A symmetry identity involving norms
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Let $x,yne 0$ in $mathbb R^n$. Give a geometrical interpretation of the identity
$Big Vertfrac xVert xVert -Vert xVert yBig Vert =
Big Vertfrac yVert yVert -Vert yVert xBig Vert $.
Here $VertcdotVert $ denotes the Euclidean norm. I can prove the identity algebraically; both sides are $1-2langle x,yrangle +Vert xVert^2+Vert yVert^2$. What is a geometrical interpretation?
linear-algebra normed-spaces
asked Aug 27 at 14:24
Shahab
4,62812474
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up vote
2
down vote
favorite
Let $x,yne 0$ in $mathbb R^n$. Give a geometrical interpretation of the identity
$Big Vertfrac xVert xVert -Vert xVert yBig Vert =
Big Vertfrac yVert yVert -Vert yVert xBig Vert $.
Here $VertcdotVert $ denotes the Euclidean norm. I can prove the identity algebraically; both sides are $1-2langle x,yrangle +Vert xVert^2+Vert yVert^2$. What is a geometrical interpretation?
linear-algebra normed-spaces
asked Aug 27 at 14:24
Shahab
4,62812474
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $x,yne 0$ in $mathbb R^n$. Give a geometrical interpretation of the identity
$Big Vertfrac xVert xVert -Vert xVert yBig Vert =
Big Vertfrac yVert yVert -Vert yVert xBig Vert $.
Here $VertcdotVert $ denotes the Euclidean norm. I can prove the identity algebraically; both sides are $1-2langle x,yrangle +Vert xVert^2+Vert yVert^2$. What is a geometrical interpretation?
linear-algebra normed-spaces
asked Aug 27 at 14:24
Shahab
4,62812474
Let $x,yne 0$ in $mathbb R^n$. Give a geometrical interpretation of the identity
$Big Vertfrac xVert xVert -Vert xVert yBig Vert =
Big Vertfrac yVert yVert -Vert yVert xBig Vert $.
Here $VertcdotVert $ denotes the Euclidean norm. I can prove the identity algebraically; both sides are $1-2langle x,yrangle +Vert xVert^2+Vert yVert^2$. What is a geometrical interpretation?
linear-algebra normed-spaces
asked Aug 27 at 14:24
Shahab
4,62812474
edited Aug 27 at 14:31
asked Aug 27 at 14:24
Shahab
4,62812474
asked Aug 27 at 14:24
Shahab
4,62812474
asked Aug 27 at 14:24
Shahab
4,62812474
4,62812474
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Let $u=x/|x|, v=y/|y|$ and $p=|x||y|$. The identity can then be rewritten as
$$
|u-pv|=|v-pu|.
$$
It's true simply because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$.
answered Aug 27 at 14:56
user1551
67.2k565123
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $u=x/|x|, v=y/|y|$ and $p=|x||y|$. The identity can then be rewritten as
$$
|u-pv|=|v-pu|.
$$
It's true simply because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$.
answered Aug 27 at 14:56
user1551
67.2k565123
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
add a comment |Â
up vote
3
down vote
accepted
Let $u=x/|x|, v=y/|y|$ and $p=|x||y|$. The identity can then be rewritten as
$$
|u-pv|=|v-pu|.
$$
It's true simply because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$.
answered Aug 27 at 14:56
user1551
67.2k565123
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $u=x/|x|, v=y/|y|$ and $p=|x||y|$. The identity can then be rewritten as
$$
|u-pv|=|v-pu|.
$$
It's true simply because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$.
answered Aug 27 at 14:56
user1551
67.2k565123
Let $u=x/|x|, v=y/|y|$ and $p=|x||y|$. The identity can then be rewritten as
$$
|u-pv|=|v-pu|.
$$
It's true simply because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$.
answered Aug 27 at 14:56
user1551
67.2k565123
answered Aug 27 at 14:56
user1551
67.2k565123
answered Aug 27 at 14:56
user1551
67.2k565123
answered Aug 27 at 14:56
user1551
67.2k565123
67.2k565123
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
add a comment |Â
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
Can you please explain why this identity is implied by the existence of Q?
– Shahab
Aug 28 at 16:59
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
@Shahab $u-pv=Q(v-pu)$.
– user1551
Aug 28 at 17:17
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
Thank you. How can we prove that Q always exists?
– Shahab
Aug 29 at 2:49
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
@Shahab Let $V=operatornamespanu,v$. Simply define $Q|_V$ as the reflection about $u+v$ (i.e. define $Q(u+v):=u+v$ and $Q(u-v):=v-u$) and $Q|_V^perp$ as the identity map.
– user1551
Aug 29 at 9:57
add a comment |Â
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