$T:V \rightarrow V$ And $U \cap Ker(T)={0}$ prove that if$ (u_1,..u_n)$ linear Independent so does $T(u_1)...T(u_n)$ 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?
 A: Let $\alpha_1,\ldots\alpha_n\in \Bbb R$ s.t.
$$\sum_{k=1}^n \alpha_k T(u_k)=0$$
and we have
$$0=\sum_{k=1}^n \alpha_k T(u_k)=\sum_{k=1}^n  T(\alpha_ku_k)=T\left(\underbrace{\sum_{k=1}^n  \alpha_ku_k}_{\in U}\right)$$
so
$$\sum_{k=1}^n  \alpha_ku_k\in U\cap \ker(T)=\{0\}$$
but since $(u_k)$ are linearly independant then $\alpha_k=0$ for all $k$ hence $(T(u_k))$ is also linearly independant.
A: You cannot deduce that $u_i$ form a basis of $U$ because $U$ may have a dimension bigger than $n$. Also, you don't need a basis of $V$. 
I'd try to proceed by contradiction:
By contradiction assume that $Tu_i$ are not linearly independent. Then there are non-zero  $a_i$ such that $\sum a_i Tu_i = 0$. But then $\sum a_i u_i \in \mathrm{ker}(T)$ which yields the desired contradiction.
A: Because T(a(1)u(1) + ...+ a(n)u(n)) = 0 ===>  c = a(1)u(1) + ... + a(n)u(n) is in Ker(T) and it is also in U since U is a subspace. So c = 0 and from this you get a(1) = ... = a(n) = 0. So T(u(1)), ..., T(u(n)) are independent.
