Proving the image of a linear map with kernel =$ \{0\}$ Question from Serge Lang's Introduction to Linear Algebra:
Let $F: V \to W$ be a linear map, whose kernel is $\{0\}$. Assume that $V$ and $W$ have both the same dimension $n$. Show that the image of $F$ is all of $W$.

Thoughts & attempt:


*

*We can express any element $w \in W$ as $\sum_{i}^n c_iw_i$, where $w_1,...,w_n$ are linearly independent.


I'm trying to reach a point where I can coherently say $\sum_i^nF(c_iv_i) = \sum_{i}^n c_iw_i$ by assuming that $v_1,..., v_n$ form a basis of $V$. 
Is it enough to say that since the ker = $\{0\}$, linear independence is preserved, and therefore $W$ also has $n$ dimension?
EDIT: I meant to consider $c_i$ on each side of the equation as not necessarily equivalent scalars. 

Let $\{v_1,\ldots,v_n\}$ be a basis for $V$. We claim that $\{T(v_1),\ldots,T(v_n)\}$, is a linearly independent set in $W$ and since $\dim W=n$ it is also a basis for $W$. Indeed suppose that
$$
c_1T(v_1)+c_1T(v_2)+\dotsb+c_nT(v_n)=0
$$
Use linearity, the fact that the kernel is trivial and the fact that the $v_i$ are linearly independent to deduce that $c_i=0$ for all $i$. It follows that $W=\text{im} T$
