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How to define angles in Hilbert spaces with field $mathbbC$?

Orthogonal projection in Hilbert spacesOrthogonal Projection on hilbert spacesClosed subspace $M=(M^perp)^perp$ in PRE hilbert spaces.(bounded linear) orthogonal projections on Hilbert spacesA statement on prehilbert spacesA property in Hilbert spacesHilbert spaces. Showing that an a linear operator forms a closed subspaceTensor products of Hilbert spaces and Hilbert-Schmidt operatorsProve that $P_W+P_V$ is orthogonal projection iff $Vperp W$ in Hillbert space.Projection on Hilbert spaces

1

$begingroup$

Suppose $H$ is a Hilbert space and the field we consider is $mathbbC$, then given two arbitrary nonzero vectors $u$ and $v$ in $H$, how to define the angle between $u$ and $v$?

I tried to define it as $cos^-1 langle u, vrangleover ||u|| cdot||v||$, but $langle u,v rangle$ might be a compex number.

This question arose from a problem I want to solve. If the definition of angles does not exist, could you help me with the proof of the third claim below?

The original problem is:

Let $H$ be a Hilbert space, $M$ be a closed linear subspace of $H$ and $P$ be the orthogonal projection from $H$ to $M$. Let $xin H$ and $ain M$. Show that

1. 
   $|langle x,arangle| le ||Px||cdot||a||$;


2. if $a in [[Px]]$, then $|langle x,arangle|=||Px||cdot||a||$;


3. if $mathbbF = mathbbC$, then the angle between $x$ and $a$ is at least $cos^-1()$.

Proof of 1:

Since $P$ is the orthogonal projection from $H$ to $M$, we know $textim P = M$, $textker P = M^perp$ and $M oplus M^perp = H$. Thus, we have

$beginalign|langle x, a rangle| & = |langle x - Px + Px, a rangle| \ & = |langle x - Px, arangle + langle Px, arangle| \ & = |langle Px,a rangle | quad text(by $x-Px in M^perp$ and $a in M$) \ &le ||Px||cdot||a|| quad text(by Cauchy-Schwarz inequality)endalign$

Proof of 2:

If the proof of 1 we have proved that $|langle x, a rangle| = |langle Px, a rangle|$. Now, if $a in [[Px]]$, then $exists k in mathbbF$ s.t. $a = k Px$. Thus, $|langle Px, arangle| = |langle Px, kPxrangle| = |k| cdot|langle Px,Pxrangle| = |k|cdot||Px||cdot ||Px|| = ||a|| cdot||Px||$.

linear-algebra functional-analysis

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asked Mar 29 at 2:39

U2647U2647

8611

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the source of this problem?
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 2:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown The original problem is in Joseph Muscat's Functional Analysis: An introduction to Metric Spaces, Hilbert Spaces, and Banach Spaces p184 8.(a). published by Springer. But my professor substituted $mathbbR$ with $mathbbC$ and I don't know if it is reasonable.
  $endgroup$
  – U2647
  Mar 29 at 2:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your professor didn't tell you their definition of angle?
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 2:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown No, he didn't. So I am wondering if there is a way to define angles in any Hilbert space with the field $mathbbC$? If not, he might make a mistake or a typo.
  $endgroup$
  – U2647
  Mar 29 at 3:13
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose $H$ is a Hilbert space and the field we consider is $mathbbC$, then given two arbitrary nonzero vectors $u$ and $v$ in $H$, how to define the angle between $u$ and $v$?

I tried to define it as $cos^-1 langle u, vrangleover ||u|| cdot||v||$, but $langle u,v rangle$ might be a compex number.

This question arose from a problem I want to solve. If the definition of angles does not exist, could you help me with the proof of the third claim below?

The original problem is:

Let $H$ be a Hilbert space, $M$ be a closed linear subspace of $H$ and $P$ be the orthogonal projection from $H$ to $M$. Let $xin H$ and $ain M$. Show that

1. 
   $|langle x,arangle| le ||Px||cdot||a||$;


2. if $a in [[Px]]$, then $|langle x,arangle|=||Px||cdot||a||$;


3. if $mathbbF = mathbbC$, then the angle between $x$ and $a$ is at least $cos^-1()$.

Proof of 1:

Since $P$ is the orthogonal projection from $H$ to $M$, we know $textim P = M$, $textker P = M^perp$ and $M oplus M^perp = H$. Thus, we have

$beginalign|langle x, a rangle| & = |langle x - Px + Px, a rangle| \ & = |langle x - Px, arangle + langle Px, arangle| \ & = |langle Px,a rangle | quad text(by $x-Px in M^perp$ and $a in M$) \ &le ||Px||cdot||a|| quad text(by Cauchy-Schwarz inequality)endalign$

Proof of 2:

If the proof of 1 we have proved that $|langle x, a rangle| = |langle Px, a rangle|$. Now, if $a in [[Px]]$, then $exists k in mathbbF$ s.t. $a = k Px$. Thus, $|langle Px, arangle| = |langle Px, kPxrangle| = |k| cdot|langle Px,Pxrangle| = |k|cdot||Px||cdot ||Px|| = ||a|| cdot||Px||$.

linear-algebra functional-analysis

share|cite|improve this question

asked Mar 29 at 2:39

U2647U2647

8611

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the source of this problem?
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 2:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown The original problem is in Joseph Muscat's Functional Analysis: An introduction to Metric Spaces, Hilbert Spaces, and Banach Spaces p184 8.(a). published by Springer. But my professor substituted $mathbbR$ with $mathbbC$ and I don't know if it is reasonable.
  $endgroup$
  – U2647
  Mar 29 at 2:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your professor didn't tell you their definition of angle?
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 2:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown No, he didn't. So I am wondering if there is a way to define angles in any Hilbert space with the field $mathbbC$? If not, he might make a mistake or a typo.
  $endgroup$
  – U2647
  Mar 29 at 3:13
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $H$ is a Hilbert space and the field we consider is $mathbbC$, then given two arbitrary nonzero vectors $u$ and $v$ in $H$, how to define the angle between $u$ and $v$?

I tried to define it as $cos^-1 langle u, vrangleover ||u|| cdot||v||$, but $langle u,v rangle$ might be a compex number.

This question arose from a problem I want to solve. If the definition of angles does not exist, could you help me with the proof of the third claim below?

The original problem is:

Let $H$ be a Hilbert space, $M$ be a closed linear subspace of $H$ and $P$ be the orthogonal projection from $H$ to $M$. Let $xin H$ and $ain M$. Show that

1. 
   $|langle x,arangle| le ||Px||cdot||a||$;


2. if $a in [[Px]]$, then $|langle x,arangle|=||Px||cdot||a||$;


3. if $mathbbF = mathbbC$, then the angle between $x$ and $a$ is at least $cos^-1()$.

Proof of 1:

Since $P$ is the orthogonal projection from $H$ to $M$, we know $textim P = M$, $textker P = M^perp$ and $M oplus M^perp = H$. Thus, we have

$beginalign|langle x, a rangle| & = |langle x - Px + Px, a rangle| \ & = |langle x - Px, arangle + langle Px, arangle| \ & = |langle Px,a rangle | quad text(by $x-Px in M^perp$ and $a in M$) \ &le ||Px||cdot||a|| quad text(by Cauchy-Schwarz inequality)endalign$

Proof of 2:

If the proof of 1 we have proved that $|langle x, a rangle| = |langle Px, a rangle|$. Now, if $a in [[Px]]$, then $exists k in mathbbF$ s.t. $a = k Px$. Thus, $|langle Px, arangle| = |langle Px, kPxrangle| = |k| cdot|langle Px,Pxrangle| = |k|cdot||Px||cdot ||Px|| = ||a|| cdot||Px||$.

linear-algebra functional-analysis

share|cite|improve this question

asked Mar 29 at 2:39

U2647U2647

8611

$endgroup$

share|cite|improve this question

asked Mar 29 at 2:39

U2647U2647

8611

share|cite|improve this question

asked Mar 29 at 2:39

U2647U2647

8611

asked Mar 29 at 2:39

U2647U2647

8611

asked Mar 29 at 2:39

U2647U2647

8611

1 Answer
1

active

oldest

votes

1

$begingroup$

If $langle cdot,cdotrangle$ is a complex inner product on a complex vector space $H$, then its real part $(x,y)=operatornameRelangle x,yrangle$ is a real inner product on $H$ (with the same induced norm as the complex inner product). So, angles can be defined using this real inner product, and this is normally what is meant by an angle in a complex inner product space.

share|cite|improve this answer

edited Mar 29 at 3:40

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what is the denominator?
  $endgroup$
  – U2647
  Mar 29 at 3:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There isn't any problem with the denominator...the formula you wrote in the question is correct if you just use the real inner product in place of $langlecdot,cdotrangle$. (Note that the real and complex inner products induce the same norm.)
  $endgroup$
  – Eric Wofsey
  Mar 29 at 3:42
  

  
  
  
  

add a comment | 

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1

$begingroup$

If $langle cdot,cdotrangle$ is a complex inner product on a complex vector space $H$, then its real part $(x,y)=operatornameRelangle x,yrangle$ is a real inner product on $H$ (with the same induced norm as the complex inner product). So, angles can be defined using this real inner product, and this is normally what is meant by an angle in a complex inner product space.

share|cite|improve this answer

edited Mar 29 at 3:40

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what is the denominator?
  $endgroup$
  – U2647
  Mar 29 at 3:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There isn't any problem with the denominator...the formula you wrote in the question is correct if you just use the real inner product in place of $langlecdot,cdotrangle$. (Note that the real and complex inner products induce the same norm.)
  $endgroup$
  – Eric Wofsey
  Mar 29 at 3:42
  

  
  
  
  

add a comment | 

1

$begingroup$

If $langle cdot,cdotrangle$ is a complex inner product on a complex vector space $H$, then its real part $(x,y)=operatornameRelangle x,yrangle$ is a real inner product on $H$ (with the same induced norm as the complex inner product). So, angles can be defined using this real inner product, and this is normally what is meant by an angle in a complex inner product space.

share|cite|improve this answer

edited Mar 29 at 3:40

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what is the denominator?
  $endgroup$
  – U2647
  Mar 29 at 3:18
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There isn't any problem with the denominator...the formula you wrote in the question is correct if you just use the real inner product in place of $langlecdot,cdotrangle$. (Note that the real and complex inner products induce the same norm.)
  $endgroup$
  – Eric Wofsey
  Mar 29 at 3:42
  

  
  
  
  

add a comment | 

$begingroup$

If $langle cdot,cdotrangle$ is a complex inner product on a complex vector space $H$, then its real part $(x,y)=operatornameRelangle x,yrangle$ is a real inner product on $H$ (with the same induced norm as the complex inner product). So, angles can be defined using this real inner product, and this is normally what is meant by an angle in a complex inner product space.

share|cite|improve this answer

edited Mar 29 at 3:40

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351

$endgroup$

share|cite|improve this answer

edited Mar 29 at 3:40

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351

answered Mar 29 at 3:14

Eric WofseyEric Wofsey

192k14217351