Vector Spaces, Bases and Dimension

If k is a field, then a vector space over k is an (additive) abelian group V equipped with a scalar multiplication; that is, there is a function k × V →V , denoted by (a, v) ↦av, such that, for all a, b, 1 ∈k and all u, v ∈V , (i) a(u + v) = au + av; (ii) (a + b)v = av + bv; (iii) (ab)v = a(bv); (iv) 1v = v. The elements of V are called vectors and the elements of k are called scalars.13

If V is a vector space over a field k, then a subspace of V is a subset U of V such that (i) 0 ∈U; (ii) u, u′ ∈U imply u + u′ ∈U; (iii) u ∈U and a ∈k imply au ∈U.

Let V be a vector space over a field k. A k-linear combination of a list v1, . . . , vn in V is a vector v of the form v = a1v1 + · · · + anvn, where ai ∈k for all i.

If X = v1, . . . , vm is a list in a vector space V , then ⟨v1, . . . , vm⟩, the set of all the k-linear combinations of v1, . . . , vm, is called the subspace spanned by X. We also say that v1, . . . , vm spans ⟨v1, . . . , vm⟩.

A list X = v1, . . . , vm in a vector space V is linearly dependent if there are scalars a1, . . . , am, not all zero, with m ℓ=1 aℓvℓ= 0; otherwise, X is called linearly independent. The empty set ∅is defined to be linearly independent (we may interpret ∅as a list of length 0).

A basis of a vector space V is a linearly independent list that spans V . Thus, bases are shortest spanning lists. Of course, all the vectors in a linearly independent list v1, . . . , vn are distinct, by Example 3.74(iv).

If V is a finite-dimensional vector space over a field k, then its dimension, denoted by dimk(V ) or dim(V ), is the number of elements in a basis of V .

A longest (or a maximal) linearly independent list u1, . . . , um is a linearly independent list for which there is no vector v ∈V such that u1, . . . , um, v is linearly independent.

Let V be a vector space over a field k. (i) Every intersection of subspaces of V is itself a subspace. (ii) If X = v1, . . . , vm is a list in V , then the intersection of all the subspaces of V containing X is ⟨v1, . . . , vm⟩, the subspace spanned by v1, . . . , vm, and so ⟨v1, . . . , vm⟩ is the smallest subspace of V containing X.

If V is a vector space, then the following conditions on a list X = v1, . . . , vm spanning V are equivalent: (i) X is not a shortest spanning list; (ii) some vi is in the subspace spanned by the others; that is, vi ∈⟨v1, . . . , vi, . . . , vm⟩; (iii) there are scalars a1, . . . , am, not all zero, with m ℓ=1 aℓvℓ= 0.

If X = v1, . . . , vm is a list spanning a vector space V , then X is a shortest spanning list if and only if X is linearly independent. Linear independence has been defined indirectly, as not being linearly dependent. Because of the importance of linear independence, let us define it directly. A list X = v1, . . . , vm is linearly independent if, whenever a k-linear combination m ℓ=1 aℓvℓ= 0, then every ai = 0. It follows that every sublist of a linearly independent list is itself linearly independent (this is one reason for decreeing that ∅be linearly independent). We have arrived at the notion we have been seeking.

Let X = v1, . . . , vn be a list in a vector space V over a field k. Then X is a basis if and only if each vector in V has a unique expression as a k-linear combination of vectors in X. Linear Algebra

If X′ is linearly independent, it is a basis; if not, X′ can be shortened to a spanning sublist X′′. Eventually, we arrive at a shortest spanning sublist, which is independent and hence is a basis. For example, it turns out that a basis of k[x] is 1, x, x2, . . . , xn, . . . . We can now prove invariance of dimension, one of the most important results about vector spaces.

Let u1, . . . , un be elements in a vector space V , and let v1, . . . , vm ∈ ⟨u1, . . . , un⟩. If m > n, then v1, . . . , vm is a linearly dependent list.

A homogeneous system of linear equations, over a field k, with more unknowns than equations has a nontrivial solution.

If X = v1, . . . , vn is a linearly dependent list of vectors in a vector space V , then there exists vr with r ≥1 with vr ∈⟨v1, v2, . . . , vr−1⟩[when r = 1, we interpret ⟨v1, . . . , vr−1⟩to mean {0}]. Remark. Let us compare Proposition 3.73 with this one. The earlier result says that if v1, v2, v3 is linearly dependent, then either v1 ∈⟨v2, v3⟩, v2 ∈⟨v1, v3⟩, or v3 ∈⟨v1, v2⟩. This lemma says that either v1 ∈{0}, v2 ∈⟨v1⟩, or v3 ∈⟨v1, v2⟩. ◀

If X = x1, . . . , xm is a basis of a vector space V and y1, . . . , yn is a linearly independent subset of V , then n ≤m.

If X = x1, . . . , xm and Y = y1, . . . , yn are bases of a vector space V , then m = n.

If V is a finite-dimensional vector space, then a longest linearly independent list v1, . . . , vn is a basis of V .

Let Z = u1, . . . , um be a linearly independent list in an n-dimensional vector space V . Then Z can be extended to a basis; i.e., there are vectors vm+1, . . . , vn so that u1, . . . , um, vm+1, . . . , vn is a basis of V .

Let V be a vector space with dim(V ) = n. (i) A list of n vectors that spans V must be linearly independent. (ii) Any linearly independent list of n vectors must span V .

Let U be a subspace of a vector space V of dimension n. (i) U is finite-dimensional and dim(U) ≤dim(V ). (ii) If dim(U) = dim(V ), then U = V .

(i) division-norm space V = Rn is a vector space over R. Vectors are n-tuples (a1, . . . , an), where ai ∈R for all i. Picture a vector v as an arrow from the origin to the point having coordinates (a1, . . . , an). Addition is given by (a1, . . . , an) + (b1, . . . , bn) = (a1 + b1, . . . , an + bn); geometrically, the sum of two vectors is described by the parallelogram law. Scalar multiplication is given by av = a(a1, . . . , an) = (aa1, . . . , aan). 13The word vector comes from the Latin word meaning “to carry”; vectors in euclidean space carry the data of length and direction. The word scalar comes from regarding v ↦av as a change of scale. The terms scale and scalar come from the Latin word meaning “ladder,” for the rungs of a ladder are evenly spaced. Scalar multiplication v ↦av “stretches” v by a factor |a|, reversing its direction when a is negative (we put quotes around stretches because av is shorter than v when |a| < 1). (ii) The example in part (i) can be generalized. If k is any field, define V = kn, the set of all n-tuples v = (a1, . . . , an), where ai ∈k for all i. Addition is given by (a1, . . . , an) + (b1, . . . , bn) = (a1 + b1, . . . , an + bn), and scalar multiplication is given by av = a(a1, . . . , an) = (aa1, . . . , aan). (iii) If R is a commutative ring and k is a subring that is a field, then R is a vector space over k. Regard the elements of R as vectors and the elements of k as scalars; define scalar multiplication av, where a ∈k and v ∈R, to be the given product of two elements in R. Notice that the axioms in the definition of vector space are just particular cases of some of the axioms holding in the commutative ring R. For example, if k is a field, then the polynomial ring R = k[x] is a vector space over k. Vectors are polynomials f (x), scalars are elements a ∈k, and scalar multiplication gives the polynomial af (x); that is, if f (x) = bnxn + · · · + b1x + b0, then af (x) = abnxn + · · · + ab1x + ab0. In particular, if a field k is a subfield of a larger field E, then E is a vector space over k. ◀ A subspace of a vector space V is a subset of V that is a vector space under the addition and scalar multiplication in V .

(i) Example 3.76 shows that kn has dimension n, which agrees with our intuition when k = R. Thus, the plane R × R is two-dimensional! (ii) If V = {0}, then dim(V ) = 0, for there are no elements in its basis ∅. (This is a good reason for defining ∅to be linearly independent.) (iii) Let X = {x1, . . . , xn} be a finite set. Define k X = {functions f : X →k}. Now k X is a vector space if we define addition f + f ′ to be f + f ′ : x ↦f (x) + f ′(x) and scalar multiplication af , for a ∈k and f : X →k, by af : x ↦af (x). It is easy to check that the set of n functions of the form fx, where x ∈X, defined by fx(y) = if y = x; if y ̸= x, form a basis, and so dim(k X) = n = |X|. The reader should note that this is not a new example: An n-tuple (a1, . . . , an) is really a function f : {1, . . . , n} →k with f (i) = ai for all i. Thus, the functions fx comprise the standard basis. We begin with a modification of the proof of Proposition 3.73.