🌐Span

Cartwright Vela sets two poles in the ground at Basis Camp and refuses to give directions any other way. Every field on the isle, she says, is "so many of this one and so many of that one" — and if a field cannot be named that way, then as far as her two poles are concerned it does not exist.

That is what a span is. Given a set of vectors, the span of the set is every vector you can build out of them using linear combinations: pick a weight for each vector, add up the results, and whatever comes out is in the span.

Span{u, v} = every vector of the form c1*u + c2*v, for any numbers c1 and c2

Three things follow straight away, and they are worth saying out loud because every question in this craft leans on one of them.

The origin is always in the span. Take every weight to be 0 and the sum is the zero vector. So a span never misses the origin — which is why "2x + 3y - z = 4" can never describe a span, however much it looks like a plane.

A span is closed. Add two things in the span and you are still in the span; scale something in the span and you are still in the span. Nothing you do with combinations can take you outside it.

A span has no edges. The weights are any numbers at all, positive, negative or zero, so a span runs forever in both directions along every one of its vectors. Span{u} is a whole line, not a segment from the origin to u.

Here is the surprise: a span is never a complicated object. Whatever vectors you start from, in R³ there are exactly four possible answers.

• ◆Just the origin. Only happens when every vector in the set is the zero vector.
• ◆A line through the origin. One independent direction. Span{u} is always this, for any non-zero u.
• ◆A plane through the origin. Two independent directions.
• ◆The whole of R³. Three independent directions.

The words that matter there are independent directions, and they are not the same thing as the number of vectors. Ten vectors that are all multiples of one another still carry only one direction, and their span is still a line.

So counting is the whole job, and row reduction is how you count. Set the vectors out as the rows of a matrix, reduce it, and count the pivots. That count — the rank — is the number of independent directions, and the shape follows from it:

0 pivots -> the origin 1 pivot -> a line 2 pivots -> a plane 3 pivots -> all of R³

The same table runs in any dimension: k pivots gives a k-dimensional flat piece through the origin, and when k reaches n the span is all of Rⁿ.

A worked feel for it: u = [1, 2, 3] and v = [2, 4, 6] look different, but v = 2u, so reduction leaves one pivot and Span{u, v} is a line. Change v to [2, 4, 7] and reduction leaves two pivots, so the same question now answers "a plane".

"Is b in Span{a1, a2}?" is a question with a completely mechanical answer, and it is the same question as "does x1*a1 + x2*a2 = b have a solution?"

Written out slot by slot, that vector equation is an ordinary system of equations with x1 and x2 as the unknowns, and the way to settle a system is to reduce it. Build the augmented matrix with a1 and a2 as columns and b on the right, and reduce.

consistent -> weights exist -> b IS in the span a row reading 0 = a non-zero number -> no weights exist -> b is NOT in the span

Nothing else in the reduction matters. In particular, whether the weights are unique is a different question altogether — it is about independence, not about membership.

For the span of a single vector there is a shortcut, because Span{u} is just the multiples of u. Divide entry by entry and see whether every slot gives the same number:

u = [4, -2, 1], v = [16, -8, 4]: 16/4 = 4, -8/-2 = 4, 4/1 = 4 -> v = 4u, so yes u = [4, -2, 1], v = [16, -9, 4]: the middle slot gives 4.5, not 4 -> no

The trap is checking one slot and stopping. A slot that agrees proves nothing on its own; only a slot that disagrees settles anything by itself.

The last kind of question turns the problem round. Instead of one target b, it asks about every b at once: do the columns of A span R^m?

Because "b is in the span" means "Ax = b is solvable", spanning all of R^m means Ax = b is solvable for every b. And that happens exactly when reduction of A leaves a pivot in every row. A row with no pivot becomes a row of zeros, and some choice of b then puts a non-zero number on the right of it, which no x can ever satisfy.

pivot in every row -> the columns span R^m a row with no pivot -> they do not, and the span is a smaller flat piece inside R^m

Notice which way round that is. Spanning is about rows; independence, in the next craft, is about columns. Mixing them up is the single most common slip in this part of the course.

One consequence worth carrying: spanning Rⁿ needs at least n vectors, because n pivots need n columns to sit in. More than n is allowed — the extra ones simply repeat directions already there — but fewer than n can never be enough, however clever the numbers look.

Finally, a plane through the origin can be described two ways, and both are useful. As a span it is "everything you can build from u and v". As an equation it is ax + by + cz = 0, where (a, b, c) is a direction perpendicular to both u and v — found by solving n·u = 0 and n·v = 0. The right-hand side is 0 and never anything else, because the origin has to be on it.