Proving Disjoint Range & Null Space of Linear Operator T

[guroten]

Homework Statement

Given a linear operator T, show that if rank(T^2)=rank(T), then the range and null space are disjoint.

So I know that I can form a the same basis for range(T^2) and range(T), but I'm not sure where to go from there.

 

[HallsofIvy]

Homework Statement

Given a linear operator T, show that if rank(T^2)=rank(T), then the range and null space are disjoint.

So I know that I can form a the same basis for range(T^2) and range(T), but I'm not sure where to go from there.

I recommend you go back and reread the problem. The range and nullspace of any linear operator are subspaces- they both include the 0 vector and so are never disjoint. Perhaps the problem is to show that the only vector in both the range and null space is the 0 vector?

If so try a proof by contradiction. Let v be a non-zero vector in both range and null space of T. Since v is in the range of T there exist u such that v= Tu. Now, what is T²u?

 

[guroten]

You're right, I meant disjoint other than 0. I get that T²u=Tv=0, but what does that say about the rank?