Norm of vectors











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Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?






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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















up vote
2
down vote

favorite












Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?






linear-algebra







share|cite|improve this question


















  • 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













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







share|cite|improve this question













Let $x$ and $y$ be two vectors. What can you say when $||x||+||y||=||x+y||$?






linear-algebra




linear-algebra






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asked 2 days ago









Maths Geek

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193








  • 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




    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










2 Answers
2






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up vote
9
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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 $kgeqslant0$.






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  • 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




















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 $x=0,lor ,y=0$).






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    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 $kgeqslant0$.






    share|cite|improve this answer























    • 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

















    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 $kgeqslant0$.






    share|cite|improve this answer























    • 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















    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 $kgeqslant0$.






    share|cite|improve this answer














    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 $kgeqslant0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    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




















    • 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












    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 $x=0,lor ,y=0$).






    share|cite|improve this answer

























      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 $x=0,lor ,y=0$).






      share|cite|improve this answer























        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 $x=0,lor ,y=0$).






        share|cite|improve this answer












        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 $x=0,lor ,y=0$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        gimusi

        86.9k74393




        86.9k74393






























             

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