f:V-->V is an idempotent (i.e. f(f(v))=f(v)) linear map of an n-dimensional F-vector space V. I need to prove that there is a basis {v1,v2,...,vn} and a natural number r: 0<=r<=n so that f(vi)=vi for 0<=i<=r and f(vi)=0 for r+1<=i<=n.
I think it has something to do with an intersection of N(f) and V having to have something other than 0, but I'm still struggling with the actual proof. I think it has something to do with the idempotency of f.
An easy to follow formalized proof would be greatly appreciated.
I am sorry I was not clear enough. I used T for linear transformation and Null(T) for the kernel of T.
Do you know what a direct sum of vector space is? If yes, it is easy. Or else you have to write a long proof...
Anyway... observe that if f is a linear transformation, then I-f is also a linear transformation, where I is the identity lin.transf. Consider the kernel(I-f). Its a finite dimensional vector space. Let its dimension be r. You can find r basis vectors for this kernel. Observe that all the elements v in that kernel satisfy f(v) = v. Now extend the basis and show that the space spanned by extended vectors is the kernel(f).
Definition of direct sum of U and W: If U and W are subspaces of V such that and ,then we say .
[Definition of U+W: (check that this is a vec. space!)]
Now we claim .
Applying the definition, we have to prove two things:
(1)
Proof: This is obvious since
(2)
Proof: is a subspace of V(Why?). Let , then . But and . Thus every vector can be written as a sum of two vectors, one from and the other from
From (1) and (2),