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Show that V is a direct sum

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##T\in L(V,V)## such that ##T^{2}=1##. Show that ##V=V_{+}\oplus V_{-}## where ##V_{+}=\{v\in V:T(v)=v\}## and ##V_{-}=\{v\in V:T(v)=-v\}##.


    3. The attempt at a solution
    I was given a theorem that said that ##V## is the direct sum if and only if every vector in ##V## can be expressed as a sum ##v=v_{1}+v_{2}## where ##v_{1}\in V_{+}## and ##v_{2}\in V_{-}## and if ##v_{1}+v_{2}=0## then ##v_{1}=v_{2}=0##.

    I was able to show that if ##v_{1}+v_{2}=0## then ##v_{1}=v_{2}=0## but I am not able to show that every vector in ##V## can be expressed as a sum ##v=v_{1}+v_{2}## where ##v_{1}\in V_{+}## and ##v_{2}\in V_{-}##.
     
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  3. Oct 5, 2013 #2

    HallsofIvy

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    Let v be any vector in V. Let u= T(v). Since [itex]T^2= 1[/itex], T(u)= v. Then T(u- v)= T(u)- T(v)= v- u= -(u- v) and T(u+ v)= T(u)+ T(v)= v+ u= u+ v.
     
  4. Oct 5, 2013 #3
    I am a little confused on how this shows that any v can be expressed as a sum of v1 and v2. It looks to me like it just shows that any vector is either in v1 or v2.
     
  5. Oct 5, 2013 #4

    Dick

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    Then you need to reread Hall's proof. v is ANY vector. u isn't any vector. u=T(v). v=(v+u)+(v-u). v+u is in ##V_{+}## and (v-u) is in ##V_{-}##.
     
  6. Oct 5, 2013 #5
    AH! Thanks! I was a little too frustrated with this problem to be able to read his proof clearly...
     
  7. Oct 5, 2013 #6

    Dick

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    Yes, but I've got a typo. v=(v+u)/2+(v-u)/2. Hope you noticed.
     
  8. Oct 5, 2013 #7
    Yea I got it. Thanks!
     
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