Can we say that $x perp y ?$ [closed]












0












$begingroup$


Suppose $left (X, left < . ,. right > right)$ be a Hilbert space over $Bbb C.$ Let $x,y in X$ be such that $|x+y|^2 = |x|^2 + |y|^2$ $($ where $|.|$ is the norm induced by the inner product on $X$ $ )$. Then can we say that $left <x, y right > = 0$ i.e. $x perp y ?$



I think it is not true in general. Would anybody please help me by giving me a counter-example? That will be really helpful for me.



Thank you very much.






functional-analysis hilbert-spaces orthogonality







share|cite|improve this question









$endgroup$



closed as off-topic by Nosrati, Cesareo, John B, José Carlos Santos, Rebellos Dec 3 '18 at 14:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Cesareo, John B, José Carlos Santos, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
    $endgroup$
    – Joel Cohen
    Dec 3 '18 at 7:13












  • $begingroup$
    @JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
    $endgroup$
    – gerw
    Dec 3 '18 at 7:14










  • $begingroup$
    Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:16












  • $begingroup$
    @gerw you are correct.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:17
















0












$begingroup$


Suppose $left (X, left < . ,. right > right)$ be a Hilbert space over $Bbb C.$ Let $x,y in X$ be such that $|x+y|^2 = |x|^2 + |y|^2$ $($ where $|.|$ is the norm induced by the inner product on $X$ $ )$. Then can we say that $left <x, y right > = 0$ i.e. $x perp y ?$



I think it is not true in general. Would anybody please help me by giving me a counter-example? That will be really helpful for me.



Thank you very much.






functional-analysis hilbert-spaces orthogonality







share|cite|improve this question









$endgroup$



closed as off-topic by Nosrati, Cesareo, John B, José Carlos Santos, Rebellos Dec 3 '18 at 14:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Cesareo, John B, José Carlos Santos, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
    $endgroup$
    – Joel Cohen
    Dec 3 '18 at 7:13












  • $begingroup$
    @JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
    $endgroup$
    – gerw
    Dec 3 '18 at 7:14










  • $begingroup$
    Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:16












  • $begingroup$
    @gerw you are correct.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:17














0












0








0





$begingroup$


Suppose $left (X, left < . ,. right > right)$ be a Hilbert space over $Bbb C.$ Let $x,y in X$ be such that $|x+y|^2 = |x|^2 + |y|^2$ $($ where $|.|$ is the norm induced by the inner product on $X$ $ )$. Then can we say that $left <x, y right > = 0$ i.e. $x perp y ?$



I think it is not true in general. Would anybody please help me by giving me a counter-example? That will be really helpful for me.



Thank you very much.






functional-analysis hilbert-spaces orthogonality







share|cite|improve this question









$endgroup$




Suppose $left (X, left < . ,. right > right)$ be a Hilbert space over $Bbb C.$ Let $x,y in X$ be such that $|x+y|^2 = |x|^2 + |y|^2$ $($ where $|.|$ is the norm induced by the inner product on $X$ $ )$. Then can we say that $left <x, y right > = 0$ i.e. $x perp y ?$



I think it is not true in general. Would anybody please help me by giving me a counter-example? That will be really helpful for me.



Thank you very much.






functional-analysis hilbert-spaces orthogonality




functional-analysis hilbert-spaces orthogonality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 7:09









Dbchatto67Dbchatto67

575116




575116




closed as off-topic by Nosrati, Cesareo, John B, José Carlos Santos, Rebellos Dec 3 '18 at 14:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Cesareo, John B, José Carlos Santos, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Nosrati, Cesareo, John B, José Carlos Santos, Rebellos Dec 3 '18 at 14:02


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Cesareo, John B, José Carlos Santos, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
    $endgroup$
    – Joel Cohen
    Dec 3 '18 at 7:13












  • $begingroup$
    @JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
    $endgroup$
    – gerw
    Dec 3 '18 at 7:14










  • $begingroup$
    Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:16












  • $begingroup$
    @gerw you are correct.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:17


















  • $begingroup$
    Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
    $endgroup$
    – Joel Cohen
    Dec 3 '18 at 7:13












  • $begingroup$
    @JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
    $endgroup$
    – gerw
    Dec 3 '18 at 7:14










  • $begingroup$
    Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:16












  • $begingroup$
    @gerw you are correct.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:17
















$begingroup$
Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
$endgroup$
– Joel Cohen
Dec 3 '18 at 7:13






$begingroup$
Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
$endgroup$
– Joel Cohen
Dec 3 '18 at 7:13














$begingroup$
@JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
$endgroup$
– gerw
Dec 3 '18 at 7:14




$begingroup$
@JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
$endgroup$
– gerw
Dec 3 '18 at 7:14












$begingroup$
Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:16






$begingroup$
Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:16














$begingroup$
@gerw you are correct.
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:17




$begingroup$
@gerw you are correct.
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:17










3 Answers
3






active

oldest

votes


















2












$begingroup$

$|x+y|^{2}=langle (x+y),(x+y)rangle$ Using properties of inner proudct this becomes $|x|^{2}+|y|^{2}+2 Re langle x,yrangle$ so the given equation holds iff $Re langle x,yrangle$ =0. If you take $x=iy$ with $yneq 0$ you get a counterexample.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:23












  • $begingroup$
    Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:35








  • 1




    $begingroup$
    @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 7:35












  • $begingroup$
    Thank you very much sir.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:37



















1












$begingroup$

Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It's $2Re langle x , y rangle$ @gerw.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:32





















0












$begingroup$

$|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. Thus $|x+y|^2-|x|^2-|y|^2=langle x,yrangle+langle y,xrangle=2Relangle x,yrangle$. Thus in $mathbb C$, $x=2,y=2i$ will work.






share|cite|improve this answer









$endgroup$




















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $|x+y|^{2}=langle (x+y),(x+y)rangle$ Using properties of inner proudct this becomes $|x|^{2}+|y|^{2}+2 Re langle x,yrangle$ so the given equation holds iff $Re langle x,yrangle$ =0. If you take $x=iy$ with $yneq 0$ you get a counterexample.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:23












    • $begingroup$
      Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:35








    • 1




      $begingroup$
      @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
      $endgroup$
      – Kavi Rama Murthy
      Dec 3 '18 at 7:35












    • $begingroup$
      Thank you very much sir.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:37
















    2












    $begingroup$

    $|x+y|^{2}=langle (x+y),(x+y)rangle$ Using properties of inner proudct this becomes $|x|^{2}+|y|^{2}+2 Re langle x,yrangle$ so the given equation holds iff $Re langle x,yrangle$ =0. If you take $x=iy$ with $yneq 0$ you get a counterexample.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:23












    • $begingroup$
      Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:35








    • 1




      $begingroup$
      @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
      $endgroup$
      – Kavi Rama Murthy
      Dec 3 '18 at 7:35












    • $begingroup$
      Thank you very much sir.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:37














    2












    2








    2





    $begingroup$

    $|x+y|^{2}=langle (x+y),(x+y)rangle$ Using properties of inner proudct this becomes $|x|^{2}+|y|^{2}+2 Re langle x,yrangle$ so the given equation holds iff $Re langle x,yrangle$ =0. If you take $x=iy$ with $yneq 0$ you get a counterexample.






    share|cite|improve this answer









    $endgroup$



    $|x+y|^{2}=langle (x+y),(x+y)rangle$ Using properties of inner proudct this becomes $|x|^{2}+|y|^{2}+2 Re langle x,yrangle$ so the given equation holds iff $Re langle x,yrangle$ =0. If you take $x=iy$ with $yneq 0$ you get a counterexample.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 7:15









    Kavi Rama MurthyKavi Rama Murthy

    56.2k42158




    56.2k42158












    • $begingroup$
      Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:23












    • $begingroup$
      Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:35








    • 1




      $begingroup$
      @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
      $endgroup$
      – Kavi Rama Murthy
      Dec 3 '18 at 7:35












    • $begingroup$
      Thank you very much sir.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:37


















    • $begingroup$
      Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:23












    • $begingroup$
      Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:35








    • 1




      $begingroup$
      @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
      $endgroup$
      – Kavi Rama Murthy
      Dec 3 '18 at 7:35












    • $begingroup$
      Thank you very much sir.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:37
















    $begingroup$
    Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:23






    $begingroup$
    Then $langle iy,y rangle = i |y|^2$ which is imaginary if $y ne 0$ i.e. it's real part vanishes though if $y ne 0$ then $langle iy,y rangle ne 0$ proving that $iy$ is not orthogonal to $y$ if $yne 0$. Am I right Sir?
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:23














    $begingroup$
    Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:35






    $begingroup$
    Not for Hilbert space, this is true for any complex inner product space. Am I right @Kavi Rama Murthy sir? Hilbert space is nothing to do with it.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:35






    1




    1




    $begingroup$
    @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 7:35






    $begingroup$
    @Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
    $endgroup$
    – Kavi Rama Murthy
    Dec 3 '18 at 7:35














    $begingroup$
    Thank you very much sir.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:37




    $begingroup$
    Thank you very much sir.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:37











    1












    $begingroup$

    Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It's $2Re langle x , y rangle$ @gerw.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:32


















    1












    $begingroup$

    Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It's $2Re langle x , y rangle$ @gerw.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:32
















    1












    1








    1





    $begingroup$

    Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?






    share|cite|improve this answer









    $endgroup$



    Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 7:13









    gerwgerw

    19.3k11334




    19.3k11334












    • $begingroup$
      It's $2Re langle x , y rangle$ @gerw.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:32




















    • $begingroup$
      It's $2Re langle x , y rangle$ @gerw.
      $endgroup$
      – Dbchatto67
      Dec 3 '18 at 7:32


















    $begingroup$
    It's $2Re langle x , y rangle$ @gerw.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:32






    $begingroup$
    It's $2Re langle x , y rangle$ @gerw.
    $endgroup$
    – Dbchatto67
    Dec 3 '18 at 7:32













    0












    $begingroup$

    $|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. Thus $|x+y|^2-|x|^2-|y|^2=langle x,yrangle+langle y,xrangle=2Relangle x,yrangle$. Thus in $mathbb C$, $x=2,y=2i$ will work.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. Thus $|x+y|^2-|x|^2-|y|^2=langle x,yrangle+langle y,xrangle=2Relangle x,yrangle$. Thus in $mathbb C$, $x=2,y=2i$ will work.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. Thus $|x+y|^2-|x|^2-|y|^2=langle x,yrangle+langle y,xrangle=2Relangle x,yrangle$. Thus in $mathbb C$, $x=2,y=2i$ will work.






        share|cite|improve this answer









        $endgroup$



        $|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. Thus $|x+y|^2-|x|^2-|y|^2=langle x,yrangle+langle y,xrangle=2Relangle x,yrangle$. Thus in $mathbb C$, $x=2,y=2i$ will work.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 7:21









        AnupamAnupam

        2,4091824




        2,4091824















            -

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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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