Two arrows are singled out: one step to the right, one step up. They are written $\hat{\imath}$ and $\hat{\jmath}$, and they are the whole of what the coordinate system is.

So $\begin{bmatrix}3\\-2\end{bmatrix}$ is an instruction: stretch $\hat{\imath}$ by three, flip $\hat{\jmath}$ and stretch it by two, then add. The green arrow is what arrives.

Nothing obliges the two arrows to be the unit ones. Scale this pair by three and by $-2$ instead and the same two numbers name a different place. The plane has not changed; what the numbers were told to scale has.

Scaling two vectors and adding the results is a linear combination of them, and the two numbers doing the scaling are its scalars. The pair being scaled is a basis, and a vector’s coordinates are exactly the scalars that basis needs in order to reach it. Everything reachable this way — every place a free choice of two scalars can put you — is the span of the pair. The word linear is there because scaling one vector on its own traces a straight line through the origin, so a combination of two is a way of joining two lines.

• Every real number is allowed as a scalar, negatives and fractions included. Allowing only whole numbers would leave a lattice of separated dots rather than a line or a plane.
• Two arrows that are not parallel span the whole plane, and each vector is reached by exactly one pair of scalars. That is the theorem below.
• Two arrows that are parallel span one line through the origin. The second reaches nothing the first could not, because scaling it is scaling the first.
• If both arrows are $\mathbf{0}$ the span is the origin alone, since every combination of them is $\mathbf{0}$.
• In three dimensions two non-parallel arrows span a flat sheet through the origin. A third arrow lying on that sheet leaves the span exactly as it was; a third arrow off it sweeps the sheet through the whole of space.
• A list from which one vector can be dropped without shrinking the span is called linearly dependent, and the vector that can go is already a combination of the others. When every vector enlarges the span, the list is linearly independent.
• Once a whole collection is in view the arrows cover each other, so each vector is drawn as the point at its tip. A span is then a line, a sheet, or all the points there are.

One combination of $\mathbf{v}$ and $\mathbf{w}$, with $a=2$ and $b=1$. Turning the two scalars freely moves the green tip, and everywhere it can be sent is the span.

Here $\mathbf{w}=-2\mathbf{v}$, so every combination $a\mathbf{v}+b\mathbf{w}$ is a multiple of $\mathbf{v}$ and lands on the dashed line. The pair is linearly dependent: drop either one and the span is the same.

The multiples of $\mathbf{v}$, drawn as the points at their tips. A single vector is an arrow; a collection of them is easier to read once the arrows are left out.

Take any vector $\mathbf{u}$ at all. From its tip, walk backwards along $\mathbf{w}$; letting the length of that walk vary traces a whole line parallel to $\mathbf{w}$. The multiples of $\mathbf{v}$ trace another line, and the two are not parallel, so they cross exactly once. The walk that reaches the crossing fixes $b$, what is left over is a multiple of $\mathbf{v}$ and fixes $a$, and putting the two back together is $\mathbf{u}$.

Scaling $\mathbf{v}$ on its own sweeps out a line through the origin. Call it $L$.

From the tip of $\mathbf{u}$, walk backwards along $\mathbf{w}$. Letting $b$ vary traces a line parallel to $\mathbf{w}$.

That line and $L$ are not parallel, because $\mathbf{v}$ and $\mathbf{w}$ are not, so they meet at exactly one place. Meeting it is this condition, and it fixes a single $b$.

Being on $L$ means what is left after the walk is some multiple of $\mathbf{v}$, and that fixes a single $a$.

Add the walk back. Every $\mathbf{u}$ in the plane has been written as a combination, so the span is the whole plane.

Four steps in the direction of $\hat{\imath}$, then three in the direction opposite to $\hat{\jmath}$.

The scalars $3$ and $-2$ against this basis reach a different vector than they would against the unit arrows.

Whatever the two scalars are, the result is one number times the first arrow.

When this can be solved, $\mathbf{u}$ is already on the sheet and adds nothing to the span.