$T(v_1), \ldots,T(v_k)$ are independent if and only if $\operatorname{span}(v_1,\ldots,v_k)\cap \ker(T)=\{0\}$ I need help with this:

If $T:V\rightarrow W$ is a linear transformation and $\{v_1,v_2,\ldots,v_k\}$ is a linearly independent set in $V$, prove that $T(v_1), T(v_2),\ldots,T(v_k)$  are independent in $W$ if and only if 
  $$\operatorname{span}(v_1,\ldots,v_k)\cap \ker(T)=\{0\}.$$

I know how to prove that a independent set transforms to a independent set, but how do I show that this intersection equals the zero vector?

Let $v=c_1v_1+\cdots+c_nv_n\in\mathrm{ker}~T$, noting that also $v\in\mathrm{span}~\{v_1,\ldots,v_n\}$, then $$T(v)=T(c_1v_1+\cdots+c_nv_n)=c_1T(v_1)+\cdots+c_nT(v_n)=0$$
implies $c_1=\cdots=c_n=0$ since $T(v_1),\ldots,T(v_n)$ are linear independent. It follows that $\mathrm{span}~\{v_1,\ldots,v_n\}\bigcap\mathrm{ker}~T=\{0\}$. The other side is immediately obtained from the equation.
