Proving that, given a surjective linear map, a set is a generator of a vector space

I know how to prove that, given $L: V rightarrow W$ to be an injective linear map and ${v_1, v_2, …, v_n}$ to be a linearly independent set in $V$, ${L(v_1), L(v_2),…,L(v_n)}$ is a linearly independent set in $W$

Proof:

We’re given that ${v_1, v_2, …, v_n}$ is a linearly independent set in $V$. Thus we have (let $a_i in Re$ where $1 leq i leq n$)

$$a_1 v_1 + a_2 v_2 + … + a_n v_n =0$$

Where $v_i in V$

Thus we know that $a_i = 0$

We want to show that ${L(v_1), L(v_2),…,L(v_n)}$ is a linearly independent set in $W$. Let us start off by writing

$$a_1 L(v_1) + a_2 L(v_2) + … + a_n L(v_n) =0$$

Where $L(v_i) in W$

As we’re given $L$ to be linear

$$a_1 L(v_1) + a_2 L(v_2) + … + a_n L(v_n) = L(a_1 v_1 + a_2 v_2 + … + a_n v_n) = 0$$

As we’re given $L$ to be injective (if $f(a) = f(a’) Rightarrow a=a’$) and we know that $0=L(0)$ (which can be easily proven using linearity) we get

$$L(a_1 v_1 + a_2 v_2 + … + a_n v_n) = L(0) Rightarrow a_1 v_1 + a_2 v_2 + … + a_n v_n =0$$

Thus $a_i = 0$ and ${L(v_1), L(v_2),…,L(v_n)}$ is linearly independent in $W$

QED.

OK, I gave the above info because is related to what I’d like to ask.

My issue is how to prove that, given $L: V rightarrow W$ to be a surjective linear map and ${v_1, v_2, …, v_n}$ to be a generator of $V$, ${L(v_1), L(v_2),…,L(v_n)}$ is a generator of $W$

My try:

We’re given that ${v_1, v_2, …, v_n}$ is a generator of $V$. Thus we have (let $v in V$ and $b_i in Re$ where $1 leq i leq n$)

$$V = text{span} (v_1, v_2, …, v_n)$$

And then

$$v = b_1 v_1 + b_2 v_2 + … + b_n v_n$$

But I do not know how to show that

$$L(v) = b_1 L(v_1) + b_2 L(v_2) + … + b_n L(v_n)$$

Based on linearity and surjectivity (where $L(v) in W$).