# Show that V is a direct sum

In summary, we are given a theorem stating that V is the direct sum if and only if every vector in V can be expressed as a sum v=v1+v2 where v1∈V+ and v2∈V−, and if v1+v2=0 then v1=v2=0. We were able to show that if v1+v2=0 then v1=v2=0, but struggled to prove that every vector in V can be expressed in this form. However, by letting u=T(v) and using T^2=1, we were able to show that any vector v can be expressed as v=(v+u)/2+(v-u)/2, where (v+u)/

## 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_{-}##.

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.

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...