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Complex conjugate on an inner product

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the set ##C^2= {x=(x_1,x_2):x_1,x_2 \in C}##.

    Prove that ##<x,y>=x_1 \overline{y_1}+x_2 \overline{y_2}## defines an inner product on ##C^2##

    2. Relevant equations
    3. The attempt at a solution

    ##<,y>=\overline {<y,x>}##

    ##= \overline {y_1x_1} + \overline {y_2x_2}##

    I think I need some other information to continue.....?

    Thanks
     
  2. jcsd
  3. May 17, 2012 #2
    verify that this satisfies the definition of an inner product:

    ##<x,y>=\overline{<y,x>}##
    ##<x+y,z>=<x,z>+<y,z>##
    ##<ax,z>=a<x,z>##
    ##<x,x> \geq 0##
    ##<x,x>=0 \iff x=0##

    for all x,y,z in ##C^2## and for all complex scalars a
     
  4. May 20, 2012 #3
    I can verify the above for x,y,z in ##R^2## but don't know how to extend to the complex field............? Especially the first axiom..

    THanks
     
  5. May 20, 2012 #4
    ##\overline{x}## means the complex conjugate of ##x##, that is if ##x=a+ib## for ##a,b \in \mathbb{R}## then ##\overline{x} = a-ib##

    find ##\overline{<y,x>}## and show that it simplifies to ##x_1\overline{y_1} + x_2\overline{y_2}=<x,y>##
     
  6. May 20, 2012 #5

    Fredrik

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    This is wrong. Use the definition you posted, and you'll see that.
     
  7. May 21, 2012 #6
    1) ##<x,y>=\overline{<y,x>}=\overline{ y_1\overline{x_1}}+\overline { y_2\overline{x_2}}=\overline{y_1}x_1+\overline{y_2}x_2##

    2) ##<x+y,z>=(x_1+y_1)\overline{z_1}+(x_2+y_2) \overline{z_2}=x_1\overline {z_1}+y_1\overline {z_1}+x_2\overline {z_2}+y_2\overline {z_2}=<x,z>+<y,z>##

    3)##<\alpha x,y>= \alpha x_1 \overline{y_1}+\alpha x_2 \overline{y_2}=\alpha(x_1 \overline{y_1}+x_2\overline{y_2})=\alpha<x,y>##

    4)##<x,x>=x_1 \overline{x_1}+x_2 \overline{x_2}=|x_1|^2+|x_2|^2>=0## since by definition ##z \overline{z}=|z|^2>=0## and ##<x,x>=0## iff x=0, that is x1=x2=0...

    thanks
     
    Last edited: May 21, 2012
  8. May 21, 2012 #7

    Fredrik

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    2-4 are fine. In part 1, the first equality you wrote down is the one you're trying to prove. This is of course not OK. Your string of equalities should end with =<x,y>, not begin with <x,y>=. The best way to do this is one step at a time, like this:
    $$\overline{\langle y,x\rangle} =\overline{y_1\overline{x_1}+y_2\overline{x_2}} =\overline{y_1\overline{x_1}} +\overline{y_2\overline{x_2}} =\overline{y_1}\,\overline{\overline{x_1}} +\overline{y_2}\,\overline{\overline{x_2}}
    = \overline{y_1}x_1+\overline{y_2}x_2 =x_1\overline{y_1}+x_2\overline{y_2} =\langle x,y\rangle$$.
     
    Last edited: May 21, 2012
  9. May 24, 2012 #8
    Thanks Fredrik for the clarification.
     
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