🧩Linear Combinations

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". That sentence is the whole craft.

A linear combination of the vectors u, v, w is anything you can build by scaling each one and adding the results:

c1*u + c2*v + c3*w

The numbers c1, c2, c3 are the weights (also called coefficients or scalars). They may be positive, negative, or zero, and each one belongs to exactly one vector.

Working one out is two jobs in order — stretch, then add:

u = [1, 3], v = [4, 1] 3u - 2v = 3[1, 3] - 2[4, 1] = [3, 9] - [8, 2] = [-5, 7]

Nothing here is new; it is the arithmetic of the last craft used deliberately. What is new is the question you ask with it. Forwards, you are given the weights and asked for the vector, and that is arithmetic. Backwards, you are given the vector and asked for the weights, and that is a whole subject.

Almost every later idea in this kingdom is a question about combinations wearing different clothes. A span is the set of all of them. Independence asks which combinations can reach the zero vector. A basis asks for a set that reaches everything exactly once. Getting comfortable here pays for all of it.

Here is the question that matters: given u, v and a target b, which weights make c1u + c2v = b?

The trick is to stop looking at it as a vector equation and read it one slot at a time. A vector equation in R² says two things at once, and each of them is an ordinary equation.

u = [2, 3], v = [3, 1], b = [11, 6]

c1[2, 3] + c2[3, 1] = [11, 6]

first entries: 2c1 + 3c2 = 11 second entries: 3c1 + c2 = 6

That is a system of two equations in two unknowns, and you already know how to solve one. From the second equation, c2 = 6 - 3c1. Substituting: 2c1 + 18 - 9c1 = 11, so -7c1 = -7 and c1 = 1, which gives c2 = 3.

Always check by putting the weights back. 1[2, 3] + 3[3, 1] = [2, 3] + [9, 3] = [11, 6]. It holds.

The same reading works in any dimension, and the count of equations comes from the length of the vectors, while the count of unknowns comes from how many vectors there are. Three vectors in R³ give three equations in three unknowns. Two vectors in R³ give three equations in only two unknowns — and then the third equation is a genuine condition, not a formality: it may or may not agree, and whether it agrees is the whole question of the next section.

Two habits save most of the marks here. Keep the weights in the order the vectors are listed, because c1 belongs to the first vector and swapping them gives a different answer to a different question. And when a system has more equations than unknowns, solve with two of them and use the rest as a check, never as an afterthought.

Not every target can be reached. Ask whether b is a combination of u and v at all, and there are exactly three possible replies — and the third one surprises people.

Exactly one choice of weights. The usual case: the system is consistent and the vectors pull in genuinely different directions, so there is one and only one way to build b.

No choice at all. The system is inconsistent. Somewhere in the reduction a row collapses to something like 0 = 1, which no weights can fix. Two vectors in R³ only ever reach the flat plane through the origin that contains them; a target off that plane is simply not reachable, however clever the weights.

Infinitely many choices. This happens when the vectors are not all pulling in different directions — when one of them is already a combination of the others. If v = 2u, then u and v reach only the one line through the origin along u, and every point of that line can be written in endlessly many ways: 5u + 0v, 3u + 1v, 1u + 2v, and so on for ever.

A system of linear equations has one solution, none, or infinitely many. It never has exactly two. If you ever find two different sets of weights that both work, you have automatically found infinitely many, because the difference between them can be added on in any amount.

The way to decide is always the same: build the augmented matrix, with the vectors as columns and the target as the last column, and reduce it.

c1[5, 3] + c2[2, 4] = [24, 20] becomes [[5, 2 | 24], [3, 4 | 20]]

Then read the pivots. A pivot in the last column means an impossible row, so no answer. A weight column with no pivot means that weight is free to be anything, so infinitely many. A pivot in every weight column and none in the last means exactly one.

This section is short and it is the most important one in the craft.

Write the vectors u and v as the columns of a matrix A, and write the weights as the entries of a vector x. Then the combination c1u + c2v is exactly what everyone calls the matrix-vector product Ax.

A = [[1, 4], [3, 1]] so the columns are [1, 3] and [4, 1] x = [2, 5]

Ax = 2*[1, 3] + 5*[4, 1] = [2, 6] + [20, 5] = [22, 11]

Ax is the combination of the columns of A, weighted by the entries of x. Read it that way and a great deal of what follows becomes obvious rather than memorised. The product only makes sense when A has as many columns as x has entries, because every column needs its own weight — and the answer has as many entries as A has rows, because that is how long the columns are.

The classic error is to read across the rows instead of down the columns. A column runs down the matrix, taking one entry from each row, and reading it the wrong way computes a different product entirely.

Once you can see it, the two big questions of this craft become one question:

• ◆Ax = ? with x given is the forwards question: work out the combination.
• ◆Ax = b with b given is the backwards question: find the weights. It is the same system of equations as before, now written in three symbols.

And that is what a word problem is too. Two seams of ore, each yielding so much iron and so much silver per crate; an order calling for exact totals. Make each seam a vector of what one crate yields, make the order the target b, and the number of crates from each seam are the weights. Mixtures, feeds, rations, portfolios, dyes — they are all Ax = b once the columns are named.