My solution:
Assume that $F(v_1),F(v_2),…F(v_n)$ are linearly independent, so $α_1F(v_1) + α_2F(v_2) + … + α_nF(v_n)=0$ only and only if $α_1=α_2=…=α_n=0$.
Linear transformation is when:
i) $F(αv)=αF(v)$
ii) $F(v+w)=F(v)+F(w)$
Then $α_1F(v_1) + α_2F(v_2) + … + α_nF(v_n) = F(α_1v_1) + … + F(α_nv_n) = 0$, I used i)
By using ii) we get $F(α_1v_1+ … + α_nv_n)=0$
We know that $0 = F(0)$ so:
$F(α_1v_1 + … + α_nv_n) = F(0) => α_1v_1 + … + α_nv_n = 0$, so $v_1, …, v_n$ are linearly independent
My teacher says it's wrong and I should do it again but I can't find a mistake 🙁
Best Answer
As pointed out in the comments already you made the mistake that you claimed the following: if for any $v \in V$ we have $F(v) =0$ then $v=0$. But this is only true if $F$ is injective, or equivalently if $\ker(F) = \{0\}$. So we cannot use this in this case.
Now we want to prove that given $a_1, \dots, a_n \in F$ (where $F$ is the underlying field of the vector spaces $V$ and $W$) we have that $$a_1v_1+ \dots + a_nv_n=0 \Rightarrow a_1= \dots = a_n=0.$$ (this is the definition of linear independence of the vectors $v_1, \dots, v_n$).
In any reasonable exercise you should somehow use the assumption. So there has to be a point in our proof where we use that $F(v_1), \dots, F(v_n)$ are linearly independent. Let's start:
For the sake of contradiction assume that there is a non-trivial linear combination (non-trivial meaning that at least one coefficient $b_i$ is non-zero) of the $v_i$ that gives zero, i.e. $$b_1v_1+ \dots b_nv_n=0 \text{ and at least one } b_i \ne 0.$$ (We now want to arrive at a contradiction because this assumption would mean that $v_1, \dots, v_n$ is linearly dependent). Now from basic facts on linear operators we know that $F(0)=0$, therefore $$0=F(0) = F(b_1v_1 + \dots b_nv_n) = b_1F(v_1) + \dots b_nF(v_n)$$ by properties 1. and 2. listed by you. But if we compare the left hand side and the right hand side we have a linear combination of the vectors $F(v_1), \dots, F(v_n)$ giving zero, namely $$0=b_1F(v_1) + \dots b_nF(v_n).$$ But remember: at least one $b_i$ is non-zero, so there is a non-trivial linear combination of $F(v_1), \dots, F(v_n)$ which gives zero. This is a contradiction of the assumption that $F(v_1), \dots, F(v_n)$ are linearly independent.
Hence our initial assumption must have been wrong and $v_1, \dots, v_n$ are linearly independent.