1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving a vector inequality in R^n

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    In R^n show that:

    [tex]\Vert\overrightarrow{x - y}\Vert \Vert\overrightarrow{x+y}\Vert \leq \Vert\overrightarrow{x}\Vert^{2} + \Vert\overrightarrow{y}\Vert^{2}[/tex]


    2. Relevant equations

    My main attempts have either been from the Triangle Inequality:
    [tex] \Vert\overrightarrow{x+y}\Vert \leq \Vert\overrightarrow{x}\Vert + \Vert\overrightarrow{y}\Vert[/tex]

    Or from attempts to implement the idea:
    sqrt(a^2 + b^2) <= |a| + |b|

    3. The attempt at a solution
    Everytime I attempt to do this algebraically, the product on the left becomes a distributive catastrophe and I can't get anything sensible out of it, and attempting to square both sides to get rid of the square root just results in the same catastrophe on the RHS. I tried to represent the left hand side as a dot product but that involved cos(theta) which just complicates the problem. If anyone could point me in the right direction it would be appreciated.
     
  2. jcsd
  3. Jan 28, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    The triangle equality is all you need. Just square both sides and you're almost there already.
     
  4. Jan 28, 2009 #3
    I feel more stuck than before! Squaring the Triangle inequality obtains:

    [tex]
    \Vert\overrightarrow{x+y}\Vert^{2} \leq \Vert\overrightarrow{x}\Vert^{2} + \Vert\overrightarrow{y}\Vert^{2} + 2\Vert\overrightarrow{x}\Vert\Vert\overrightarrow{y}\Vert[/tex]

    So do I now need to show that:

    [tex]
    \Vert\overrightarrow{x - y}\Vert \Vert\overrightarrow{x+y}\Vert \leq \Vert\overrightarrow{x+y}\Vert^{2} - 2\Vert\overrightarrow{x}\Vert\Vert\overrightarrow{y}\Vert[/tex]
    ? EDIT: I found a counterexample to this equation, so mark it off the list

    Even more frustrating is this easy result from the Triangle Inequality:

    [tex]
    \Vert\overrightarrow{x+y}\Vert\Vert\overrightarrow{x-y}\Vert \leq \Vert\overrightarrow{x}\Vert^{2} + \Vert\overrightarrow{y}\Vert^{2} + 2\Vert\overrightarrow{x}\Vert\Vert\overrightarrow{y}\Vert[/tex]

    I just can't get rid of the 2xy term!
     
    Last edited: Jan 28, 2009
  5. Jan 28, 2009 #4

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Oohhh, now I see, there is a minus sign in one of them.
    Then I think the triangle inequality doesn't help you much... sorry.

    I did get it to work (more or less) through the dot product. You have to think about cos(theta) a bit, but eventually you will be able to use that a - b <= a if b >= 0 and sqrt(a^2 + b^2) <= |a| + |b| to get where you want.
    I hope that's a better hint and you'll get there now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Proving a vector inequality in R^n
  1. Adding vectors in R^n (Replies: 1)

  2. Vector Basis of R^n (Replies: 0)

Loading...