Mixing
Reasoning for equality of Cauchy Schwarz inequality holds
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I am trying to understand the proof of the Cauchy-Schwarz Inequality. I understand that, in addition to that, there is a remark that the equality holds if and only if one of term is scalar multiple of the other. And in the book, this is the proof given.
I know there are many alternative proof on MSE but I wanted to understand argument in book.
It is as follows :
$$langle x,yrangle=Vert xVert ,Vert yVertimpliesleftlanglefrac xVert xVert ,fracyVert yVertrightrangle=1impliesfrac x=fracy$$
Thus,$$x= Vert xVert,fracyVert yVert.$$
I don't understand second to last line
Any help will be appreciated.
I had associated screenshot of proof
proof-explanation cauchy-schwarz-inequality
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
asked Aug 10 at 12:11
SRJ
1,374417
 |Â
show 2 more comments
up vote
4
down vote
favorite
I am trying to understand the proof of the Cauchy-Schwarz Inequality. I understand that, in addition to that, there is a remark that the equality holds if and only if one of term is scalar multiple of the other. And in the book, this is the proof given.
I know there are many alternative proof on MSE but I wanted to understand argument in book.
It is as follows :
$$langle x,yrangle=Vert xVert ,Vert yVertimpliesleftlanglefrac xVert xVert ,fracyVert yVertrightrangle=1impliesfrac x=fracy$$
Thus,$$x= Vert xVert,fracyVert yVert.$$
I don't understand second to last line
Any help will be appreciated.
I had associated screenshot of proof
proof-explanation cauchy-schwarz-inequality
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
asked Aug 10 at 12:11
SRJ
1,374417
The second line is unclear, as you say. I suggest you post the whole proof of the CS inequality, including the part where the inequality is shown. Then, we can see if any inequalities/results from that proof were used implicitly in this second line. The answer below uses the dot product of vectors with the angular interpretation, but the issue is that the CS inequality is valid for all inner products, not just those associated with a dot product.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 10 at 12:18
$x/||x||$ and $y/||y||$ are unit vectors with inner product 1. It follows that they are equal, but this is not obvious (that is, it is where the content of the argument lies). Perhaps this assertion is proved somewhere else?
– Daniel Littlewood
Aug 10 at 12:20
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó I had uploaded screen shot of proof.
– SRJ
Aug 10 at 12:27
Have you read the earlier part of the equality proof for unit vectors?
– Calvin Khor
Aug 10 at 12:30
@CalvinKhor .Yes I had read the proof.But I don't understand equality argument.
– SRJ
Aug 10 at 12:35
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to understand the proof of the Cauchy-Schwarz Inequality. I understand that, in addition to that, there is a remark that the equality holds if and only if one of term is scalar multiple of the other. And in the book, this is the proof given.
I know there are many alternative proof on MSE but I wanted to understand argument in book.
It is as follows :
$$langle x,yrangle=Vert xVert ,Vert yVertimpliesleftlanglefrac xVert xVert ,fracyVert yVertrightrangle=1impliesfrac x=fracy$$
Thus,$$x= Vert xVert,fracyVert yVert.$$
I don't understand second to last line
Any help will be appreciated.
I had associated screenshot of proof
proof-explanation cauchy-schwarz-inequality
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
asked Aug 10 at 12:11
SRJ
1,374417
I am trying to understand the proof of the Cauchy-Schwarz Inequality. I understand that, in addition to that, there is a remark that the equality holds if and only if one of term is scalar multiple of the other. And in the book, this is the proof given.
I know there are many alternative proof on MSE but I wanted to understand argument in book.
It is as follows :
$$langle x,yrangle=Vert xVert ,Vert yVertimpliesleftlanglefrac xVert xVert ,fracyVert yVertrightrangle=1impliesfrac x=fracy$$
Thus,$$x= Vert xVert,fracyVert yVert.$$
I don't understand second to last line
Any help will be appreciated.
I had associated screenshot of proof
proof-explanation cauchy-schwarz-inequality
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
asked Aug 10 at 12:11
SRJ
1,374417
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
edited Aug 10 at 12:30
José Carlos Santos
119k16101182
119k16101182
asked Aug 10 at 12:11
SRJ
1,374417
asked Aug 10 at 12:11
SRJ
1,374417
asked Aug 10 at 12:11
SRJ
1,374417
1,374417
The second line is unclear, as you say. I suggest you post the whole proof of the CS inequality, including the part where the inequality is shown. Then, we can see if any inequalities/results from that proof were used implicitly in this second line. The answer below uses the dot product of vectors with the angular interpretation, but the issue is that the CS inequality is valid for all inner products, not just those associated with a dot product.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 10 at 12:18
$x/||x||$ and $y/||y||$ are unit vectors with inner product 1. It follows that they are equal, but this is not obvious (that is, it is where the content of the argument lies). Perhaps this assertion is proved somewhere else?
– Daniel Littlewood
Aug 10 at 12:20
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó I had uploaded screen shot of proof.
– SRJ
Aug 10 at 12:27
Have you read the earlier part of the equality proof for unit vectors?
– Calvin Khor
Aug 10 at 12:30
@CalvinKhor .Yes I had read the proof.But I don't understand equality argument.
– SRJ
Aug 10 at 12:35
 |Â
show 2 more comments
The second line is unclear, as you say. I suggest you post the whole proof of the CS inequality, including the part where the inequality is shown. Then, we can see if any inequalities/results from that proof were used implicitly in this second line. The answer below uses the dot product of vectors with the angular interpretation, but the issue is that the CS inequality is valid for all inner products, not just those associated with a dot product.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 10 at 12:18
$x/||x||$ and $y/||y||$ are unit vectors with inner product 1. It follows that they are equal, but this is not obvious (that is, it is where the content of the argument lies). Perhaps this assertion is proved somewhere else?
– Daniel Littlewood
Aug 10 at 12:20
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó I had uploaded screen shot of proof.
– SRJ
Aug 10 at 12:27
Have you read the earlier part of the equality proof for unit vectors?
– Calvin Khor
Aug 10 at 12:30
@CalvinKhor .Yes I had read the proof.But I don't understand equality argument.
– SRJ
Aug 10 at 12:35
The second line is unclear, as you say. I suggest you post the whole proof of the CS inequality, including the part where the inequality is shown. Then, we can see if any inequalities/results from that proof were used implicitly in this second line. The answer below uses the dot product of vectors with the angular interpretation, but the issue is that the CS inequality is valid for all inner products, not just those associated with a dot product.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 10 at 12:18
The second line is unclear, as you say. I suggest you post the whole proof of the CS inequality, including the part where the inequality is shown. Then, we can see if any inequalities/results from that proof were used implicitly in this second line. The answer below uses the dot product of vectors with the angular interpretation, but the issue is that the CS inequality is valid for all inner products, not just those associated with a dot product.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 10 at 12:18
$x/||x||$ and $y/||y||$ are unit vectors with inner product 1. It follows that they are equal, but this is not obvious (that is, it is where the content of the argument lies). Perhaps this assertion is proved somewhere else?
– Daniel Littlewood
Aug 10 at 12:20
$x/||x||$ and $y/||y||$ are unit vectors with inner product 1. It follows that they are equal, but this is not obvious (that is, it is where the content of the argument lies). Perhaps this assertion is proved somewhere else?
– Daniel Littlewood
Aug 10 at 12:20
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó I had uploaded screen shot of proof.
– SRJ
Aug 10 at 12:27
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó I had uploaded screen shot of proof.
– SRJ
Aug 10 at 12:27
Have you read the earlier part of the equality proof for unit vectors?
– Calvin Khor
Aug 10 at 12:30
Have you read the earlier part of the equality proof for unit vectors?
– Calvin Khor
Aug 10 at 12:30
@CalvinKhor .Yes I had read the proof.But I don't understand equality argument.
– SRJ
Aug 10 at 12:35
@CalvinKhor .Yes I had read the proof.But I don't understand equality argument.
– SRJ
Aug 10 at 12:35
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Near the start of the proof, they show that for unit length vectors
$$langle x-y, x-yrangle = 2 - 2langle x,yrangle.$$
But recall that $langle u,urangle = 0 iff u = 0$. Thus if $langle x,yrangle = 1$, then
$$langle x-y, x-yrangle = 2 - 2 = 0 implies x -y = 0 implies x = y$$
Instead of assuming $x$ and $y$ are unit length, if we repeat these calculations with $fracxVert xVert$ and $fracyVert yVert$, you obtain precisely their proof.
answered Aug 10 at 12:36
Harambe
5,80121843
add a comment |Â
up vote
6
down vote
That step was explained way before in the proof:
If $langle x,yrangle=1$, from the above chain of inequalities we deduce that $langle x-y,x-yrangle=0$.
(of course, this in the case $|x|=|y|=1$, as stated in the book).
answered Aug 10 at 12:31
Martin Argerami
117k1071165
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Near the start of the proof, they show that for unit length vectors
$$langle x-y, x-yrangle = 2 - 2langle x,yrangle.$$
But recall that $langle u,urangle = 0 iff u = 0$. Thus if $langle x,yrangle = 1$, then
$$langle x-y, x-yrangle = 2 - 2 = 0 implies x -y = 0 implies x = y$$
Instead of assuming $x$ and $y$ are unit length, if we repeat these calculations with $fracxVert xVert$ and $fracyVert yVert$, you obtain precisely their proof.
answered Aug 10 at 12:36
Harambe
5,80121843
add a comment |Â
up vote
6
down vote
accepted
Near the start of the proof, they show that for unit length vectors
$$langle x-y, x-yrangle = 2 - 2langle x,yrangle.$$
But recall that $langle u,urangle = 0 iff u = 0$. Thus if $langle x,yrangle = 1$, then
$$langle x-y, x-yrangle = 2 - 2 = 0 implies x -y = 0 implies x = y$$
Instead of assuming $x$ and $y$ are unit length, if we repeat these calculations with $fracxVert xVert$ and $fracyVert yVert$, you obtain precisely their proof.
answered Aug 10 at 12:36
Harambe
5,80121843
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Near the start of the proof, they show that for unit length vectors
$$langle x-y, x-yrangle = 2 - 2langle x,yrangle.$$
But recall that $langle u,urangle = 0 iff u = 0$. Thus if $langle x,yrangle = 1$, then
$$langle x-y, x-yrangle = 2 - 2 = 0 implies x -y = 0 implies x = y$$
Instead of assuming $x$ and $y$ are unit length, if we repeat these calculations with $fracxVert xVert$ and $fracyVert yVert$, you obtain precisely their proof.
answered Aug 10 at 12:36
Harambe
5,80121843
Near the start of the proof, they show that for unit length vectors
$$langle x-y, x-yrangle = 2 - 2langle x,yrangle.$$
But recall that $langle u,urangle = 0 iff u = 0$. Thus if $langle x,yrangle = 1$, then
$$langle x-y, x-yrangle = 2 - 2 = 0 implies x -y = 0 implies x = y$$
Instead of assuming $x$ and $y$ are unit length, if we repeat these calculations with $fracxVert xVert$ and $fracyVert yVert$, you obtain precisely their proof.
answered Aug 10 at 12:36
Harambe
5,80121843
answered Aug 10 at 12:36
Harambe
5,80121843
answered Aug 10 at 12:36
Harambe
5,80121843
answered Aug 10 at 12:36
Harambe
5,80121843
5,80121843
add a comment |Â
add a comment |Â
up vote
6
down vote
That step was explained way before in the proof:
If $langle x,yrangle=1$, from the above chain of inequalities we deduce that $langle x-y,x-yrangle=0$.
(of course, this in the case $|x|=|y|=1$, as stated in the book).
answered Aug 10 at 12:31
Martin Argerami
117k1071165
add a comment |Â
up vote
6
down vote
That step was explained way before in the proof:
If $langle x,yrangle=1$, from the above chain of inequalities we deduce that $langle x-y,x-yrangle=0$.
(of course, this in the case $|x|=|y|=1$, as stated in the book).
answered Aug 10 at 12:31
Martin Argerami
117k1071165
add a comment |Â
up vote
6
down vote
up vote
6
down vote
That step was explained way before in the proof:
If $langle x,yrangle=1$, from the above chain of inequalities we deduce that $langle x-y,x-yrangle=0$.
(of course, this in the case $|x|=|y|=1$, as stated in the book).
answered Aug 10 at 12:31
Martin Argerami
117k1071165
That step was explained way before in the proof:
If $langle x,yrangle=1$, from the above chain of inequalities we deduce that $langle x-y,x-yrangle=0$.
(of course, this in the case $|x|=|y|=1$, as stated in the book).
answered Aug 10 at 12:31
Martin Argerami
117k1071165
edited Aug 10 at 18:49
answered Aug 10 at 12:31
Martin Argerami
117k1071165
answered Aug 10 at 12:31
Martin Argerami
117k1071165
answered Aug 10 at 12:31
Martin Argerami
117k1071165
117k1071165
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2878304%2freasoning-for-equality-of-cauchy-schwarz-inequality-holds%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
The second line is unclear, as you say. I suggest you post the whole proof of the CS inequality, including the part where the inequality is shown. Then, we can see if any inequalities/results from that proof were used implicitly in this second line. The answer below uses the dot product of vectors with the angular interpretation, but the issue is that the CS inequality is valid for all inner products, not just those associated with a dot product.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 10 at 12:18
$x/||x||$ and $y/||y||$ are unit vectors with inner product 1. It follows that they are equal, but this is not obvious (that is, it is where the content of the argument lies). Perhaps this assertion is proved somewhere else?
– Daniel Littlewood
Aug 10 at 12:20
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó I had uploaded screen shot of proof.
– SRJ
Aug 10 at 12:27
Have you read the earlier part of the equality proof for unit vectors?
– Calvin Khor
Aug 10 at 12:30
@CalvinKhor .Yes I had read the proof.But I don't understand equality argument.
– SRJ
Aug 10 at 12:35