Is V a Direct Sum of V+ and V-?

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SUMMARY

The discussion centers on proving that a vector space \( V \) can be expressed as a direct sum of two subspaces \( V_{+} \) and \( V_{-} \) under the linear transformation \( T \) where \( T^{2} = 1 \). The subspace \( V_{+} \) consists of vectors \( v \) such that \( T(v) = v \), while \( V_{-} \) consists of vectors \( v \) such that \( T(v) = -v \). The proof hinges on demonstrating that any vector \( v \) in \( V \) can be represented as \( v = (v + T(v))/2 + (v - T(v))/2 \), with the first term in \( V_{+} \) and the second in \( V_{-} \).

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Homework Statement


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\}##.

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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Let v be any vector in V. Let u= T(v). Since T^2= 1, 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.
 
HallsofIvy said:
Let v be any vector in V. Let u= T(v). Since T^2= 1, 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.

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.
 
DeadOriginal said:
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.

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_{-}##.
 
Dick said:
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_{-}##.

AH! Thanks! I was a little too frustrated with this problem to be able to read his proof clearly...
 
DeadOriginal said:
AH! Thanks! I was a little too frustrated with this problem to be able to read his proof clearly...

Yes, but I've got a typo. v=(v+u)/2+(v-u)/2. Hope you noticed.
 
Dick said:
Yes, but I've got a typo. v=(v+u)/2+(v-u)/2. Hope you noticed.

Yea I got it. Thanks!
 

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