Scale a few fixed vectors, add the results, and you reach a new vector. Everything reachable that way from one fixed list is called its span, and it is usually far larger than the list.

With coins of two values you can pay a whole range of amounts. Which amounts you can reach depends on the two values, not on how many coins you happen to hold.

A linear combination weights each vector of a fixed list by a scalar and adds the results. The scalars may be any real numbers at all, negative and zero included. The span of the list is the set of every vector that some choice of scalars produces.

• Write $V=\operatorname{span}\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ for the span of the list.
• Taking every $c_i=0$ shows that $\mathbf{0}$ belongs to the span of any list, so a span is never empty.
• Each $\mathbf{v}_i$ belongs to its own span: take $c_i=1$ and all the other scalars zero.
• The span of one non-zero vector $\mathbf{v}$ of $\mathbb{R}^2$ is the line of all its multiples, not the whole of $\mathbb{R}^2$.
• Requiring the scalars to add up to 1 defines a different and smaller object, an affine combination. A linear combination places no such condition.

Each dot is $a\mathbf{v}+b\mathbf{w}$ for some pair of whole numbers, and they are only a sample: the scalars may be any real numbers at all. Given that freedom this pair reaches every point of the plane, which is why a span is usually far larger than the list that makes it.

Membership of the span means that some scalars build the vector, so name the scalars that build each of $\mathbf{x}$ and $\mathbf{y}$. Adding the two expressions and gathering the terms belonging to each $\mathbf{v}_i$ leaves an expression of exactly the same shape, with the sums of the scalars as its weights.

Being in the span means some scalars build the vector. Name the ones that build $\mathbf{x}$.

Do the same for $\mathbf{y}$, with its own scalars, over the very same list of vectors.

This is $\mathbf{x}+\mathbf{y}$ with the terms belonging to each $\mathbf{v}_i$ gathered together, which addition of vectors permits.

Each bracket collapses to a single term by the rule for scaling by a sum.

The result is a linear combination of the same list, so it lies in the span.

The same scalar 3 works in both positions, so $(9,6)$ is a combination of $(3,2)$.

A negative fraction and a positive whole number are both legitimate weights.

Every reachable vector has its two components equal, which is a line and not a plane.

Whatever a and b are, the result is a multiple of $(1,1)$.