📐Basis, Dimension & Coordinates

Cartwright Vela describes the whole isle from two poles driven into the stone: so far along this one, so far along that one. Every field has an address, and the address is a pair of numbers.

A set B of vectors is a basis for a space when it does two jobs at once.

• ◆It spans. Every vector of the space can be written as a combination of the vectors in B, so nowhere is out of reach.
• ◆It is independent. No vector in B is a combination of the others, so nothing in the set is wasted.

Those two demands pull in opposite directions. Adding more vectors makes spanning easier and independence harder; removing them does the reverse. A basis is exactly where the two meet — the smallest set that still reaches everywhere, and equally the largest set that is still free of repetition.

That tug-of-war has a consequence worth knowing before you calculate anything. In Rⁿ, a basis always has exactly n vectors. Fewer than n can never span, and more than n can never be independent. So a set of the wrong size is settled by counting alone, with no arithmetic at all:

• ◆fewer than n vectors: too few, it cannot reach everywhere;
• ◆more than n vectors: too many, one of them must repeat a direction;
• ◆exactly n vectors: count is right, and now you have to check independence.

For that last case, put the vectors in as the rows of a matrix and reduce. If every row survives — n non-zero rows — the set is independent and therefore a basis. If a row reduces to zero, one vector was a combination of the others and the set is not a basis, however many vectors there are.

A space has many bases. In R² you could use { [1, 0], [0, 1] }, or { [2, 1], [-1, 1] }, or a thousand other pairs. What every single one of them shares is how many vectors it holds, and that shared number is the dimension.

This is the quiet theorem the whole topic rests on: any two bases of the same space have the same number of vectors. Because of it, "dimension" is a property of the space, not of the person who chose the basis.

To find the dimension of a span, list the vectors as rows, reduce, and count the rows that survive. The vectors that reduce away were already reachable from the others, so they add nothing:

span of [1, 2], [2, 4], [0, 1] -> [2, 4] is 2 times [1, 2], so it adds nothing -> two directions survive, dimension 2

Note that the answer is not the number of vectors written down, and it is not the dimension of the space they live in. Three vectors in R⁴ can span something of dimension 3, 2 or 1, depending entirely on how much they repeat one another.

To find the dimension of a subspace given by a homogeneous system, count the unknowns and take one away for each genuinely different equation. One equation in R³ leaves a plane, of dimension 2. Two real equations leave a line, of dimension 1.

For the null space of a matrix A — every x with Ax = 0 — the same count is usually phrased through pivots. Reduce A, count the pivot columns, and every column without a pivot belongs to a free unknown. A basis for the null space carries exactly one vector for each free unknown, so:

number of vectors in a basis for the null space = (columns of A) - (pivots of A)

Three phrasings, one idea: the dimension counts the directions that are still free once every condition has had its say.

Once a basis B = { b1, b2 } is fixed, every vector v can be written in exactly one way as

v = c1·b1 + c2·b2

and that pair of weights is the coordinate vector of v relative to B, written [v]_B = (c1, c2). Uniqueness is what makes it an address rather than a guess: if two different pairs of weights gave the same v, subtracting them would produce a dependence relation, and a basis has none.

The ordinary way of writing a vector is itself a coordinate vector — it is the address relative to the standard basis { [1, 0], [0, 1] }, and it looks like the vector because [3, -2] really is 3 of the first standard vector and -2 of the second. Relative to any other basis, the numbers change while the vector itself sits exactly where it was. The field has not moved; Vela has moved her poles.

Finding coordinates means solving. Write v = c1·b1 + c2·b2, compare slots, and you have a small system in c1 and c2. For B = { b1 = [2, 1], b2 = [-1, 1] } and v = [-7, -2]:

slot 1: 2c1 - c2 = -7 slot 2: c1 + c2 = -2

Adding the two equations gives 3c1 = -9, so c1 = -3, and then c2 = 1. So [v]_B = (-3, 1). Check it: -3[2, 1] + 1[-1, 1] = [-6 - 1, -3 + 1] = [-7, -2].

Going back is the easy direction, with no solving at all. Given [v]_B = (4, 2) for the same basis, just build the combination: 4[2, 1] + 2[-1, 1] = [8 - 2, 4 + 2] = [6, 6].

Two habits are worth forming. Keep the order of the basis, because the first weight always belongs to the first basis vector, and (3, -1) is a different address from (-1, 3). And check by rebuilding: it costs one line and catches almost every slip.

Building v out of its coordinates is a matrix product waiting to happen. Put the basis vectors in as the columns of a matrix P — first vector, first column — and then

v = P·[v]_B

For B = { [2, 1], [-1, 1] }, that matrix is P = [[2, -1], [1, 1]], and P·(4, 2) = [6, 6], which is the same arithmetic as before, packed up. P is called the change-of-coordinates matrix for B; it takes an address in Vela's poles and hands back the ordinary one.

The single most common error here is laying the basis vectors in as rows. They are columns, and the whole formula depends on it: reading the entry in row 1, column 2 of P means looking at the second basis vector and taking its first entry, not the other way round.

Running the map backwards means inverting it:

[v]_B = P^(-1)·v

which is why finding coordinates was the harder direction. For a 2 by 2 matrix the inverse has a formula: swap the two entries on the main diagonal, change the sign of the other two, and divide everything by the determinant.

P = [[a, b], [c, d]] -> P^-1 = (1 / (ad - bc))·[[d, -b], [-c, a]]

For P = [[3, -1], [1, 2]] the determinant is (3)(2) - (-1)(1) = 7, the swapped-and-negated matrix is [[2, 1], [-1, 3]], and the inverse is that with every entry over 7. Its entry in row 1, column 2 is 1/7. Entries of an inverse are often fractions, and they should be left exact rather than rounded.

P is invertible precisely because B is a basis: the columns are independent, so the determinant is not 0. If someone hands you a set whose matrix has determinant 0, you have been handed a dependent set, and it was never a basis in the first place.