Norm of vectors
up vote
1
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Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
asked 2 days ago
Maths Geek
173
add a comment |
up vote
1
down vote
favorite
Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
asked 2 days ago
Maths Geek
173
2
What is your space and what norm are you using? If you are using the standard norm on an Euclidean space then $x=ay$ for some $a geq 0$ or $y=ax$ for some $a geq 0$.
– Kavi Rama Murthy
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
asked 2 days ago
Maths Geek
173
Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
linear-algebra
asked 2 days ago
Maths Geek
173
asked 2 days ago
Maths Geek
173
asked 2 days ago
Maths Geek
173
asked 2 days ago
Maths Geek
173
asked 2 days ago
Maths Geek
173
173
2
What is your space and what norm are you using? If you are using the standard norm on an Euclidean space then $x=ay$ for some $a geq 0$ or $y=ax$ for some $a geq 0$.
– Kavi Rama Murthy
2 days ago
add a comment |
2
What is your space and what norm are you using? If you are using the standard norm on an Euclidean space then $x=ay$ for some $a geq 0$ or $y=ax$ for some $a geq 0$.
– Kavi Rama Murthy
2 days ago
2
2
What is your space and what norm are you using? If you are using the standard norm on an Euclidean space then $x=ay$ for some $a geq 0$ or $y=ax$ for some $a geq 0$.
– Kavi Rama Murthy
2 days ago
What is your space and what norm are you using? If you are using the standard norm on an Euclidean space then $x=ay$ for some $a geq 0$ or $y=ax$ for some $a geq 0$.
– Kavi Rama Murthy
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
9
down vote
In $mathbb R^n$ you may consider geometric view of the relation
$$||x||+||y||=||x+y||$$
it says you have a triangle with sides $x$, $y$ and $x+y$. Such triangle trivially is a segment and therefore $x$, $y$ lie on $x+y$. This shows that $x$ and $y$ are a positive multiplier of each other, that is $x=ky$ or $y=kx$, where $kgeqslant0$.
answered 2 days ago
Nosrati
26.2k62352
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
add a comment |
up vote
5
down vote
This is the equality case for triangular inequality which holds if and only if $x$ and $y$ are multiple vectors with the same direction, that is $y=kx$ with $k>0$ (excluding trivial cases $x=0,lor ,y=0$).
answered 2 days ago
gimusi
86.9k74393
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
In $mathbb R^n$ you may consider geometric view of the relation
$$||x||+||y||=||x+y||$$
it says you have a triangle with sides $x$, $y$ and $x+y$. Such triangle trivially is a segment and therefore $x$, $y$ lie on $x+y$. This shows that $x$ and $y$ are a positive multiplier of each other, that is $x=ky$ or $y=kx$, where $kgeqslant0$.
answered 2 days ago
Nosrati
26.2k62352
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
add a comment |
up vote
9
down vote
In $mathbb R^n$ you may consider geometric view of the relation
$$||x||+||y||=||x+y||$$
it says you have a triangle with sides $x$, $y$ and $x+y$. Such triangle trivially is a segment and therefore $x$, $y$ lie on $x+y$. This shows that $x$ and $y$ are a positive multiplier of each other, that is $x=ky$ or $y=kx$, where $kgeqslant0$.
answered 2 days ago
Nosrati
26.2k62352
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
add a comment |
up vote
9
down vote
up vote
9
down vote
In $mathbb R^n$ you may consider geometric view of the relation
$$||x||+||y||=||x+y||$$
it says you have a triangle with sides $x$, $y$ and $x+y$. Such triangle trivially is a segment and therefore $x$, $y$ lie on $x+y$. This shows that $x$ and $y$ are a positive multiplier of each other, that is $x=ky$ or $y=kx$, where $kgeqslant0$.
answered 2 days ago
Nosrati
26.2k62352
In $mathbb R^n$ you may consider geometric view of the relation
$$||x||+||y||=||x+y||$$
it says you have a triangle with sides $x$, $y$ and $x+y$. Such triangle trivially is a segment and therefore $x$, $y$ lie on $x+y$. This shows that $x$ and $y$ are a positive multiplier of each other, that is $x=ky$ or $y=kx$, where $kgeqslant0$.
answered 2 days ago
Nosrati
26.2k62352
edited 2 days ago
answered 2 days ago
Nosrati
26.2k62352
answered 2 days ago
Nosrati
26.2k62352
answered 2 days ago
Nosrati
26.2k62352
26.2k62352
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
add a comment |
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
– Nosrati
2 days ago
add a comment |
up vote
5
down vote
This is the equality case for triangular inequality which holds if and only if $x$ and $y$ are multiple vectors with the same direction, that is $y=kx$ with $k>0$ (excluding trivial cases $x=0,lor ,y=0$).
answered 2 days ago
gimusi
86.9k74393
add a comment |
up vote
5
down vote
This is the equality case for triangular inequality which holds if and only if $x$ and $y$ are multiple vectors with the same direction, that is $y=kx$ with $k>0$ (excluding trivial cases $x=0,lor ,y=0$).
answered 2 days ago
gimusi
86.9k74393
add a comment |
up vote
5
down vote
up vote
5
down vote
This is the equality case for triangular inequality which holds if and only if $x$ and $y$ are multiple vectors with the same direction, that is $y=kx$ with $k>0$ (excluding trivial cases $x=0,lor ,y=0$).
answered 2 days ago
gimusi
86.9k74393
This is the equality case for triangular inequality which holds if and only if $x$ and $y$ are multiple vectors with the same direction, that is $y=kx$ with $k>0$ (excluding trivial cases $x=0,lor ,y=0$).
answered 2 days ago
gimusi
86.9k74393
answered 2 days ago
gimusi
86.9k74393
answered 2 days ago
gimusi
86.9k74393
answered 2 days ago
gimusi
86.9k74393
86.9k74393
add a comment |
add a comment |
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2
What is your space and what norm are you using? If you are using the standard norm on an Euclidean space then $x=ay$ for some $a geq 0$ or $y=ax$ for some $a geq 0$.
– Kavi Rama Murthy
2 days ago