Norm of vectors [closed]











up vote
-1
down vote

favorite












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










share|cite|improve this question













closed 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

If this question can be reworded to fit the rules in the help center, please edit the 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
    Nov 22 at 6:41















up vote
-1
down vote

favorite












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










share|cite|improve this question













closed 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

If this question can be reworded to fit the rules in the help center, please edit the 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
    Nov 22 at 6:41













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











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










share|cite|improve this question













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







linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 6:38









Maths Geek

163




163




closed 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

If this question can be reworded to fit the rules in the help center, please edit the question.




closed 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

If this question can be reworded to fit the rules in the help center, please edit the 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
    Nov 22 at 6:41














  • 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










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 $kgeqslant0$.






share|cite|improve this answer























  • 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 $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
    – Nosrati
    Nov 22 at 8:22




















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




























    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 $kgeqslant0$.






    share|cite|improve this answer























    • 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 $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
      – Nosrati
      Nov 22 at 8:22

















    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
      Nov 22 at 7:08












    • Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
      – Nosrati
      Nov 22 at 8:22















    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 Nov 22 at 8:28

























    answered Nov 22 at 6:48









    Nosrati

    26.3k62353




    26.3k62353












    • 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 $x=vec{i}$ and $y=-vec{i}$ 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












    • Of course. Simply $x=vec{i}$ and $y=-vec{i}$ 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 $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
    – Nosrati
    Nov 22 at 8:22






    Of course. Simply $x=vec{i}$ and $y=-vec{i}$ in $mathbb R^2$.
    – Nosrati
    Nov 22 at 8:22












    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 Nov 22 at 8:05









        gimusi

        91.9k84495




        91.9k84495















            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Aardman Animations

            Are they similar matrix