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Annihilator of a direct sum

  1. May 13, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose V=U⊕W. Prove that V0=U0⊕W0. (V0= annihilator of V).

    2. Relevant equations
    (U+W)0=U0∩W0

    3. The attempt at a solution
    Well, I don't see how this is possible. If V0=U0⊕W0, then U0∩W0={0}, and since (U+W)0=U0∩W0, it means (U+W)0={0}, but V=U⊕W, so V0={0}. I don't think this is the desirable result so it means I misunderstood something along the way, clarification would be appreciated.
     
  2. jcsd
  3. May 13, 2017 #2

    andrewkirk

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    This part is not relevant. It is not the case that ##V=U+W##. We have ##V=U\oplus W## which has a different meaning.

    To get a better handle on this, think about the following: say ##u\in U^0-\{0_U\}##. What is a corresponding element of ##V## that is in ##V^0##? Then do the same for ##w\in W^0-\{0_W\}##.
     
  4. May 14, 2017 #3
    that is wrong.
    the correct conclusion must be ##V^*=U^0\oplus W^0##
    Indeed,
    1) it is clear that ##U^0\cap W^0=\{0\}##; it also follows from
    2) take any function ##f\in V^*## and define a linear function ##f_U## is follows ##f_U(x):=f(x)## provided ##x\in U## and ##f_U(x)=0## provided ##x\in W##;
    similarly ##f_W(x):=f(x)## provided ##x\in W## and ##f_W(x)=0## provided ##x\in U##.
    Obviously it follows that ##f=f_U+f_W,\quad f_U\in W^0,\quad f_W\in U^0##
     
    Last edited: May 14, 2017
  5. May 15, 2017 #4
    But V=U⊕W means V=U+W and U∩W={0}, meaning the theorem still applies, doesn't it?
    Also I am not familiar with the notation 0u and 0w, so i'd appreciate an explanation about that.
     
  6. May 15, 2017 #5
    Well this does make sense, if it is indeed a misprint in the book it would not be the first one...
     
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