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:


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*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. 

What you are really trying to show is the following:
If $T:V\rightarrow W$ is an injective linear map, where $V$ and $W$ have the same dimension, then $T$ is surjective.
By the rank nullity theorem, $dimimT=dimW$. As $imT$ is a subspace of $W$, the result follows.
