Hey Stormey.

For your problem can you show that if u is non-zero (in general) then since <u,v> = 0 implies u = 0 or v = 0, then if v is not necessarily zero then u = 0.

if u = Tx and x is not necessarily zero, then you need to show T = 0 but for all x.

So the proof now becomes showing that T = 0 satisfies Tx = 0 for all valid vectors in x. You could try proof by contradiction by assuming that T is non-zero and find a value for some realization of x' where Tx = 0 but Tx' != 0 for x' != x. This will prove the statement that T must be zero in order for Tx = 0 for all valid x in the vector space.