Given that $f:V\rightarrow W$, prove that $v_1,..,v_n$ are linearly independent Let $V , W$ be two vector spaces and $f : V \rightarrow W$ a linear map. Let $w_1,...,w_n$ be elements of $W$ that are
linearly independent and let $v_1,...,v_n$ be elements of V such that $f(v_i) = w_i$ for $i=1,...,n.$ Prove that
$v_1,...,v_n$ are linearly independent.
This is the actual proof, but i'm not sure if its complete:
I will prove this by contradiction.
Suppose $V_a,V_b\in V $ such that they are linearly dependent vectors.
then $F(V_a)=W_b$
$V_b=\alpha V_a$
then:
$f(V_b)=f(\alpha V_a)=\alpha f(V_a)=\alpha W_a$
Since $W$ has independent  vectors, this results as False. So by contradiction we know that $V$ has independent vectors
 A: You are on the right track, but you have to generalize from two vectors to $n$. But the same idea can be used: Suppose, that $v_1, \ldots, v_n$ are linearly dependent, then for some $\lambda_i\in k$ ($k$ the ground field), not all $\lambda_i = 0$, we have $\sum_i \lambda_i v_i = 0$. Applying $f$, we have by linearity
$$ 0 = f(0) = f\left(\sum_i \lambda_i v_i\right) = \sum_i \lambda_i f(v_i) = \sum_i \lambda_i w_i $$
As not all $\lambda_i$ are 0, this contradicts the idependence of the $w_i$.
A: $f(v_i)=w_i, \;i=1,...,n.$
Choose any $v=\sum_{i=1}^{n} \alpha_i v_i \in{V}, f(v)=\sum_{i=1}^{n} \alpha_i w_i \in{W}.$
Now, $ f(v)=0 \;(v \in{ker(f)})\;$ if and only if $\alpha_i=0 \;\forall{1 \le i \le n}. \;$ But $ v=\sum_{i=1}^{n} \alpha_i v_i=0,$ hence $ker(f)=\{0\},$ and $v=0$ if and only if $\alpha_i=0 \;\forall{1 \le i \le n}.$ Therefore, $i=1,...,n, v_i$ are linearly independent.
A: Consider $T(a_1v_1+a_2v_2....+a_nv_n)=0$ Then by linearity you get $a_1w_1+.....+a_nw_n=0$. As $w_1,....,w_n$ are linearly independent, you get $a_1=a_2....=a_n=0$
