# Intuition for “linearly independent mod $W_1$”

Definition. If $W_1 \subseteq W_2$ are subspaces of vector space $V$, then we say that $v_1, \ldots, v_m \in W_2$ are linearly indepdent mod $W_1$ if $$a_1 v_1 + \cdots a_m v_m \in W_1 \text{ implies } a_i=0, \forall i$$

I'm having some trouble getting the intuition for this definition. I know that one way to think of it is "linear independence, but with $W_1$ as the zero vector," but this still isn't very clear for me.

I tried doing a toy example with $W_2 =\mathbb{R}^3$ and the $x$-, $y$-, and $z$-axes, where $W_1$ is the $x$-axis. I think that any two vectors will be linearly independent mod $W_1$ iff their projections onto the $yz$-plane are linearly independent in that subspace. (Can someone confirm this?)

In fact, it seems we can replace the $yz$-plane with any subspace $W'_1$ such that $W_2 = W_1 \oplus W'_1$ (which can be found by "extending" a basis for $W_1$ to one for $W_2$).

So in general, my "intuition" for linear independence mod $W_1$ is that the projections of $v_1, \ldots, v_m$ onto $W'_1$ are linearly independent. Does this hold?

Thanks in advance!

• Yes, you are right. – Boris Novikov Aug 28 '13 at 6:48

## 1 Answer

You are right. But a better statement would be that linear independence mod $W_1$ means that the (canonical) images of the vectors in the quotient space $W_2/W_1$ are linearly independent.

Any complementary subspace $W_1'$ projects isomorphically ot the quotient $W_2/W_1$, which explains why it makes no difference whether projecting to $W_1'$ (parallel to$~W_1$) or to $W_2/W_1$. But the quotient space is defined without needing the choice of a complementary subspace.