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Inner products and adjoints

  1. Apr 6, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=57712&stc=1&d=1365596947.png


    2. Relevant equations


    3. The attempt at a solution

    So I know that ##T^*## will be from F to V and we have a fixed scalar a in F. So :

    ##<u, T^*(a)> \space = \space <T(u), a> \space = \space <<u,v>,a>##

    Now, I am unsure how to continue this. What should I do with <u,v>? I know that <u,v> is in F, but that doesn't seem to help me much?
     

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    Last edited by a moderator: Apr 10, 2013
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  3. Apr 6, 2013 #2

    Zondrina

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    Sorry for the double, but I stared at this for a second and thought of something.

    ##<<u,v>, a> \\
    = < u_1 \overline{v_1} + ... + u_n \overline{v_n}, a> \\
    = < u_1 \overline{v_1}, a> + ... + < u_n \overline{v_n}, a>##

    I'm not quite sure this helps either though.
     
  4. Apr 6, 2013 #3

    micromass

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    You know that ##<u,v>## and ##a## are scalars. So what is ##<<u,v>,a>##? What is the inner product in ##\mathbb{R}##?
     
  5. Apr 6, 2013 #4

    Zondrina

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    Hmm... ##<<u,v>,a> = a<<u,v>,1>\space##

    its just a real number when we work over ##ℝ## I believe.
     
    Last edited: Apr 6, 2013
  6. Apr 6, 2013 #5

    Dick

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    I don't get this whole question. V is a vector space. It doesn't say that it is F or R. If T maps V to F. Then T* doesn't map F to V. It maps the dual space V* to F. It doesn't act on scalars.

    I'm confused. Or is there some interpretation of adjoint I don't know.
     
  7. Apr 7, 2013 #6

    Fredrik

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    Suppose that H and K are Hilbert spaces over F, where F is either ℝ or ℂ. I will use the convention that when F=ℂ, the inner product is linear in the second variable. Suppose that ##T:H\to K## is linear and bounded. Let ##x\in K## be arbitrary. Define ##\phi_x:H\to F## by ##\phi_x(y)=\langle x,Ty\rangle_K## for all ##y\in H##. This map is linear and bounded, so ##\phi_x\in H^*##. The Riesz representation theorem for Hilbert spaces tells us that this implies that there's a unique ##z\in H## such that ##\phi_x=\langle z,\cdot\rangle_H##. (The right-hand side denotes the map that takes an arbitrary ##w\in H## to ##\langle z,w\rangle_H##). The map ##x\mapsto z## is denoted by ##T^*## and is called the adjoint of T. Note that ##x\in K## and ##z\in H##, so ##T^*:K\to H##.

    Since ##\phi_x=\langle T^*x,\cdot\rangle_H##, we have
    $$\langle T^*x,y\rangle_H =\phi_x(y) =\langle x,Ty\rangle_K$$ for all ##y\in H##.

    The field of scalars F can be interpreted as a Hilbert space over F, with addition being the field's addition operation, and both scalar multiplication and the inner product being the field's multiplication operation. So the definition of "adjoint" above implies that if ##T:V\to F##, then ##T^*:F\to V##.
     
    Last edited: Apr 7, 2013
  8. Apr 7, 2013 #7

    Zondrina

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    This is from the book linear algebra done right. So throughout the course of the chapter, V denotes a finite dimensional inner product space over F.

    I was confused about it acting on scalars as well.
     
  9. Apr 7, 2013 #8

    Dick

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    Certainly. The only problem I have with that is that the problem doesn't say that V is F regarded as a vector space over itself. Unless this is in the context of other problems which say it is. And it's pretty sloppy to say ##T^*:F\to V## in that case. V and F have the same elements but V is F regarded as a vector space and F is a field. To say ##T^*:F\to F## is less sloppy. I don't see why you would want to reverse the symbols F and V.
     
    Last edited: Apr 7, 2013
  10. Apr 7, 2013 #9

    Fredrik

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    V can be any Hilbert space over F in this problem. It doesn't have to be equal to F.
     
  11. Apr 7, 2013 #10

    Zondrina

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    I don't believe we've covered Hilbert Spaces yet, so I don't think I can use anything pertaining to it.

    V is an inner product space over a field F, but I don't think I've ever seen an inner product for scalars only which is why I'm having so much trouble manipulating the equation in my first post.
     
  12. Apr 7, 2013 #11

    micromass

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    The space ##\mathbb{C}## is always an inner-product space over itself. The inner product is given by ##<x,y> = \overline{x}y##. You should check that this is indeed an inner product.
    What will the inner product be of the space ##\mathbb{R}## over itself?
     
  13. Apr 7, 2013 #12

    Zondrina

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    It's the same as the complex one, except complex conjugation is not needed.

    <x,y> = xy
     
  14. Apr 7, 2013 #13

    micromass

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    Right! So the product in ##\mathbb{R}## is a special case of the inproduct in an inner-product space! It's good to keep this in mind!

    Can you solve the exercise now?
     
  15. Apr 7, 2013 #14

    Zondrina

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    ##<u, T^*(a)> \space = \space <T(u), a> \space = \space <<u,v>,a> = \space <u \overline{v},a> = u \overline{va}##

    Still can't see where this is going.
     
  16. Apr 7, 2013 #15

    micromass

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    No!! You can't say ##<u,v> = \overline{u}v##!! That is only true in the special case ##V=\mathbb{C}##. But your ##u## and ##v## are in a general space, so you can't use that expression.

    The only thing you can conclude is

    [tex]<u,T*(a)> = \overline{a}<u,v>[/tex]

    Now, can you use this to write ##T*(a)## in another form?? That is, do you know some other vector ##x## such that

    [tex]<u,x> = \overline{a} <u,v>[/tex]
     
  17. Apr 7, 2013 #16

    Fredrik

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    Every finite-dimensional inner product space is a Hilbert space, so everything I said holds for this V as well. One thing is easier when we're dealing with a finite-dimensional space: We don't have to use the Riesz representation theorem.

    Note however that what I wrote was an explanation meant for Dick. It may be useful for you too, but I was just talking about the definition of the adjoint in general. I haven't said anything about this specific problem.
     
  18. Apr 7, 2013 #17

    Zondrina

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    I just noticed my first post was wrong by a bit. I'll start over here and see what happens.

    Okay so ##T : V → F \Rightarrow T^*: F → V##. Fix a scalar ##a \in F## and we know ##v \in V##. Also ##T(u) = <u,v>##.

    ##<v, T^*(a)> \space = \space <T(v), a> \space = \space <<v,v>,a> \space##

    That's better. I don't know if V is ℝ or ℂ in particular, we're supposed to assume it can be ℂ unless stated otherwise I believe.

    So ##<<v,v>,a> = \overline{a}<<v,v>,1> = \overline{a}<v,v> = <v, av> = <v, T^*(a)>##

    ##∴T^*(a) = av##

    EDIT : If anyone sees this, I believe this is a Linear functional?
     
    Last edited: Apr 7, 2013
  19. Apr 8, 2013 #18

    Fredrik

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    <v,T*a>=<v,av> doesn't imply that T*a=av. Can you think of something you can do just a little bit differently that will allow you to conclude that T*a=av at the end of the calculation? (That is the correct formula for T*a).

    A linear functional is a linear transformation into the field of scalars. Since the codomain of T* is V, not F, it's not a linear functional. It's just a linear transformation.
     
  20. Apr 8, 2013 #19

    Zondrina

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    This confused me, how does that not imply it? What could I have possibly done differently to arrive at the general formula for ##T^*(a)##?
     
  21. Apr 8, 2013 #20

    Fredrik

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    Take the vector space ##\mathbb R^2## for example. We have $$\langle (1,0),(2,0)\rangle=2 =\langle (1,0),(2,5)\rangle$$ but ##(2,0)\neq (2,5)##. So it's not always true that if ##\langle x,y\rangle=\langle x,z\rangle##, then ##y=z##.

    Also note that for all w such that <w,v>=0, we have <v,T*a+w>=<v,T*a>=<v,av>. This observation should be enough to cast some doubt on your derivation of T*a=av.
     
    Last edited: Apr 8, 2013
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