Can we say that $x perp y ?$ [closed]
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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
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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.
add a comment |
$begingroup$
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
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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.
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Yes, this is true. Remember $|u|^2=langle u, urangle$, so $|x+y|^2=|x|^2+|y|^2+2langle x ,yrangle$.
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– Joel Cohen
Dec 3 '18 at 7:13
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@JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
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– gerw
Dec 3 '18 at 7:14
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Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
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– Dbchatto67
Dec 3 '18 at 7:16
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@gerw you are correct.
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– Dbchatto67
Dec 3 '18 at 7:17
add a comment |
$begingroup$
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
$endgroup$
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
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
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@JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
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– gerw
Dec 3 '18 at 7:14
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Then how do I conclude that $mathcal Re langle x,y rangle = 0 implies langle x,y rangle = 0$?
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– Dbchatto67
Dec 3 '18 at 7:16
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@gerw you are correct.
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:17
add a comment |
$begingroup$
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
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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
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@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
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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.
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– Dbchatto67
Dec 3 '18 at 7:17
add a comment |
3 Answers
3
active
oldest
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$|x+y|^{2}=langle (x+y),(x+y)rangle$ 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.
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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?
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– Dbchatto67
Dec 3 '18 at 7:23
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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.
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– Dbchatto67
Dec 3 '18 at 7:35
1
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@Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
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– Kavi Rama Murthy
Dec 3 '18 at 7:35
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Thank you very much sir.
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– Dbchatto67
Dec 3 '18 at 7:37
add a comment |
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Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?
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It's $2Re langle x , y rangle$ @gerw.
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– Dbchatto67
Dec 3 '18 at 7:32
add a comment |
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$|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. 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.
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$|x+y|^{2}=langle (x+y),(x+y)rangle$ 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.
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$begingroup$
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
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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
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@Dbchatto67 For your first comment: yes, that is exactly what I am saying. Second comment: yes, completeness is not necessary here.
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– Kavi Rama Murthy
Dec 3 '18 at 7:35
$begingroup$
Thank you very much sir.
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:37
add a comment |
$begingroup$
$|x+y|^{2}=langle (x+y),(x+y)rangle$ 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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$|x+y|^{2}=langle (x+y),(x+y)rangle$ 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.
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$|x+y|^{2}=langle (x+y),(x+y)rangle$ 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.
answered Dec 3 '18 at 7:15
Kavi Rama MurthyKavi Rama Murthy
56.2k42158
56.2k42158
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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
add a comment |
$begingroup$
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
add a comment |
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Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?
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It's $2Re langle x , y rangle$ @gerw.
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– Dbchatto67
Dec 3 '18 at 7:32
add a comment |
$begingroup$
Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?
$endgroup$
$begingroup$
It's $2Re langle x , y rangle$ @gerw.
$endgroup$
– Dbchatto67
Dec 3 '18 at 7:32
add a comment |
$begingroup$
Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?
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Hint: What is $|x+y|^2 - |x|^2 - |y|^2$ for general vectors $x,y in X$?
answered Dec 3 '18 at 7:13
gerwgerw
19.3k11334
19.3k11334
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It's $2Re langle x , y rangle$ @gerw.
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– Dbchatto67
Dec 3 '18 at 7:32
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. 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.
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add a comment |
$begingroup$
$|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. 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.
$endgroup$
add a comment |
$begingroup$
$|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. 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.
$endgroup$
$|x+y|^2=langle x+y, x+yrangle=langle x, xrangle+langle x, yrangle+langle y, xrangle+langle y, yrangle=|x|^2+|y|^2+langle x,yrangle+langle y,xrangle$. 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.
answered Dec 3 '18 at 7:21
AnupamAnupam
2,4091824
2,4091824
add a comment |
add a comment |
$begingroup$
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
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@JoelCohen: This is wrong since we are dealing with a complex Hilbert space.
$endgroup$
– gerw
Dec 3 '18 at 7:14
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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