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Confusing Cauchy-Schwarz Inequality Proof

1

1

$begingroup$

Teacher proved it like this:

Very elegant (way simpler than most of the ones I find online), but I'm still not convinced -- particularly the last two steps or so, where the absolute value on the left-hand side seems to disappear.

Any thoughts?

linear-algebra algebra-precalculus vector-spaces vectors cauchy-schwarz-inequality

share|cite|improve this question

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

asked Apr 2 at 17:30

Will Will

565

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: What's the absolute value of a nonnegative number?
  $endgroup$
  – Ennar
  Apr 2 at 17:36
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  It is so elegant that it is wrong. If $cos theta$ is negative, then $fracncos thetage n$ is false. Besides, the fact that you can write $$vecacdotvecb=|a||b|cos theta$$ relies on the Cauchy-Schwarz inequality, so this reasoning (if it were correct) would be circular.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ennar: But how do we know the dot product of vectors a and b IS nonnegative? Dot products can sometimes be negative, after all.
  $endgroup$
  – Will
  Apr 2 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @GiuseppeNegro: Let's be charitable and assume that there might have been a transcription error, and the actual claim is that $fracn ge n$ if $n ge 0$. Then the remainder of the argument follows, although the reasoning is, as you point out, fundamentally circular.
  $endgroup$
  – John Hughes
  Apr 2 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did the teacher define $vec acdotvec b $, $lVertvec arVert $ and $theta $?
  $endgroup$
  – user
  Apr 3 at 5:22
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Teacher proved it like this:

Very elegant (way simpler than most of the ones I find online), but I'm still not convinced -- particularly the last two steps or so, where the absolute value on the left-hand side seems to disappear.

Any thoughts?

linear-algebra algebra-precalculus vector-spaces vectors cauchy-schwarz-inequality

share|cite|improve this question

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

asked Apr 2 at 17:30

Will Will

565

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: What's the absolute value of a nonnegative number?
  $endgroup$
  – Ennar
  Apr 2 at 17:36
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  It is so elegant that it is wrong. If $cos theta$ is negative, then $fracncos thetage n$ is false. Besides, the fact that you can write $$vecacdotvecb=|a||b|cos theta$$ relies on the Cauchy-Schwarz inequality, so this reasoning (if it were correct) would be circular.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Ennar: But how do we know the dot product of vectors a and b IS nonnegative? Dot products can sometimes be negative, after all.
  $endgroup$
  – Will
  Apr 2 at 17:38
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @GiuseppeNegro: Let's be charitable and assume that there might have been a transcription error, and the actual claim is that $fracn ge n$ if $n ge 0$. Then the remainder of the argument follows, although the reasoning is, as you point out, fundamentally circular.
  $endgroup$
  – John Hughes
  Apr 2 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did the teacher define $vec acdotvec b $, $lVertvec arVert $ and $theta $?
  $endgroup$
  – user
  Apr 3 at 5:22
  

  
  
  
  

add a comment | 

$begingroup$

Teacher proved it like this:

Very elegant (way simpler than most of the ones I find online), but I'm still not convinced -- particularly the last two steps or so, where the absolute value on the left-hand side seems to disappear.

Any thoughts?

linear-algebra algebra-precalculus vector-spaces vectors cauchy-schwarz-inequality

share|cite|improve this question

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

asked Apr 2 at 17:30

Will Will

565

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

asked Apr 2 at 17:30

Will Will

565

share|cite|improve this question

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

asked Apr 2 at 17:30

Will Will

565

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

edited Apr 2 at 17:45

José Carlos Santos

177k24138248

asked Apr 2 at 17:30

Will Will

565

asked Apr 2 at 17:30

Will Will

565

2 Answers
2

active

oldest

votes

1

$begingroup$

given that $acdot b= |a|cdot |b|cos(theta)$ we have that
$$|acdot b|= |a|cdot |b||cos(theta)|$$ (since the length of a vector is always non-negative).

Noting that $|cos(theta)|leq 1$ we can have that:

1. 
   $|cos(theta)|=1$ which implies that $|acdot b|leq|a|cdot |b|$ is satisfied with equality.


2. 
   $|cos(theta)|<1$ which implies that $|acdot b|<|a|cdot |b|$
   

The latter proves the statement.

$$|acdot b| leq |a|cdot |b|$$

---

EDIT: As pointed out by @user in the comments down below, there's no sense in moving the $|cos(theta)|$ to the denominator as your teacher suggested, cause nothing prevents you to have $theta=fracpi2$ implying that $cos(theta)=0$. It's better to keep it on the other side and deriving the conclusion from there.

Disclaimer I think you should,nevertheless, consider what @jose-carlos-santos and @giuseppe-negro are pointing out, that actually this is a circular reasoning, and hence not a valid proof.

share|cite|improve this answer

edited Apr 3 at 6:35

answered Apr 2 at 17:42

RScrlliRScrlli

777114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why does the fact that cosine is less than or equal to one imply the final statement? I don't quite follow.
  $endgroup$
  – Will
  Apr 2 at 17:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited the answer to be more clear
  $endgroup$
  – RScrlli
  Apr 2 at 17:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why should one divide both sides of the equality by $|costheta|$ introducing possible singularity? This step is absolutely excessive even assuming that both expressions $vec acdotvec b=lVertvec arVert,lVertvec brVertcostheta $ and $-1lecosthetale1$ are "given".
  $endgroup$
  – user
  Apr 3 at 5:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just replicating the proof given in the sketch of his professor.
  $endgroup$
  – RScrlli
  Apr 3 at 5:55
  

  
  
  
  

add a comment | 

8

$begingroup$

This proof assumes that $vec a.vec b$ can be written as $leftlVertvec arightrVert.leftlVertvec brightrVert.costheta$ for some number $theta$. This is the same thing as asserting that $leftlvertvec a.vec brightrvertleqslantleftlVertvec arightrVert.leftlVertvec brightrVert$, and this is precisely the Cauchy-Schwarz inequality, which is what you want to prove. There is therefore a circular reasoning here.

share|cite|improve this answer

edited Apr 3 at 9:43

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

add a comment | 

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1

$begingroup$

given that $acdot b= |a|cdot |b|cos(theta)$ we have that
$$|acdot b|= |a|cdot |b||cos(theta)|$$ (since the length of a vector is always non-negative).

Noting that $|cos(theta)|leq 1$ we can have that:

1. 
   $|cos(theta)|=1$ which implies that $|acdot b|leq|a|cdot |b|$ is satisfied with equality.


2. 
   $|cos(theta)|<1$ which implies that $|acdot b|<|a|cdot |b|$
   

The latter proves the statement.

$$|acdot b| leq |a|cdot |b|$$

---

EDIT: As pointed out by @user in the comments down below, there's no sense in moving the $|cos(theta)|$ to the denominator as your teacher suggested, cause nothing prevents you to have $theta=fracpi2$ implying that $cos(theta)=0$. It's better to keep it on the other side and deriving the conclusion from there.

Disclaimer I think you should,nevertheless, consider what @jose-carlos-santos and @giuseppe-negro are pointing out, that actually this is a circular reasoning, and hence not a valid proof.

share|cite|improve this answer

edited Apr 3 at 6:35

answered Apr 2 at 17:42

RScrlliRScrlli

777114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why does the fact that cosine is less than or equal to one imply the final statement? I don't quite follow.
  $endgroup$
  – Will
  Apr 2 at 17:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited the answer to be more clear
  $endgroup$
  – RScrlli
  Apr 2 at 17:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why should one divide both sides of the equality by $|costheta|$ introducing possible singularity? This step is absolutely excessive even assuming that both expressions $vec acdotvec b=lVertvec arVert,lVertvec brVertcostheta $ and $-1lecosthetale1$ are "given".
  $endgroup$
  – user
  Apr 3 at 5:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just replicating the proof given in the sketch of his professor.
  $endgroup$
  – RScrlli
  Apr 3 at 5:55
  

  
  
  
  

add a comment | 

1

$begingroup$

given that $acdot b= |a|cdot |b|cos(theta)$ we have that
$$|acdot b|= |a|cdot |b||cos(theta)|$$ (since the length of a vector is always non-negative).

Noting that $|cos(theta)|leq 1$ we can have that:

1. 
   $|cos(theta)|=1$ which implies that $|acdot b|leq|a|cdot |b|$ is satisfied with equality.


2. 
   $|cos(theta)|<1$ which implies that $|acdot b|<|a|cdot |b|$
   

The latter proves the statement.

$$|acdot b| leq |a|cdot |b|$$

---

EDIT: As pointed out by @user in the comments down below, there's no sense in moving the $|cos(theta)|$ to the denominator as your teacher suggested, cause nothing prevents you to have $theta=fracpi2$ implying that $cos(theta)=0$. It's better to keep it on the other side and deriving the conclusion from there.

Disclaimer I think you should,nevertheless, consider what @jose-carlos-santos and @giuseppe-negro are pointing out, that actually this is a circular reasoning, and hence not a valid proof.

share|cite|improve this answer

edited Apr 3 at 6:35

answered Apr 2 at 17:42

RScrlliRScrlli

777114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why does the fact that cosine is less than or equal to one imply the final statement? I don't quite follow.
  $endgroup$
  – Will
  Apr 2 at 17:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited the answer to be more clear
  $endgroup$
  – RScrlli
  Apr 2 at 17:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why should one divide both sides of the equality by $|costheta|$ introducing possible singularity? This step is absolutely excessive even assuming that both expressions $vec acdotvec b=lVertvec arVert,lVertvec brVertcostheta $ and $-1lecosthetale1$ are "given".
  $endgroup$
  – user
  Apr 3 at 5:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just replicating the proof given in the sketch of his professor.
  $endgroup$
  – RScrlli
  Apr 3 at 5:55
  

  
  
  
  

add a comment | 

$begingroup$

given that $acdot b= |a|cdot |b|cos(theta)$ we have that
$$|acdot b|= |a|cdot |b||cos(theta)|$$ (since the length of a vector is always non-negative).

Noting that $|cos(theta)|leq 1$ we can have that:

1. 
   $|cos(theta)|=1$ which implies that $|acdot b|leq|a|cdot |b|$ is satisfied with equality.


2. 
   $|cos(theta)|<1$ which implies that $|acdot b|<|a|cdot |b|$
   

The latter proves the statement.

$$|acdot b| leq |a|cdot |b|$$

---

EDIT: As pointed out by @user in the comments down below, there's no sense in moving the $|cos(theta)|$ to the denominator as your teacher suggested, cause nothing prevents you to have $theta=fracpi2$ implying that $cos(theta)=0$. It's better to keep it on the other side and deriving the conclusion from there.

Disclaimer I think you should,nevertheless, consider what @jose-carlos-santos and @giuseppe-negro are pointing out, that actually this is a circular reasoning, and hence not a valid proof.

share|cite|improve this answer

edited Apr 3 at 6:35

answered Apr 2 at 17:42

RScrlliRScrlli

777114

$endgroup$

share|cite|improve this answer

edited Apr 3 at 6:35

answered Apr 2 at 17:42

RScrlliRScrlli

777114

answered Apr 2 at 17:42

RScrlliRScrlli

777114

answered Apr 2 at 17:42

RScrlliRScrlli

777114

8

$begingroup$

This proof assumes that $vec a.vec b$ can be written as $leftlVertvec arightrVert.leftlVertvec brightrVert.costheta$ for some number $theta$. This is the same thing as asserting that $leftlvertvec a.vec brightrvertleqslantleftlVertvec arightrVert.leftlVertvec brightrVert$, and this is precisely the Cauchy-Schwarz inequality, which is what you want to prove. There is therefore a circular reasoning here.

share|cite|improve this answer

edited Apr 3 at 9:43

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

add a comment | 

8

$begingroup$

This proof assumes that $vec a.vec b$ can be written as $leftlVertvec arightrVert.leftlVertvec brightrVert.costheta$ for some number $theta$. This is the same thing as asserting that $leftlvertvec a.vec brightrvertleqslantleftlVertvec arightrVert.leftlVertvec brightrVert$, and this is precisely the Cauchy-Schwarz inequality, which is what you want to prove. There is therefore a circular reasoning here.

share|cite|improve this answer

edited Apr 3 at 9:43

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

add a comment | 

$begingroup$

This proof assumes that $vec a.vec b$ can be written as $leftlVertvec arightrVert.leftlVertvec brightrVert.costheta$ for some number $theta$. This is the same thing as asserting that $leftlvertvec a.vec brightrvertleqslantleftlVertvec arightrVert.leftlVertvec brightrVert$, and this is precisely the Cauchy-Schwarz inequality, which is what you want to prove. There is therefore a circular reasoning here.

share|cite|improve this answer

edited Apr 3 at 9:43

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

share|cite|improve this answer

edited Apr 3 at 9:43

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248

answered Apr 2 at 17:40

José Carlos SantosJosé Carlos Santos

177k24138248