Can Disjoint Nullspace and Range Exist if Rank T Equals Rank T^2?

[lineintegral1]

Homework Statement

I'm trying to prove the following:
Let V be a finite dimensional vector space and let T be a linear operator on V. Suppose that the rank of T is equivalent to the rank of T^2. Then the range and the nullspace of T are disjoint. 2. The attempt at a solution

I've played around with a few things so far. Clearly, the nullity of T and the nullity of T^2 are the same by the rank-nullity theorem. I'm thinking that the way to start this is to show that the range of T and T^2 are the same. But I'm not sure exactly how to use the given rank information to my advantage.

I was also playing with a theorem that states that, given subspaces W_1 and W_2, then,

dim(W_1) + dim(W_2) = dim(W_1 intersection W_2) + dim(W_1 + W_2).

I was thinking that I could show that the dimension of the intersection is zero since the intersection is zero (and, therefore, the intersection won't have a basis). But again, I'm a little uncertain as to how to use the rank to my advantage.

A small hint would be appreciated. :)

Thanks!

 

[lanedance]

assume you had a vector in both the range and nullspace of T, what would happen when operated on again by T?