How to prove that these 2 statements regarding Linear Transformations are equivalent? Let $V$ and $W$ be vector spaces, and $T:V\to W$ a linear transformation. 
Prove that the following statements are equivalent : 
$1)$ T is injective (one-to-one)
$2)$ If S is a linearly independent subset of V, then its image, $T(S) = \{ T(x) | x ∈ S \}$ , is a linearly independent subset of $W.$
I know that to prove equivalence I have to prove that both $1$ implies $2$ and $2$ implies $1.$ I think I can prove that $1$ implies $2,$ but I'm having trouble proving the converse.
To prove $1$ implies $2:$ Let $S= \{x_{1},...,x_{n} \}$, and assume $c_1T(x_1)+...+c_nT(x_n) = 0$. Since T is linear, $T(c_1x_1+...+c_nx_n) = T(0),$ and since it's one-to-one, $c_1x_1+...+c_nx_n = 0$. Since $S$ is linearly independent, $c_1 = ... = c_n = 0,$ therefore $T(S)$ is linearly independent as well.
Is this a good proof? and how would you prove that $2$ implies $1?$
 A: Hint
For the converse statement, pick a basis $\{v_1, \ldots, v_n\}$ of $V$ (wich is a linearly independent subset) and prove that any vector $w$ in the range of $T$ can be uniquely decomposed into 
$$w = \sum_{i=1}^n \lambda_i T(v_i)$$
A: Your (1) implies (2) is correct. For (2) implies (1) you may want to go with proof by contradiction, i.e. you assume that (2) holds but (1) doesn't.
So assume that every independent set maps to an independent set but $T$ is not one-one, this means there exists a non-zero vector $u \in V$ such that $T(u)=T(0)=0$. 
Then consider the independent set $S=\{u\}$. Now $T(S)=\{0\}$ which is NOT a linearly independent set. Hence it contradicts the hypothesis. So $T$ must be one-one.
A: I would prefer a proof by contraposition: if $f$ is not injective, there exists a linearly independent subset $S$ such that its image is linearly dependent. 
Indeed , if $f$ is not injective, its kernel is not zero. Pick a non zero  $x$ in $\ker T$; then $T(x)=0$, which is linearly dependent!
