Chapter 1
Vector Spaces

1.1 Vector Spaces

The definition of a vector space over a field and that of a subspace of a vector space are given in Section B.6. Our focus in this book is exclusively on vector spaces over the real numbers (as opposed to the complex numbers or some other field).

Throughout, all vector spaces are over , the field of real numbers

For brevity, we will drop the reference to R whenever possible and write, for example, "linear" instead of "R-linear".

Of particular importance is the vector space R m , but many other examples of vector spaces will be encountered. It is easily shown that the intersection of any collection of subspaces of a vector space is itself a subspace. The zero vector of a vector space is denoted by 0, and the zero subspace of a vector space by {0}. The zero vector space, also denoted by {0}, is the vector space consisting only of the zero vector. We will generally avoid explicit consideration of the zero vector space. Most of the results on vector spaces either apply directly to the zero vector space or can be made applicable with a minor reworking of definitions and proofs. The details are usually left to the reader.

Example 1.1.1

Let V and W be vector spaces. Following Section B.5 and Section B.6, we denote by Lin(V, W) the vector space of linear maps from V to W, where addition and scalar multiplication are defined as follows: for all maps A, B in Lin (V, W) and all real numbers c,

and

for all vectors V in V. The zero element of Lin (V, W), denoted by 0, is the zero map, that is, the map that sends all vectors in V to the zero vector 0 in W. When V = W, we make Lin(V, V) into a ring by defining multiplication to be composition of maps: for all maps A, B in Lin(V, V), let

for all vectors v in V. The identity element of the ring Lin (V, V) is the identity map on V, denoted by id V .

A linear combination of vectors in a vector space V is defined to be a finite sum of the form a 1 v 1 + . + a k v k , where a 1, ., a k are real numbers and v 1, ., v k are vectors in V. The possibility that some (or all) of a 1, ., a k equal zero is not excluded.

Let us pause here to comment on an aspect of notation. Following the usual convention in differential geometry, we index the scalars and vectors in a linear combination with superscripts and subscripts, respectively. This opens the door to the Einstein summation convention, according to which, for example, a 1 v 1?+??.??+?a k v k and are abbreviated as a i v i . The logic is that when an expression has a superscript and subscript in common, it is understood that the index is being summed over. Despite the potential advantages of this notation, especially when multiple indices involved, the Einstein summation convention will not be adopted here.

Let S be a (nonempty and not necessarily finite) subset of V. The span of S is denoted by span (S) and defined to be the set of linear combinations of vectors in S:

For a vector v in V, let us denote

For example, in 2, we have

and

It is easily shown that span (S) is a subspace of V. In fact, span (S) is the smallest subspace of V containing S, in the sense that any subspace of V containing S also contains span (S). When span (S) = V, it is said that S spans V or that the vectors in S span V , and that each vector in V is in the span of S .

We say that S is linearly independent or that the vectors in S are linearly independent if the only linear combination of distinct vectors in S that equals the zero vector is the one with all coefficients equal to 0. That is, if v 1,., v k are distinct vectors in S and a 1, ., a k are real numbers such that a 1 v 1 + . + a k v k = 0, then a 1 = . = a = 0. Evidently, any subset of a linearly independent set is linearly independent. When S is not linearly independent, it is said to be linearly dependent. In particular, the zero vector in any vector space is linearly dependent. As further examples, the vectors (1, 0), (0, 1) in 2 are linearly independent, whereas (0, 0), (1, 0) and (1, 0), (2, 0) are linearly dependent.

The next result shows that when a linearly independent set does not span a vector space, it has a linearly independent extension.

Theorem 1.1.2

Let V be a vector space, let S be a nonempty subset of V such that span(S) ? V, and let V be a vector in V\span(S). Then S is linearly independent if and only if S U {v} is linearly independent.

(?): Suppose av + b 1 s 1 + . + b k s k = 0 for distinct vectors s 1, ., s k in S and real numbers a, b 1, ., b k . Then a = 0; for if not, then

hence v is in span (V), which is a contradiction. Thus, b 1 s 1 + . + b k s k = 0, and since S is linearly independent, we have b 1 = . = b k = 0.

(?): As remarked above, any subset of a linearly independent set is linearly independent.

A (not necessarily finite) subset H of a vector space V is said to be an unordered basis for V if it spans V and is linearly independent.

Theorem 1.1.3

If V is a vector space and H is an unordered basis for V, then each vector in V can be expressed uniquely (up to order of terms) as a linear combination of vectors in H.

Since H spans V, each vector in V can be expressed as a linear combination of vectors in H. Suppose a vector v in V can be expressed as a linear combination in two ways. Let h 1, . , h k be the distinct vectors in the linear combinations. Then

for some real numbers a 1, . , a k , b 1,., b k , hence

Since H is linearly independent, a i ?-?b i  = 0  for i = 1,?.?, k.

Theorem 1.1.4

Let V be a vector space, and let S and T be nonempty subsets of V, where S is linearly independent, and T is finite and spans V. Then S is finite and card (S) = card (T), where card denotes cardinality.

Since S is linearly independent, it does not contain the zero vector. Let card (T) = m and T = {t 1,?.?, t m } . We proceed in steps. For the first step, let S1 be a vector in S. Since V = span (T), s 1 is a linear combination of t 1,?.?, t m .Because s 1 is not the zero vector, at least one of the coefficients in the linear combination must be nonzero. Renumbering t 1,?.?, t m if necessary, suppose it is the coefficient of t 1, and let S 1 = {s 1, t 2,?.?, t m }. Then t 1 can be expressed as a linear combination of the vectors in S 1, hence V = span(Si). For the second step, let s 2 be a vector in Since V = span (S 1), s 2 is a linear combination of s 1, t 2,?.?, t m . Because s 1, s 2 are linearly independent, at least one of the coefficients of t 2, ., t m in the linear combination is nonzero. Renumbering t 2, ., t m if necessary, suppose it is the coefficient of t 2, and let S 2 = {s 1, s 2, t 3,?.?, t m }. Then t 2 can be expressed as a linear combination of the vectors in S 2, hence V = span (S 2)....