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Inner Product Space, quick questions.

  1. Apr 7, 2013 #1
    1) Using the indentity:||u+v||^2=||u||^2+||v||^2+(u,v)+(v,u) - where (x,y) denotes the inner product .
    Now if we let the second term = iv, my book gives the identity as then:
    ||u+iv||^2=||u||^2+||v||^2-i(u,v)+i(v,u) [1].

    And I am struggling to derive this myself, here is my working:
    ||u+iv||^2=(u,u)+(u,iv)+(iv,u)+(iv,iv)
    = ||u||^2+(u,iv)+i(v,u)+||iv||^2
    Now the two terms I am struggling with are why ||iv||^2=||v||^2 and trying to get the (u,iv) to -i(u,v) as it is in [1].
    My attempt was to use the hermiticity property to give (u,iv)=(iv,u)* - letting * denote the complex conugate - gives (u,iv)=(-iv,u)=-i(v,u), where the last equivalence follws from linearity in the first factor, rather than -i(u,v).

    2) Is to find an orthornomal basis for the orthogonal complement of the subspace spanned by (2,1-i,0,1) and (1,0,i,3), under the standard inner product. Now I am ok with the main concepts of the strategy needed here: the orthogonal complement of the subspace is given by the space of solutions to
    2a+(1-i)b+d=0 and,
    a+ic+3d=0,
    obtained by ensuring that any arbitarty vector that lies in this space (a,b,c,d) is orthogonal to the spanning vectors given of the subspace.
    HOWEVER, my book gives these equations (1) and (2) as 2a+(1+i)b+d=0 and a-i+3d=0, and I can not see why the 'i's have been multipled by -1 compared to my workng.

    Many Thanks for any assistance.
     
  2. jcsd
  3. Apr 7, 2013 #2

    BruceW

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    well, effectively, the idea is to find the components of a vector which gives zero inner product with the vectors that you have been given, right? So what is the proper way to do an inner product?
     
  4. Apr 8, 2013 #3
    Ahh thanks , the complex inner product of x and y = (x,y*), where y* is the complex conjugate . so letting a=(2,1-i,0,1) and b=(1,0,i,3) and z=(a,b,c,d) for a,b,c,d in ℝ, I can see where this comes from taking the inner products: (z,a*) and (z,b*), however not if I take the other order of these, i.e. (a*,z) and (b*,z) as z*=(a,b,c,d).
     
    Last edited: Apr 8, 2013
  5. Apr 8, 2013 #4

    BruceW

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    Well, you've got (a*,z)=0 or (z*,a)=0 Do these two statements disagree with each other?
     
  6. Apr 8, 2013 #5
    Sorry, the post above should read the other order as (a,z*) or (b,z*).
    (z,a*) gives me the equation in line witht the equation: 2a+b(1+i)+d=0, and (a,z*) gives 2a+b(1-i)+d=0, which as far as I can see are not equivalent?
     
  7. Apr 8, 2013 #6

    BruceW

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    well, you need to do a proper calculation to see if they are equivalent. The two equations are:
    (z,a*) = 0 and (a,z*) = 0
    Is there a way to re-write one equation, as the other equation?
     
  8. Apr 9, 2013 #7
    Ahh I see, thanks, by hermiticity the two are equivalent.
     
  9. Apr 9, 2013 #8

    BruceW

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    yep, no worries, man
     
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