When a linearly independent set is taken to a linearly independent set on the quotient space?

Question

It is known that, if a set $\{v_1 + W, \dots, v_n + W \}$ is linearly independent in a quotient space $V/W$, then the set $\{v_1, \dots, v_n \}$ is linearly independent. I wonder under what condition the converse is true: maybe would be enough to demand that every equivalence class $v + W$ is disjoint, so that you can't mix elements of different classes and therefore the set is linearly independent, but it seems that I'm missing something. Is there a condition for that to be true?

Answer 1

As I said in my comment the set of $\{v_1+W,\dots,v_n+W\}$ is linearly independent iff for any $w\in W$ the set $\{v_1,\dots,v_n,w\}$ is linearly independent, this you can show by putting all $w_i$ from the $\bar{v_i}=v_i+w_i$ together in the definition of linear dependence.

Assuming thaat the $v_i$'s are independent, the relation for linear dependence is $$\lambda_1v_1+\cdots+\lambda_nv_n-w=0\iff \lambda_1v_1+\cdots+\lambda_nv_n=w,$$ for $w\neq 0$.

So the above holds for non-trivial $w$ iff $w$ lies in the linear span of $v_1,\dots,v_n,$ which can be set-theoretically rephrased as $$ \langle v_1,\dots,v_n\rangle \cap W= \{0\}\iff \langle v_1,\dots,v_n\rangle \subseteq W^c $$