Normal operator in Hilbert Complex space share an eigenvalue.












0














Everyone, I get stuck in an exercise of Functional Analysis.



Let $T in B(H)$ (H a complex Hilbert space) and $T^*$ adjoint of $T$. Supose $T$ is a normal operator.



1) Prove that $Ker(T)= Ker(T^*) = R(T)^perp$ - I've finished this.



2) Using previous proof. If $alpha$ is an eigenvalue of $T$ then the conjugate $baralpha$ is an eigenvalue of $T^*$.



3) If $alpha neq beta $ are eigenvectors of T, then the associated eigenspaces are orthogonal between them.



My try:



2) Let $x in H$ such that $Tx = alpha x $.



$$<x,Tx> = <T^*x,x> = alpha<x,x> = <baralpha x,x>$$



Then we have that



$$<T^*x - bar alpha x,x>=0$$



Here I don't know if above implies that $T^*x - bar alpha x =0$.



3) I didn't make a great progress here.






functional-analysis hilbert-spaces adjoint-operators







share|cite|improve this question




















  • 1




    Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing.
    – DisintegratingByParts
    Nov 29 at 19:05


















0














Everyone, I get stuck in an exercise of Functional Analysis.



Let $T in B(H)$ (H a complex Hilbert space) and $T^*$ adjoint of $T$. Supose $T$ is a normal operator.



1) Prove that $Ker(T)= Ker(T^*) = R(T)^perp$ - I've finished this.



2) Using previous proof. If $alpha$ is an eigenvalue of $T$ then the conjugate $baralpha$ is an eigenvalue of $T^*$.



3) If $alpha neq beta $ are eigenvectors of T, then the associated eigenspaces are orthogonal between them.



My try:



2) Let $x in H$ such that $Tx = alpha x $.



$$<x,Tx> = <T^*x,x> = alpha<x,x> = <baralpha x,x>$$



Then we have that



$$<T^*x - bar alpha x,x>=0$$



Here I don't know if above implies that $T^*x - bar alpha x =0$.



3) I didn't make a great progress here.






functional-analysis hilbert-spaces adjoint-operators







share|cite|improve this question




















  • 1




    Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing.
    – DisintegratingByParts
    Nov 29 at 19:05
















0












0








0







Everyone, I get stuck in an exercise of Functional Analysis.



Let $T in B(H)$ (H a complex Hilbert space) and $T^*$ adjoint of $T$. Supose $T$ is a normal operator.



1) Prove that $Ker(T)= Ker(T^*) = R(T)^perp$ - I've finished this.



2) Using previous proof. If $alpha$ is an eigenvalue of $T$ then the conjugate $baralpha$ is an eigenvalue of $T^*$.



3) If $alpha neq beta $ are eigenvectors of T, then the associated eigenspaces are orthogonal between them.



My try:



2) Let $x in H$ such that $Tx = alpha x $.



$$<x,Tx> = <T^*x,x> = alpha<x,x> = <baralpha x,x>$$



Then we have that



$$<T^*x - bar alpha x,x>=0$$



Here I don't know if above implies that $T^*x - bar alpha x =0$.



3) I didn't make a great progress here.






functional-analysis hilbert-spaces adjoint-operators







share|cite|improve this question















Everyone, I get stuck in an exercise of Functional Analysis.



Let $T in B(H)$ (H a complex Hilbert space) and $T^*$ adjoint of $T$. Supose $T$ is a normal operator.



1) Prove that $Ker(T)= Ker(T^*) = R(T)^perp$ - I've finished this.



2) Using previous proof. If $alpha$ is an eigenvalue of $T$ then the conjugate $baralpha$ is an eigenvalue of $T^*$.



3) If $alpha neq beta $ are eigenvectors of T, then the associated eigenspaces are orthogonal between them.



My try:



2) Let $x in H$ such that $Tx = alpha x $.



$$<x,Tx> = <T^*x,x> = alpha<x,x> = <baralpha x,x>$$



Then we have that



$$<T^*x - bar alpha x,x>=0$$



Here I don't know if above implies that $T^*x - bar alpha x =0$.



3) I didn't make a great progress here.






functional-analysis hilbert-spaces adjoint-operators




functional-analysis hilbert-spaces adjoint-operators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 19:12

























asked Nov 29 at 18:33









Kutz

327




327








  • 1




    Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing.
    – DisintegratingByParts
    Nov 29 at 19:05
















  • 1




    Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing.
    – DisintegratingByParts
    Nov 29 at 19:05










1




1




Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing.
– DisintegratingByParts
Nov 29 at 19:05






Use T^* to get $T^*$. Your notation without the * in the exponent can be confusing.
– DisintegratingByParts
Nov 29 at 19:05












2 Answers
2






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1














For part 2, using the proof of part 1 (that $langle Tx, Txrangle = langle T^*x, T^*x rangle$), we see$$||T^*x-overline{alpha}x||^2 = langle T^*x-overline{alpha}x, T^*x-overline{alpha}xrangle = langle T^*x, T^*xrangle - langle overline{alpha}x, T^*xrangle - langle T^*x, overline{alpha}xrangle+langle overline{alpha} x, overline{alpha} xrangle$$ $$ = langle Tx, Tx rangle-langle Tx, alpha xrangle-langle alpha x, Txrangle + langle alpha x, alpha xrangle = 0.$$



For part 3, if $Tx = alpha x, Ty = beta y$, then $$alphalangle x,y rangle = langle Tx,y rangle = langle x,T^*y rangle = langle x, overline{beta} y rangle = beta langle x,y rangle$$ implies $langle x,y rangle = 0$. This means that the eigenspaces are orthogonal.






share|cite|improve this answer























  • Thank you! for this part.
    – Kutz
    Nov 29 at 19:10










  • @Kutz I added part 3
    – mathworker21
    Nov 29 at 19:13










  • Thank you for answer.
    – Kutz
    Nov 29 at 19:22



















0














$T$ is normal means $T^*T=TT^*$, which is equivalent to
$$
|Tx| = |T^*x|,;;; xin H.
$$
So $mathcal{N}(T)=mathcal{N}(T^*)$ follows. The sum of normal operators is normal, and any scalar times a normal operator is normal. And the identity $I$ is normal. So, if $T$ is normal, then so is $alpha I-T$ for any scalar $alpha$. Therefore $mathcal{N}(T-alpha I)=mathcal{N}(T^*-overline{alpha}I)$, and any eigenvector of a normal $T$ with eigenvalue $alpha$ is an eigenvector of $T^*$ with eigenvalue $overline{alpha}$. If $Tx=alpha x$ and $Ty=beta y$, then



begin{align}
(alpha-beta)langle x,yrangle
& = langle alpha x,yrangle-langle x,overline{beta}yrangle \
& = langle Tx,yrangle-langle x,T^*yrangle \
& = langle Tx,yrangle-langle Tx,yrangle =0.
end{align}
Therefore, if $alphane beta$, it follows that $xperp y$.






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    2 Answers
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    1














    For part 2, using the proof of part 1 (that $langle Tx, Txrangle = langle T^*x, T^*x rangle$), we see$$||T^*x-overline{alpha}x||^2 = langle T^*x-overline{alpha}x, T^*x-overline{alpha}xrangle = langle T^*x, T^*xrangle - langle overline{alpha}x, T^*xrangle - langle T^*x, overline{alpha}xrangle+langle overline{alpha} x, overline{alpha} xrangle$$ $$ = langle Tx, Tx rangle-langle Tx, alpha xrangle-langle alpha x, Txrangle + langle alpha x, alpha xrangle = 0.$$



    For part 3, if $Tx = alpha x, Ty = beta y$, then $$alphalangle x,y rangle = langle Tx,y rangle = langle x,T^*y rangle = langle x, overline{beta} y rangle = beta langle x,y rangle$$ implies $langle x,y rangle = 0$. This means that the eigenspaces are orthogonal.






    share|cite|improve this answer























    • Thank you! for this part.
      – Kutz
      Nov 29 at 19:10










    • @Kutz I added part 3
      – mathworker21
      Nov 29 at 19:13










    • Thank you for answer.
      – Kutz
      Nov 29 at 19:22
















    1














    For part 2, using the proof of part 1 (that $langle Tx, Txrangle = langle T^*x, T^*x rangle$), we see$$||T^*x-overline{alpha}x||^2 = langle T^*x-overline{alpha}x, T^*x-overline{alpha}xrangle = langle T^*x, T^*xrangle - langle overline{alpha}x, T^*xrangle - langle T^*x, overline{alpha}xrangle+langle overline{alpha} x, overline{alpha} xrangle$$ $$ = langle Tx, Tx rangle-langle Tx, alpha xrangle-langle alpha x, Txrangle + langle alpha x, alpha xrangle = 0.$$



    For part 3, if $Tx = alpha x, Ty = beta y$, then $$alphalangle x,y rangle = langle Tx,y rangle = langle x,T^*y rangle = langle x, overline{beta} y rangle = beta langle x,y rangle$$ implies $langle x,y rangle = 0$. This means that the eigenspaces are orthogonal.






    share|cite|improve this answer























    • Thank you! for this part.
      – Kutz
      Nov 29 at 19:10










    • @Kutz I added part 3
      – mathworker21
      Nov 29 at 19:13










    • Thank you for answer.
      – Kutz
      Nov 29 at 19:22














    1












    1








    1






    For part 2, using the proof of part 1 (that $langle Tx, Txrangle = langle T^*x, T^*x rangle$), we see$$||T^*x-overline{alpha}x||^2 = langle T^*x-overline{alpha}x, T^*x-overline{alpha}xrangle = langle T^*x, T^*xrangle - langle overline{alpha}x, T^*xrangle - langle T^*x, overline{alpha}xrangle+langle overline{alpha} x, overline{alpha} xrangle$$ $$ = langle Tx, Tx rangle-langle Tx, alpha xrangle-langle alpha x, Txrangle + langle alpha x, alpha xrangle = 0.$$



    For part 3, if $Tx = alpha x, Ty = beta y$, then $$alphalangle x,y rangle = langle Tx,y rangle = langle x,T^*y rangle = langle x, overline{beta} y rangle = beta langle x,y rangle$$ implies $langle x,y rangle = 0$. This means that the eigenspaces are orthogonal.






    share|cite|improve this answer














    For part 2, using the proof of part 1 (that $langle Tx, Txrangle = langle T^*x, T^*x rangle$), we see$$||T^*x-overline{alpha}x||^2 = langle T^*x-overline{alpha}x, T^*x-overline{alpha}xrangle = langle T^*x, T^*xrangle - langle overline{alpha}x, T^*xrangle - langle T^*x, overline{alpha}xrangle+langle overline{alpha} x, overline{alpha} xrangle$$ $$ = langle Tx, Tx rangle-langle Tx, alpha xrangle-langle alpha x, Txrangle + langle alpha x, alpha xrangle = 0.$$



    For part 3, if $Tx = alpha x, Ty = beta y$, then $$alphalangle x,y rangle = langle Tx,y rangle = langle x,T^*y rangle = langle x, overline{beta} y rangle = beta langle x,y rangle$$ implies $langle x,y rangle = 0$. This means that the eigenspaces are orthogonal.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 29 at 19:12

























    answered Nov 29 at 18:54









    mathworker21

    8,5161928




    8,5161928












    • Thank you! for this part.
      – Kutz
      Nov 29 at 19:10










    • @Kutz I added part 3
      – mathworker21
      Nov 29 at 19:13










    • Thank you for answer.
      – Kutz
      Nov 29 at 19:22


















    • Thank you! for this part.
      – Kutz
      Nov 29 at 19:10










    • @Kutz I added part 3
      – mathworker21
      Nov 29 at 19:13










    • Thank you for answer.
      – Kutz
      Nov 29 at 19:22
















    Thank you! for this part.
    – Kutz
    Nov 29 at 19:10




    Thank you! for this part.
    – Kutz
    Nov 29 at 19:10












    @Kutz I added part 3
    – mathworker21
    Nov 29 at 19:13




    @Kutz I added part 3
    – mathworker21
    Nov 29 at 19:13












    Thank you for answer.
    – Kutz
    Nov 29 at 19:22




    Thank you for answer.
    – Kutz
    Nov 29 at 19:22











    0














    $T$ is normal means $T^*T=TT^*$, which is equivalent to
    $$
    |Tx| = |T^*x|,;;; xin H.
    $$
    So $mathcal{N}(T)=mathcal{N}(T^*)$ follows. The sum of normal operators is normal, and any scalar times a normal operator is normal. And the identity $I$ is normal. So, if $T$ is normal, then so is $alpha I-T$ for any scalar $alpha$. Therefore $mathcal{N}(T-alpha I)=mathcal{N}(T^*-overline{alpha}I)$, and any eigenvector of a normal $T$ with eigenvalue $alpha$ is an eigenvector of $T^*$ with eigenvalue $overline{alpha}$. If $Tx=alpha x$ and $Ty=beta y$, then



    begin{align}
    (alpha-beta)langle x,yrangle
    & = langle alpha x,yrangle-langle x,overline{beta}yrangle \
    & = langle Tx,yrangle-langle x,T^*yrangle \
    & = langle Tx,yrangle-langle Tx,yrangle =0.
    end{align}
    Therefore, if $alphane beta$, it follows that $xperp y$.






    share|cite|improve this answer


























      0














      $T$ is normal means $T^*T=TT^*$, which is equivalent to
      $$
      |Tx| = |T^*x|,;;; xin H.
      $$
      So $mathcal{N}(T)=mathcal{N}(T^*)$ follows. The sum of normal operators is normal, and any scalar times a normal operator is normal. And the identity $I$ is normal. So, if $T$ is normal, then so is $alpha I-T$ for any scalar $alpha$. Therefore $mathcal{N}(T-alpha I)=mathcal{N}(T^*-overline{alpha}I)$, and any eigenvector of a normal $T$ with eigenvalue $alpha$ is an eigenvector of $T^*$ with eigenvalue $overline{alpha}$. If $Tx=alpha x$ and $Ty=beta y$, then



      begin{align}
      (alpha-beta)langle x,yrangle
      & = langle alpha x,yrangle-langle x,overline{beta}yrangle \
      & = langle Tx,yrangle-langle x,T^*yrangle \
      & = langle Tx,yrangle-langle Tx,yrangle =0.
      end{align}
      Therefore, if $alphane beta$, it follows that $xperp y$.






      share|cite|improve this answer
























        0












        0








        0






        $T$ is normal means $T^*T=TT^*$, which is equivalent to
        $$
        |Tx| = |T^*x|,;;; xin H.
        $$
        So $mathcal{N}(T)=mathcal{N}(T^*)$ follows. The sum of normal operators is normal, and any scalar times a normal operator is normal. And the identity $I$ is normal. So, if $T$ is normal, then so is $alpha I-T$ for any scalar $alpha$. Therefore $mathcal{N}(T-alpha I)=mathcal{N}(T^*-overline{alpha}I)$, and any eigenvector of a normal $T$ with eigenvalue $alpha$ is an eigenvector of $T^*$ with eigenvalue $overline{alpha}$. If $Tx=alpha x$ and $Ty=beta y$, then



        begin{align}
        (alpha-beta)langle x,yrangle
        & = langle alpha x,yrangle-langle x,overline{beta}yrangle \
        & = langle Tx,yrangle-langle x,T^*yrangle \
        & = langle Tx,yrangle-langle Tx,yrangle =0.
        end{align}
        Therefore, if $alphane beta$, it follows that $xperp y$.






        share|cite|improve this answer












        $T$ is normal means $T^*T=TT^*$, which is equivalent to
        $$
        |Tx| = |T^*x|,;;; xin H.
        $$
        So $mathcal{N}(T)=mathcal{N}(T^*)$ follows. The sum of normal operators is normal, and any scalar times a normal operator is normal. And the identity $I$ is normal. So, if $T$ is normal, then so is $alpha I-T$ for any scalar $alpha$. Therefore $mathcal{N}(T-alpha I)=mathcal{N}(T^*-overline{alpha}I)$, and any eigenvector of a normal $T$ with eigenvalue $alpha$ is an eigenvector of $T^*$ with eigenvalue $overline{alpha}$. If $Tx=alpha x$ and $Ty=beta y$, then



        begin{align}
        (alpha-beta)langle x,yrangle
        & = langle alpha x,yrangle-langle x,overline{beta}yrangle \
        & = langle Tx,yrangle-langle x,T^*yrangle \
        & = langle Tx,yrangle-langle Tx,yrangle =0.
        end{align}
        Therefore, if $alphane beta$, it follows that $xperp y$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 19:23









        DisintegratingByParts

        58.4k42579




        58.4k42579






























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