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

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Homework Help Overview

The discussion revolves around the linear operator T on a vector space V, specifically examining the conditions under which V can be expressed as a direct sum of two subspaces, V+ and V-. The original poster attempts to demonstrate that every vector in V can be represented as a sum of vectors from these subspaces, based on the properties of T.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the operator T and its properties, particularly T^2 = 1. Questions arise regarding the representation of any vector in V as a sum of vectors from V+ and V-, with some participants expressing confusion about how certain manipulations demonstrate this representation.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications and references to existing proofs. There is acknowledgment of confusion and the need for careful reading of the proof, indicating a productive exchange of ideas without reaching a definitive conclusion.

Contextual Notes

There is mention of a theorem regarding direct sums and the conditions that must be satisfied for V to be expressed as V+ ⊕ V-. Some participants note a typo in the expressions used, which may affect clarity in the discussion.

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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 [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.
 
HallsofIvy said:
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.

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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