev Ban

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          3 Answers
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          Let $H_1 = ell^2(mathbbN)$ and $H_2 = ell^2(mathbbN)$, with standard basis elements $e_0,e_1,e_2,cdots$. Then $U : H_1 rightarrow H_2$ defined by $Ue_n = e_2n$ satisfies $langle Ux,Uyrangle_2 = langle x,yrangle_1$, but $U$ is not surjective.






          share|cite|improve this answer





















          • Thanks for simple and clear example!
            – Lev Ban
            Jul 26 at 4:31

















          up vote
          2
          down vote













          Consider the space $ell^2=L^2(Bbb N)$ (where $Bbb N$ carries the counting measure) and the right-shift operator that sends a sequence $(a_n)_n in Bbb N$ to the sequence $(b_n)_n in Bbb N$ given by $b_0=0$ and $b_n=a_n-1$ for $ngeq 1$.






          share|cite|improve this answer




























            up vote
            2
            down vote













            Counterexamples are in order here. One is:



            Take



            $Bbb H_2 = Bbb H_1 oplus Bbb H_1, tag 1$



            and let the inner product on $Bbb H_2$ be given by



            $langle (x_1, y_1), (x_2, y_2) rangle_2 = langle x_1, x_2 rangle_1 + langle y_1, y_2 rangle_1, tag 2$



            where $(x_1, y_1), (x_2, y_2) in Bbb H_1 oplus Bbb H_1$; now take



            $U:Bbb H_1 to Bbb H_2, ; Ux = (x, 0); tag 3$



            then



            $langle Ux, Uy rangle_2 = langle (x, 0), (y, 0) rangle_2 = langle x, y rangle_1 + langle 0, 0 rangle_1 = langle x, y rangle_1; tag 4$



            but it is easy to see that our map $U$ is not surjective, since for $0 ne z in Bbb H_1$, $(0, z) in Bbb H_2$ is not in the image of $U$.






            share|cite|improve this answer





















            • Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
              – Lev Ban
              Jul 26 at 4:37











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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

            oldest

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            active

            oldest

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            up vote
            4
            down vote



            accepted










            Let $H_1 = ell^2(mathbbN)$ and $H_2 = ell^2(mathbbN)$, with standard basis elements $e_0,e_1,e_2,cdots$. Then $U : H_1 rightarrow H_2$ defined by $Ue_n = e_2n$ satisfies $langle Ux,Uyrangle_2 = langle x,yrangle_1$, but $U$ is not surjective.






            share|cite|improve this answer





















            • Thanks for simple and clear example!
              – Lev Ban
              Jul 26 at 4:31














            up vote
            4
            down vote



            accepted










            Let $H_1 = ell^2(mathbbN)$ and $H_2 = ell^2(mathbbN)$, with standard basis elements $e_0,e_1,e_2,cdots$. Then $U : H_1 rightarrow H_2$ defined by $Ue_n = e_2n$ satisfies $langle Ux,Uyrangle_2 = langle x,yrangle_1$, but $U$ is not surjective.






            share|cite|improve this answer





















            • Thanks for simple and clear example!
              – Lev Ban
              Jul 26 at 4:31












            up vote
            4
            down vote



            accepted







            up vote
            4
            down vote



            accepted






            Let $H_1 = ell^2(mathbbN)$ and $H_2 = ell^2(mathbbN)$, with standard basis elements $e_0,e_1,e_2,cdots$. Then $U : H_1 rightarrow H_2$ defined by $Ue_n = e_2n$ satisfies $langle Ux,Uyrangle_2 = langle x,yrangle_1$, but $U$ is not surjective.






            share|cite|improve this answer













            Let $H_1 = ell^2(mathbbN)$ and $H_2 = ell^2(mathbbN)$, with standard basis elements $e_0,e_1,e_2,cdots$. Then $U : H_1 rightarrow H_2$ defined by $Ue_n = e_2n$ satisfies $langle Ux,Uyrangle_2 = langle x,yrangle_1$, but $U$ is not surjective.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 26 at 3:07









            DisintegratingByParts

            55.5k42273




            55.5k42273











            • Thanks for simple and clear example!
              – Lev Ban
              Jul 26 at 4:31
















            • Thanks for simple and clear example!
              – Lev Ban
              Jul 26 at 4:31















            Thanks for simple and clear example!
            – Lev Ban
            Jul 26 at 4:31




            Thanks for simple and clear example!
            – Lev Ban
            Jul 26 at 4:31










            up vote
            2
            down vote













            Consider the space $ell^2=L^2(Bbb N)$ (where $Bbb N$ carries the counting measure) and the right-shift operator that sends a sequence $(a_n)_n in Bbb N$ to the sequence $(b_n)_n in Bbb N$ given by $b_0=0$ and $b_n=a_n-1$ for $ngeq 1$.






            share|cite|improve this answer

























              up vote
              2
              down vote













              Consider the space $ell^2=L^2(Bbb N)$ (where $Bbb N$ carries the counting measure) and the right-shift operator that sends a sequence $(a_n)_n in Bbb N$ to the sequence $(b_n)_n in Bbb N$ given by $b_0=0$ and $b_n=a_n-1$ for $ngeq 1$.






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                Consider the space $ell^2=L^2(Bbb N)$ (where $Bbb N$ carries the counting measure) and the right-shift operator that sends a sequence $(a_n)_n in Bbb N$ to the sequence $(b_n)_n in Bbb N$ given by $b_0=0$ and $b_n=a_n-1$ for $ngeq 1$.






                share|cite|improve this answer













                Consider the space $ell^2=L^2(Bbb N)$ (where $Bbb N$ carries the counting measure) and the right-shift operator that sends a sequence $(a_n)_n in Bbb N$ to the sequence $(b_n)_n in Bbb N$ given by $b_0=0$ and $b_n=a_n-1$ for $ngeq 1$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 26 at 2:59









                MatheinBoulomenos

                7,7211934




                7,7211934




















                    up vote
                    2
                    down vote













                    Counterexamples are in order here. One is:



                    Take



                    $Bbb H_2 = Bbb H_1 oplus Bbb H_1, tag 1$



                    and let the inner product on $Bbb H_2$ be given by



                    $langle (x_1, y_1), (x_2, y_2) rangle_2 = langle x_1, x_2 rangle_1 + langle y_1, y_2 rangle_1, tag 2$



                    where $(x_1, y_1), (x_2, y_2) in Bbb H_1 oplus Bbb H_1$; now take



                    $U:Bbb H_1 to Bbb H_2, ; Ux = (x, 0); tag 3$



                    then



                    $langle Ux, Uy rangle_2 = langle (x, 0), (y, 0) rangle_2 = langle x, y rangle_1 + langle 0, 0 rangle_1 = langle x, y rangle_1; tag 4$



                    but it is easy to see that our map $U$ is not surjective, since for $0 ne z in Bbb H_1$, $(0, z) in Bbb H_2$ is not in the image of $U$.






                    share|cite|improve this answer





















                    • Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
                      – Lev Ban
                      Jul 26 at 4:37















                    up vote
                    2
                    down vote













                    Counterexamples are in order here. One is:



                    Take



                    $Bbb H_2 = Bbb H_1 oplus Bbb H_1, tag 1$



                    and let the inner product on $Bbb H_2$ be given by



                    $langle (x_1, y_1), (x_2, y_2) rangle_2 = langle x_1, x_2 rangle_1 + langle y_1, y_2 rangle_1, tag 2$



                    where $(x_1, y_1), (x_2, y_2) in Bbb H_1 oplus Bbb H_1$; now take



                    $U:Bbb H_1 to Bbb H_2, ; Ux = (x, 0); tag 3$



                    then



                    $langle Ux, Uy rangle_2 = langle (x, 0), (y, 0) rangle_2 = langle x, y rangle_1 + langle 0, 0 rangle_1 = langle x, y rangle_1; tag 4$



                    but it is easy to see that our map $U$ is not surjective, since for $0 ne z in Bbb H_1$, $(0, z) in Bbb H_2$ is not in the image of $U$.






                    share|cite|improve this answer





















                    • Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
                      – Lev Ban
                      Jul 26 at 4:37













                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Counterexamples are in order here. One is:



                    Take



                    $Bbb H_2 = Bbb H_1 oplus Bbb H_1, tag 1$



                    and let the inner product on $Bbb H_2$ be given by



                    $langle (x_1, y_1), (x_2, y_2) rangle_2 = langle x_1, x_2 rangle_1 + langle y_1, y_2 rangle_1, tag 2$



                    where $(x_1, y_1), (x_2, y_2) in Bbb H_1 oplus Bbb H_1$; now take



                    $U:Bbb H_1 to Bbb H_2, ; Ux = (x, 0); tag 3$



                    then



                    $langle Ux, Uy rangle_2 = langle (x, 0), (y, 0) rangle_2 = langle x, y rangle_1 + langle 0, 0 rangle_1 = langle x, y rangle_1; tag 4$



                    but it is easy to see that our map $U$ is not surjective, since for $0 ne z in Bbb H_1$, $(0, z) in Bbb H_2$ is not in the image of $U$.






                    share|cite|improve this answer













                    Counterexamples are in order here. One is:



                    Take



                    $Bbb H_2 = Bbb H_1 oplus Bbb H_1, tag 1$



                    and let the inner product on $Bbb H_2$ be given by



                    $langle (x_1, y_1), (x_2, y_2) rangle_2 = langle x_1, x_2 rangle_1 + langle y_1, y_2 rangle_1, tag 2$



                    where $(x_1, y_1), (x_2, y_2) in Bbb H_1 oplus Bbb H_1$; now take



                    $U:Bbb H_1 to Bbb H_2, ; Ux = (x, 0); tag 3$



                    then



                    $langle Ux, Uy rangle_2 = langle (x, 0), (y, 0) rangle_2 = langle x, y rangle_1 + langle 0, 0 rangle_1 = langle x, y rangle_1; tag 4$



                    but it is easy to see that our map $U$ is not surjective, since for $0 ne z in Bbb H_1$, $(0, z) in Bbb H_2$ is not in the image of $U$.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Jul 26 at 3:07









                    Robert Lewis

                    36.8k22155




                    36.8k22155











                    • Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
                      – Lev Ban
                      Jul 26 at 4:37

















                    • Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
                      – Lev Ban
                      Jul 26 at 4:37
















                    Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
                    – Lev Ban
                    Jul 26 at 4:37





                    Consider that for any $x_1,x_2in mathbbHsetminus 0 $ , $ left<(x_1,x_1),(x_2,-x_2) right>_2= left< x_1,x_2 right>_1+left< x_1,-x_2 right>_1=0$ even if $(x_1,x_2)neq 0$. So, it seems the defined inner product is not correct one.
                    – Lev Ban
                    Jul 26 at 4:37













                     

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