# Show that V is a direct sum

1. Oct 5, 2013

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 $v_{2}\in V_{-}$.

2. Oct 5, 2013

### HallsofIvy

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.

3. Oct 5, 2013

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.

4. Oct 5, 2013

### Dick

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 $V_{-}$.

5. Oct 5, 2013

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

6. Oct 5, 2013

### Dick

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

7. Oct 5, 2013