Let T:V->W be a linear transformation with NS(T)={0}. Suppose {u1,u2} is a linearly independent subset of V. Show that {T(u1), T(u2)} is linearly independent.

My attempt:

Suppose {T(u1), T(u2)} is linearly dependent, then there exsists c1, c2 =/ 0 (can't find the "is not equal" symbol, sorry) such that:

c1T(u1) + c2T(u2) = 0

c1u1 + c2u2 = 0 (as T(0)=0 ) **

c1=c2=0, as {u1, u2} are linearly indepedent

Then {T(u1), T(u2)} is also linearly independent as the only solution of c1T(u1) + c2T(u2) = 0 is the trivial solution.

The part I'm wondering if I've done wrong is at the **. Given NS(T)={0}, instead of having c1u1 + c2u2 = 0, I could have c1u1 + c2u2 = x where x V : T(x) =0, couldn't I? Then I haven't proved anything at all...