Quotient spaces in linear algebra: proof of linear independence Let $V$ be a vector space and $F$ a vector subspace of $E$. Proof that, for $v_1, \dots, v_k \in E$, the classes $v_1 + F, \dots, v_k + F \in E/F$ are linearly independent if, and only if, dim(span($v_1, \dots,v_k$)) $=k$ and span$(v_1, \dots, v_k) \cap F = \{0\}$.
$(\implies)$ I wanted to use the fact that they are linearly independent, thus $\lambda_1 (v_1 + F) + \dots + \lambda_k (v_k + F) = \lambda_1 v_1 + \dots + \lambda_k v_k + F = F$. Therefore, $\lambda_1 v_1 + \dots + \lambda_k v_k \in F$. But now I don't know how to proceed
$(\Longleftarrow)$ I have no clue how to even start.
Can someone help me?
 A: $(\implies)$ First, we need to show that $\dim(\text{span}(v_1,...,v_k))=k$, i.e that the vectors are linearly independent. So suppose $\lambda_1 v_1+...+\lambda_k v_k=0$. Since $0\in F$ this tells that $\lambda_1 v_1+...+\lambda_k v_k\in F$, and so:
$F=\lambda_1 v_1+...+\lambda_k v_k+F=\lambda_1(v_1+F)+...+\lambda_k(v_k+F)$
Since $v_1+F,...,v_k+F$ are linearly independent it follows that $\lambda_1=...=\lambda_k=0$.
Note that during the proof we also showed that $\text{span}(v_1,...,v_k)\cap F=\{0\}$, as we have shown that $\lambda_1 v_1+...+\lambda_k v_k\in F$ implies $\lambda_1=...=\lambda_k=0$.
$(\Longleftarrow)$ Suppose $\lambda_1(v_1+F)+...+\lambda_k(v_k+F)=F$. This means that $\lambda_1 v_1+...+\lambda_k v_k\in F$. Since $\text{span}(v_1,...,v_k)\cap F=\{0\}$ it follows that $\lambda_1 v_1+...+\lambda_k v_k=0$. Since $v_1,...,v_k$ are linearly independent (as $\dim(\text{span}(v_1,...,v_k))=k$) it follows that $\lambda_1=...=\lambda_k=0$. So $v_1+F,...,v_k+F$ are linearly independent, as required.
