Understanding linearly independent vectors modulo $W$ We've learned in class:  

Let $W \subseteq V$, a subspace. $v_1, \ldots, v_k \in V$ are said to be linearly independent modulo $W$ if for all $\alpha_1, \ldots, \alpha_k: \sum_{k=1}^n \alpha_i v_i \in W \implies \alpha_1 = \ldots = \alpha_k = 0$

Can you explain it to me / give some intuitions or an example?
Thanks. 
 A: Consider the quotient space $V/W.$ Then the given condition says that the image of the vectors $v_1, v_2, \cdots , v_n \in V,$ under the natural map $V \rightarrow V/W$ is also linearly ondependent.
A: Example
Let $V=\Bbb R[x]$ the vector space of the real polynomials and $W=(x^3)=\{x^3P(x)\mid P(x)\in V\}$. Then using the Euclidean division we see that $(1,x,x^2)$ is a basis for $V/W$ where we mean by $1,x,x^2$ the cosets $1+W,x+W,x^2+W$ respectively.
A: As an example, consider $V=\mathbb R^3$ - Euclidean space with coordinates $(x, y, z)$. Let $W$ be the line $x=y=0$ ($Z$ axis). Then you can prove by definition that the set of vectors $(x_i, y_i, z_i), i=1,\ldots,k$ is linearly independent modulo $W$ if and only if the set of 2-vectors $(x_i, y_i), i=1,\ldots,k$ is linearly independent. In other words, their projections on $xy$ plane are linearly independent.
More generally, we can always select a subspace $W'$ such that $V=W\oplus W'$. (If you have dot product defined on your space, you can take $W'=$ orthogonal complement of $W$). Then linear independence modulo $W$ is equivalent to linear independence of projections on $W'$. This geometric transformation (projection) is equivalent to the natural map onto a factor-space mentioned in other answers.
