There will be $T:V \rightarrow V $ Linear Transformation

U is sub-space of V so that $U \cap Ker(T)={0}$ 

Prove that if $(u\_1,u\_2,...,u\_n)$ are linear independent so does $T(u\_1),T(u\_2),...,T(u\_n)$.

Attempt: assume $(u\_1,u\_2,...,u\_n)$ are linear independent so they are the basis of the sub-space U, we can add $(v\_1,...,v\_k)$ that are linear independent to complete it to a basis for V.
Because T is Linear Transformation if we have a group of vectors that make up V so does the vectors that went trough the transformation, they also build up the space and in our case it is V too.
Therefore $T(u\_1),...,T(u\_n)$ must be linear independent.

But how does $U \cap Ker(T)={0}$ help us?