Linear Independence kept through a Linear Map - Proof Attempt I wanted to check whether my proof was correct and thorough since my book offers a different solution.

Let $V, W$ be vector spaces, and $F: V \to W$ a linear map. Let $w_1, ..., w_n$ be elements of $W$ which are linearly independent, and let $v_1, ..., v_n$ be elements of $V$ such that $F(v_i) = w_i$ for $i = 1,...,n$. Show that $v_1,..., v_n$ are linearly independent.


Attempt:
For two scalars $t, k$ and for any $v_a, v_b, v_c \in V$ and $w_a, w_b, w_c \in W$:
$F(tv_a + kv_b) = tw_a + kw_b$
If $tv_a + tv_b = v_c$, i.e. $v_c$ is a linear combination of $v_a$ and $v_b$,
then we would have a contradiction because $tw_a + kw_b = w_c$, and thus elements of $W$ would be linearly dependent.
Therefore the arbitrary elements of $V$ must be linearly independent.
Thanks!
 A: By contradiction if the $v_i$'s are linearly dependent then there are $t_1,\ldots,t_n$ not all $0$ such that $\sum_{i=1}^n t_iv_i=0$ and if we apply $F$ we find that $\sum_{i=1}^n t_iw_i=0$ which means that the $w_i$'s are linearly dependent and this contradicts the hypothesis.
A: I think you're on the right track, but it's not enough to consider just three elements of $V$.
Instead, suppose that there are scalars $c_1,\dots,c_n$ such that $c_1v_1+\dots+c_nv_n=0$. What happens when you apply $F$ to this equation, and what does this tell you about the $c_i$?
A: It is not enough to consider linear combinations of only two vectors.  For example, the vectors
$$
v_1 = (1,0,0), v_2 = (0,1,0), v_3 = (0,0,1), v_4 = (1,2,3)
$$
are linearly dependent, but there are no scalars or vectors for which $tv_a + kv_b = v_c$
Instead, we use the general definition of linear dependence, namely: there exist scalars $c_i$ such that
$$
c_1 v_1 + c_2v_2 + c_3 v_3 + c_4 v_4 = 0
$$
A: Suppose $\alpha_1v_1+\alpha_2v_2+\dots + \alpha v_n=0$, then $T(\alpha_1v_1+\alpha_2v_2+\dots + \alpha v_n)=\alpha_1w_1+\alpha_2w_2+\dots+ \alpha w_n=0$.
So $\alpha_1=\alpha_2=\dots = \alpha_n=0$, since the $w_i$ are L.I.
