Linear span

The span of some vectors is the set of all their linear combinations, with any real numbers as coefficients. It always contains the origin O, and in space it is O alone, a line or a plane through O, or all of space.

In the opening view, the glass disc through O is $W$, the span of $\vec v$ and $\vec w$. The crossings of its engraved grid are the combinations with integer coefficients $c_1\vec v + c_2\vec w$, and its rim fades because the plane goes on forever. The fainter discs are the layers $W + k\vec u$ for integers $k$, and the blue disc is the plane $W + s\vec u$ through the target point T. A chain of thin blue arrows, $c_1\vec v$, then $c_2\vec w$, then $s\vec u$, walks from O to T. The red segment from the tip of $\vec u$ down to $W$ has length $|h|$, where $h$ is the signed height of $\vec u$ off the plane.

Why can a third vector add a whole dimension? Every combination of $\vec v$, $\vec w$ and $\vec u$ is a point of $W$ moved by some multiple $s$ of $\vec u$, so the span is the union of the planes $W + s\vec u$ over all real $s$. If $\vec u$ is off $W$, the planes at $s$ and $s + 1$ are $|h|$ apart, and as $s$ runs through the real numbers the planes sweep through all of space.

Now lower $\vec u$ until it is only a hair off the plane. With the default $\vec v = (2, 1, 0)$ and $\vec w = (0, 1, 2)$, $\vec u = (1, 1, 1.01)$ is about 0.004 off $W$, and the span is still all of space. Reaching $(0, 0, 1)$ takes $s = 100$: $(0, 0, 1) = -50\vec v - 50\vec w + 100\vec u$. For a fixed target off $W$, the smaller $|h|$ is, the larger $|s|$ has to be, but there is always an answer. Only when $h$ is exactly 0 does every plane $W + s\vec u$ stay $W$, and the span of $\vec v$, $\vec w$ and $\vec u$ drops back to $W$.

Vectors are linearly dependent if some combination of them with coefficients not all zero is the zero vector. When the vectors are nonzero, the page draws this as a closed loop: multiples of the vectors, laid tip to tail, walk back to O. Equivalently, one of them is a combination of the others. No two of $\vec v$, $\vec w$ and $\vec u = (1, 1, 1)$ are parallel, yet the three are dependent, because $\vec u$ lies in $W$:

A loop also means that a point has many recipes. Adding a nonzero multiple of the loop to a recipe for T changes the coefficients but not the point.

The dimension of the span is the largest number of independent vectors among them.