Norm of vectors [on hold]
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Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
asked Nov 22 at 6:38
Maths Geek
113
put on hold as off-topic by user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did Nov 25 at 9:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did
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up vote
-2
down vote
favorite
Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
asked Nov 22 at 6:38
Maths Geek
113
put on hold as off-topic by user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did Nov 25 at 9:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did
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
Nov 22 at 6:41
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
asked Nov 22 at 6:38
Maths Geek
113
Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?
linear-algebra
linear-algebra
asked Nov 22 at 6:38
Maths Geek
113
asked Nov 22 at 6:38
Maths Geek
113
asked Nov 22 at 6:38
Maths Geek
113
asked Nov 22 at 6:38
Maths Geek
113
asked Nov 22 at 6:38
Maths Geek
113
113
put on hold as off-topic by user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did Nov 25 at 9:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did
put on hold as off-topic by user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did Nov 25 at 9:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Jyrki Lahtonen, TheSimpliFire, Holo, Did
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
Nov 22 at 6:41
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
Nov 22 at 6:41
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
Nov 22 at 6:41
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
Nov 22 at 6:41
add a comment |
2 Answers
2
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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 Nov 22 at 6:48
Nosrati
1
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
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 Nov 22 at 8:05
gimusi
88k74393
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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 Nov 22 at 6:48
Nosrati
1
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
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 Nov 22 at 6:48
Nosrati
1
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
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 Nov 22 at 6:48
Nosrati
1
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 Nov 22 at 6:48
Nosrati
1
edited Nov 22 at 8:28
answered Nov 22 at 6:48
Nosrati
1
answered Nov 22 at 6:48
Nosrati
1
answered Nov 22 at 6:48
Nosrati
1
1
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
add a comment |
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
I don't care the vote, but I like to know where of my answer is wrong.
– Nosrati
Nov 22 at 7:08
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
Of course. Simply $x=veci$ and $y=-veci$ in $mathbb R^2$.
– Nosrati
Nov 22 at 8:22
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 Nov 22 at 8:05
gimusi
88k74393
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 Nov 22 at 8:05
gimusi
88k74393
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 Nov 22 at 8:05
gimusi
88k74393
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 Nov 22 at 8:05
gimusi
88k74393
answered Nov 22 at 8:05
gimusi
88k74393
answered Nov 22 at 8:05
gimusi
88k74393
answered Nov 22 at 8:05
gimusi
88k74393
88k74393
add a comment |
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
Nov 22 at 6:41