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For a linear transformation T: V->V

If Rank(T)=Rank(T^2), then prove that Range(T) U Null(T) = {0}.

I don't know how to get started, I tried initiating a variable x=a_1*x_1 + ...+ a_n*x_n as a linear combination of a basis of V, that is an element of both the range and the null space, but am stuck after that.

I have the idea that using the dimension theorem (rank nullity theorem), The equation Rank(T^2) + Nullity(T) = Dim(V) will be used in a suitable contradiction.

I am confused, as I thought as the basis for the range and null space are always disjoint (how the rank nullity theorem is proved), then there's no way Range(T)U Null(T) != {0}.

This problem made my question my understanding of linear transformations and their ranges and null spaces...