Where are you having trouble?
show that if f is in Hom(U,U) such that f^{2}=f then V= ker f + ker(f - id_{v}). also let B_{0 }be an ordered basis For ker f and B1 be ordered basis for ker(f-idv). if B is an ordered basis and B=B_{0} + B_{1} for V. what is the matrix representation of f given B,B?
Do you know what the kernel of a homomorphism is? If v is in the kernel of f, then f(v) = 0. If v is in the kernel of (f-id_V), then (f-id_V)(v)=0. So, the first thing you want to do is show that V=ker f (+) ker (f-id_V) is a subspace of U (where (+) is direct sum since I am having trouble with LaTeX).
I just told you. There is nothing magic about it. Suppose v is in ker f and v is in ker(f-id_V). Show that v is the zero vector. Since v is in ker f, you have f(v)=0. Since v is in ker(f-id_V), (f-id_V)(v) = 0. For the second one, you have (f-id_V)(v) = f(v) - id_V(v) = f(v)-v = 0. If f(v)-v=0 then f(v)=v. If f(v)=v and f(v)=0 then by transitivity, v=0.
Let v be any element of V.
Then if w = f(v), we have:
f(w) = f(f(v)) = f^{2}(v) = f(v) = w.
Thus f(w) - w = 0, so:
(f - id_{V})(w) = 0, that is: f(v) is in ker(f - id_{V}).
Write:
v = f(v) + v - f(v) (which is always trivially true).
We have, by the above, that f(v) is in ker(f - id_{V}).
Now let's compute f(v - f(v)):
f(v - f(v)) = f(v) - f(f(v)) = f(v) - f^{2}(v) = f(v) - f(v) = 0, so for ANY v, v - f(v) is in ker(f).
So this tells us that we can write any element of V as the sum of something in ker(f) (namely: v - f(v)) and something in ker(f - id_{V}) (namely: f(v)).
Hence V = ker(f) + ker(f - id_{V}).
Now SlipEternal has already shown that the intersection of these two sets is {0}, hence the sum is direct.
As for the matrix representation, we have:
[f]_{B} = [0 0...0 b_{1} b_{2}....b_{k}] (these are column vectors) where B_{1} = {b_{1},...,b_{k}}
Here is a simple example:
Let V = R^{2}, and let f(x,y) = (x,0).
Then f^{2}(x,y) = f(f(x,y)) = f(x,0) = (x,0), so we see that f^{2} = f.
Clearly, a basis for ker(f) is {(0,1)}. let's find a basis for ker(f - id_{V}).
If f(x,y) = (x,y), then:
(x,0) = (x,y), so y = 0. Thus a basis is: {(1,0)}. In the ordered basis B = {(0,1),(1,0)} (note this is not the USUAL ordered basis), we have
[f]_{B} =
[0 1]
[0 0].
(Slight note: usually we write the ordered basis the "other way around" as B = B_{1} + B_{0}, so the "0's come last").