Prove that T:V->V T^2=T ker(T)+im(T)=V

[specialnlovin]

Let V be a vector space in R and T : V −> V be a linear map such that
T2 = T. Show that V = ker(T)+im(T).
does V = ker(T)  im(T) imply T2 = T? Give a
proof or counter-example.

I really just have no idea how to start this

 

[Tedjn]

In the first direction, if something in V is not in im(T), can you break it up into a part that is in im(T) and a part that is in ker(T)?

 

[micromass]

I think you need to prove that V=\ker T \oplus I am T.

This consists out of two parts:

V=\ker T + I am T

I'll give you a hint for this: v=T(v)+(v-T(v))

And you also need to prove

\ker T \cap I am T = \{0\}

This shouldn't prove so hard, I think...

 

[specialnlovin]

yes i did mean that, i just have no idea to add the (+) symbol on a computer.
In any case. v=T(v)+(v-T(v))
i know that v-T(v) is in ker(T). is that because v-T(v)=0? But how does this help?

 

[micromass]

v-T(v) is in ker T since T(v-T(v))=0.

v=T(v)+(v-T(v)) shows that V=ker T + I am T. Now you only need to show that the intersection in zero.

 

[wisvuze]

show that the intersection is 0 and then use the rank-nullity theorem

 

[micromass]

show that the intersection is 0 and then use the rank-nullity theorem

Yes, but I don't think he has seen the rank-nullity theorem yet. Since this would make it really trivial...

 

[jaimerobin]

show that the intersection is 0

How do you do that?