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.

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.

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.

We say that a vector space is finite‐dimensional if it has a finite unordered basis. Finite‐dimensional vector spaces have an associated invariant that, as we will see, largely characterizes them.

For completeness, we assign the zero vector space the dimension 0:

Throughout the remainder of Part I, unless stated otherwise, all vector spaces are finite‐dimensional

Let V be a vector space, and let {h ₁, … , h _m } be an unordered basis for V. The m‐tuple (h ₁, … , h _m ) is said to be an ordered basis for V, as is any m‐tuple derived from (h ₁, … , h _m ) by permuting h ₁, … , h _m . For example, (h ₁, h ₂, … , h _m ) and (h ₂, h ₁, … , h _m ) are distinct ordered bases for V.

Throughout the remainder of Part I, unless stated otherwise, all bases are ordered

Accordingly, we now refer to (e ₁, … , e _m ) as the standard basis for ^m . Let V and W be vector spaces. A map A: V → W is said to be linear if

for all vectors v, w in V and all real numbers c. Thus, a linear map respects vector space structure. Suppose A is in fact a linear map. Given a basis ℋ = (h ₁, … , h _m ) for V, let us denote

We say that A is a linear isomorphism, and that V and W are isomorphic, if A is bijective. To illustrate, let x be an indeterminate, and let

be the set of real polynomials of degree at most m. From the properties of polynomials, it is easily shown that is a vector space of dimension m + 1, and that the map A : ℙ ^m + 1 → ℙ _m given by A(a ₀, … , a _m ) = a ₀ + a ₁ x +  …  + a _m  x ^m for all vectors (a ₀, … , a _m ) in R ^{m + 1} is a linear isomorphism. Following Section B.5, we denote the existence of an isomorphism by R ^{m  +  1} ≈ P _m .

Since a linear isomorphism is a bijective map, it has an inverse map. The next result shows that the inverse of a linear isomorphism is automatically a linear isomorphism.

A linear map is completely determined by its values on a basis, as we now show.

From the point of view of linear structure, isomorphic vector spaces are indistinguishable. In fact, it is easily shown using Theorem 1.1.10 that all m‐dimensional vector space are isomorphic. More than that, they are all isomorphic to ^m . The isomorphism constructed with the help of Theorem 1.1.10 depends on the choice of bases for the vector spaces. However, we will see an instance in Section 1.2 where an isomorphism can be defined without having to resort to such an arbitrary choice.

Let V and W be vector spaces, and let A : V → W be a linear map. The kernel of A is defined by

and the image of A by

It is easily shown that ker (A) is a subspace of V, and im (A) is a subspace of W. The nullity of A is defined by

and the rank of A by

The nullity and rank of a linear map satisfy an important identity.

As an example of the rank‐nullity identity, consider the linear map A : ℝ³ → ℝ² given by A(x, y, z) = (x + y, 0). Then

and

In geometric terms, ker (A) is a plane in ³ and im (A) is a line in ². Thus, null (A) = 2 and rank (A) = 1, which agrees with Theorem 1.1.11.

In the notation of Theorem 1.1.11, we observe from (1.1.2) that rank (A) ≤ dim (V). Thus, a linear map at best “preserves” dimension, but never increases it.

We pause here to comment on the way proofs are presented when there is an equation or other type of display that stretches over several lines of text. The necessary justification for logical steps in such displays, whether it be equation numbers, theorem numbers, example numbers, and so on, are often provided in brackets at the end of corresponding lines. In order to economize on space, “[Theorem x.y.z]” and “[Example x.y.z]” are abbreviated to “[Th x.y.z]” and “[Ex x.y.z]”. The proof of the next result illustrates these conventions.

Let V be a vector space, and let U ₁,…,U _k be subspaces. The sum of U ₁ ,…,U _k is denoted by U ₁ + … + U _k and defined by

For example, (1, 0) + (0, 1) = ². It is easily shown that

from which it follows that U ₁ + … + U _k is the smallest subspace of V containing each of U ₁,…, U _k , in the sense that any subspace containing each of U ₁,…, U _k also contains U ₁ + … + U _k . We observe that

which shows that adding the zero vector spaces does not change a sum. For vectors v ₁,…,v _k in V, we have the following connection between spans and sums:

Let V be a vector space, and let U ₁,…,U _k be subspaces of V. We say that the subspace U ₁ +  …  + U _k of V is a direct sum, and write

if each vector V in U ₁ +  …  + U _k can be expressed uniquely (up to order of terms) in the form v = u ₁ +  …  + u _k for some vectors u _i in U _i for i = 1, … , k. As a matter of notation, writing V = U ₁ ⊕  …  ⊕ U _k is shorthand for

For example, ℝ² = ℝ(1, 0) ⊕ ℝ(0, 1).

Let V ₁, … , V _k be vector spaces. Following Section B.5, we make V ₁ ×  …  × V _k into a vector space, called the product of V ₁ ,…, V _k , as follows: for all vectors (v ₁, … , v _k), (w ₁, … , w _k ) in V ₁ ×  …  × V _k and all real numbers c, let

and

When V ₁ =  …  = V _k  = V, we denote

We close this section with two definitions that have obvious geometric content. Let V be a vector space. A subset S of V is said to be star‐shaped if there is a vector v₀ in S such that for all vectors v in S and all real numbers 0 ≤ t ≤ 1, the vector tv + (1 − t)v ₀ is in S. In that case, we say that S is star‐shaped about v ₀ . Since tv + (1 − t)v ₀ = v ₀ + t(v − v ₀), we can think of {tv + (1 − t)v ₀ : 0 ≤ t ≤ 1} as the “line segment” joining v ₀ to V. A subset C of V is said to be cone‐shaped if for all vectors v, v ₁,v ₂ in C and all real numbers c > 0, the vectors cv and v ₁ + v ₂ are in C. It is easily shown that if C is cone‐shaped, then it is star‐shaped about any vector it contains. For example, the closed cell {(x, y) : x, y ∈ [−1, 1]} in ² is star‐shaped about (0, 0), but not cone‐shaped; whereas the half‐plane {(x, y) ∈ ℝ² : x > 0} in ² is cone‐shaped, hence star‐shaped about (0, 0).

Throughout, we identify V** with V, and write V** = V

Let V be a vector in V, and let η be a covector in V*. Having made the identification V** = V, we henceforth denote

Thus, (1.2.1) and (1.2.2) both become

(1.2.3)

In particular, we have

for i, j = 1,…, m.