Results 1 to 4 of 4

Thread: Prove <x,y>=0 iff llx+yll=llx-yll

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    14

    Prove <x,y>=0 iff llx+yll=llx-yll

    I'm revising for my final and I can't seem to prove this.

    If llx+yll=llx-yll then
    then Real parts are 0
    hence <x,y>=0

    But I can't prove it starting with <x,y>
    any help appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Roland25 View Post
    I'm revising for my final and I can't seem to prove this.

    If llx+yll=llx-yll then
    then Real parts are 0
    hence <x,y>=0

    But I can't prove it starting with <x,y>
    any help appreciated
    Since you mention the real part is 0 I assume that x,y \in \mathbb{C}

    Remember that
    |z|^2=z \bar{z}

    so if we square both sides we get

    |x+y|^2=|x-y|^2 and using the above identity we get

    (x+y)\overline{(x+y)}= (x-y)\overline{(x-y)}

    (x+y)(\bar {x}+\bar{ y})= (x-y)(\bar{ x}- \bar {y})

    |x|^2+x\bar{y}+y\bar{x}+|y|^2=|x|^2-x\bar{y}-y\bar{x}+|y|^2

    x\bar{y}+y\bar{x}=0 \iff x\bar{y}+\overline{x\bar{y}}=0 \iff 2Re[{x\bar{y}}]=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    8
    Quote Originally Posted by Roland25 View Post
    Prove <x,y>=0 iff llx+yll=llx-yll
    This is true if the scalar field is the real numbers, but not if it is the complex numbers. For example, |1+i| = |1i|, but \langle1,i\rangle = \overline{i}\ne0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2009
    Posts
    11
    Quote Originally Posted by Roland25 View Post
    I'm revising for my final and I can't seem to prove this.

    If llx+yll=llx-yll then
    then Real parts are 0
    hence <x,y>=0

    But I can't prove it starting with <x,y>
    any help appreciated
    Suppose Re<x,y>=0
    ||x+y||^2=||x||^2+||y||^2+2Re<x,y>=||x||^2+||y||^2-2Re<x,y>=||x-y||^2

    Suppose ||x+y||=||x-y||
    <x+y,x+y>=<x-y,x-y>
    ....
    4Re<x,y>=0

    Quote Originally Posted by Opalg View Post
    This is true if the scalar field is the real numbers, but not if it is the complex numbers. For example, |1+i| = |1i|, but \langle1,i\rangle = \overline{i}\ne0.
    You're right, it is not true that the <x,y>=0, but it is always true that the Real part of <x,y> is zero.
    Last edited by mr fantastic; May 9th 2009 at 06:38 PM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a/b and a/c then a/ (3b-7c)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 23rd 2010, 05:20 PM
  2. prove,,,
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 1st 2010, 09:02 AM
  3. Prove |w + z| <= |w| +|z|
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 28th 2010, 05:44 AM
  4. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  5. How to prove that n^2 + n + 2 is even??
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 30th 2008, 01:24 PM

Search Tags


/mathhelpforum @mathhelpforum