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 $L(v) \in W$).
Best Answer
Let $w\in W$. Since $L$ is surjective, there is some $v\in V$ such that $L(v)=w$. By your observation, we may write $v=b_1v_1+\cdots + b_nv_n$ for some scalars $b_1,\dots, b_n$. Hence $$ w=L(v)=L(b_1v_1+\cdots + b_nv_n)=b_1L(v_1)+\cdots + b_nL(v_n).$$ So $W=\mathrm{span}\{L(v_1),\dots,L(v_n)\}$.