Basis and dimension

Intuitive problem

On a city grid every corner has an address of the same shape: so many blocks east, so many blocks north of a reference corner. Three blocks east and four blocks north names one corner and no other. Two directions are enough to name every corner, and each corner gets exactly one pair of numbers. Now add a third street direction, a diagonal avenue, and nothing improves: every corner the diagonal can reach already had an east-and-north address, so the third direction adds no corner the first two could not name, and it creates duplicate addresses for corners that had one. Drop down to a single direction and the failure flips: east alone reaches the corners of one avenue and strands you there. The analogy breaks in two places, and both matter. A grid counts whole blocks from a fixed corner, and its directions meet at a right angle; the spaces this entry works in allow any real weights, and the directions need no right angle between them, only a genuine difference in heading. What survives the break is the question: why exactly two?

Definition

Work in a space of vectors, quantities that can be scaled and added, in the sense of the entry on vectors as directed quantities. A combination of vectors v1 through vk means a sum c1 v1 + ... + ck vk, where each weight ci is a real number. A set of vectors spans the space when every vector in the space is some combination of the set: the set builds everything. A set is linearly independent when it carries no redundancy: no member is a combination of the others, or equivalently, the zero vector has exactly one recipe from the set, the one with every weight zero. A basis is a set that does both jobs at once, spanning with independence. The dimension of the space is the number of vectors in a basis; the derivation below shows that number is forced, the same for every basis the space admits.

Each half can fail alone. In the plane, the three vectors (1, 0), (0, 1), and (1, 1) span, since the first two already do, but they are dependent: (1, 1) is (1, 0) plus (0, 1). The single vector (1, 0) is independent, yet it spans nothing beyond the horizontal line through the origin. Spanning without independence, and independence without spanning.

One demarcation. The entry on coordinates and reference frames establishes the informal act: choosing directions in which a point's numbers are measured. This entry supplies the algebra of that choice, what qualifies a set of directions to serve, and why their count is fixed by the space rather than by the chooser.

Derivation

The load-bearing fact: in a space spanned by n vectors, any set of more than n vectors is dependent. Here n is the size of some spanning set. Take spanning vectors v1 through vn and candidate vectors w1 through wm with m greater than n. Look for weights c1 through cm, not all zero, making c1 w1 + ... + cm wm equal the zero vector. Because the v's span, each wj is a combination of them; substitute those combinations into the sum and collect the total coefficient sitting on each vi. The sum is certainly the zero vector if each of those n collected coefficients is zero, and that requirement is a system of n linear equations in the m unknowns c1 through cm, with zero on every right side. The entry on systems of linear equations supplies the finish: elimination on a system with more unknowns than equations leaves at least one unknown free, so solutions beyond the all-zero one exist. Any such solution is a dependence among the w's. The count alone forces it; nothing about the particular vectors entered the argument.

Two consequences fall out at once.

Equal size of any two bases. Suppose one basis has p vectors and another has q. The first basis spans, and the second is independent, so by the fact above q is at most p. Swap the roles: the second spans and the first is independent, so p is at most q. Hence p equals q. The size of a basis is therefore a property of the space, and dimension is well defined.

Unique coordinates. Let b1 through bn be a basis, and suppose a vector v has two recipes: v = a1 b1 + ... + an bn and also v = a1' b1 + ... + an' bn. Subtracting, (a1 - a1') b1 + ... + (an - an') bn is the zero vector, and independence forces every difference to zero, so the two recipes agree. The weights a1 through an are thus pinned down; call them the coordinates of v in the basis. Coordinates are a name for the vector, and the name depends on the basis chosen.

Change of basis is then nothing more than re-expressing the same vector. Write each vector of a new basis in the old basis and collect those expressions as the columns of a matrix; by the reading in matrices as transformations, a matrix acts by sending coordinates to the corresponding combination of its columns, so this matrix converts a vector's new-basis coordinates into its old-basis coordinates. The vector itself sits still; only its name is translated.

One further consequence, used below: n independent vectors in a space of dimension n span it. If they failed to span, some unreached vector could be appended to give n + 1 independent vectors inside a space spanned by n, which the count fact forbids.

Worked example

Take the plane. One basis is the standard pair e1 = (1, 0) and e2 = (0, 1). Another is b1 = (1, 1) and b2 = (1, -1): neither is a multiple of the other, so the pair is independent, and two independent vectors in a two-dimensional space span it by the derivation's closing consequence.

The vector v = (3, 1) has coordinates 3 and 1 in the standard basis. For its coordinates in the second basis, seek weights a and b with a(1, 1) + b(1, -1) = (3, 1). The two components give a + b = 3 and a - b = 1, so a = 2 and b = 1. Check: 2 times (1, 1) is (2, 2), and adding (1, -1) gives (3, 1). One vector, two names: (3, 1) in the standard basis, (2, 1) in the other, and both names are two numbers long. The change between them is the matrix with columns (1, 1) and (1, -1): fed the coordinates (2, 1), it returns 2 times its first column plus 1 times its second, which is (3, 1).

Limits and boundary conditions

Everything above is finite-dimensional: the counting argument leans on a finite spanning set and on elimination over finitely many equations. Within that scope, a basis is far from unique; the plane has infinitely many, and the worked example used two of them. What the derivation fixes is the size, and only the size. Infinite-dimensional spaces, spaces of functions for instance, need more care: an infinite sum is not a combination in the sense defined here, so spanning has to be reworded before the questions even make sense, and the theory that results is genuinely harder.

Common mistakes

Three recur. Expecting more basis vectors to give a richer description: in a space of dimension n, any additional vector is already a combination of a basis, so it breaks independence and buys duplicate names instead of new reach, the diagonal avenue again. Treating the coordinates as the vector itself: the numbers are a basis-dependent name, and the worked example shows one arrow answering to (3, 1) and to (2, 1) depending on who is asking. Treating any spanning set as a basis: spanning is half the job, and the dependent triple in the definition spans the plane without qualifying.

Build with it

Two candidate sets of plane vectors. Set A: (2, 1) and (4, 2). Set B: (1, 2) and (3, 1). Test each for independence and for spanning, and decide which is a basis, with reasons. Then take the target vector (5, 5): write its coordinates in the qualifying basis and in the standard basis, and state what changed between the two descriptions and what did not. Finally, a colleague proposes (1, 0), (0, 1), (1, 1) as a basis of the plane. Rule the proposal out from the dimension count alone, without examining the particular vectors.

Success looks like this. Set A is rejected because (4, 2) is 2 times (2, 1), a dependence, and a dependent set fails the definition; it also spans a line rather than the plane. Set B is certified: neither vector is a multiple of the other, so the pair is independent, and two independent vectors span a two-dimensional space by the derivation's closing consequence. The coordinates of (5, 5) come out as 2 and 1 in set B, since 2 times (1, 2) plus 1 times (3, 1) is (5, 5), and as 5 and 5 in the standard basis. The invariant is stated correctly: the numbers changed, while the vector and the count of numbers did not. And the three-vector proposal is rejected from dimension: the plane has dimension two, so any three vectors in it are dependent by the count fact, and no dependent set is a basis, whichever vectors were offered.