Real Analysis with Economic Applications - Chapter F pot

introduc-After going through a number of basic definitions and examples where infinitedimensional spaces are given a bit more emphasis than usual, we review the notions of basis and dime

Chapter F

Linear Spaces

The main goal of this chapter is to provide a foundation for our subsequent tion to linear functional analysis The latter is a vast subject, and there are manydifferent ways in which one can provide a first pass at it We mostly adopt a geometricviewpoint in this book Indeed, we will later spend quite a bit of time covering therudiments of (infinite dimensional) convex analysis The present chapter introducesthe elementary theory of linear spaces with this objective in mind

introduc-After going through a number of basic definitions and examples (where infinitedimensional spaces are given a bit more emphasis than usual), we review the notions

of basis and dimension, and talk about linear operators and functionals.1 Keeping aneye on the convex analysis to come, we also discuss here the notion of affinity at somelength In addition, we conclude an unfinished business by proving Carathéodory’sTheorem, characterize the finite dimensional linear spaces, and explore the connec-tion between hyperplanes and linear functionals in some detail On the whole, ourexposition is fairly elementary, the only minor exception being the proof of the factthat every linear space has a basis — this proof is based on the Axiom of Choice Aseconomic applications, we prove some basic results of expected utility theory in thecontext of finite prize spaces, and introduce the elements of cooperative game theory.These applications illustrate well what a little linear algebra can do for you

Recall that Rn is naturally endowed with three basic mathematical structures: anorder structure, a metric structure and a linear structure In the previous chapters

we have studied the generalizations of the first two of these structures, which led

to the formulation of posets and metric spaces, respectively In this chapter we willstudy how such a generalization can be carried out in the case of the linear structure

of Rn which, among other things, allows us to “add” any two n-vectors The idea isthat Rn is naturally equipped with an addition operation, and our immediate goal is

to obtain a suitable axiomatization of this operation

1 You can consult on any one of the numerous texts on linear algebra for more detailed treatments

of these topics and related matters My favorite is Hoffman and Kunze (1971), but this may be due to the fact that I learned this stuff from that book first Among the more recent and popular expositions are Broida and Williamson (1989) and Strang (1988).

1.1 Abelian Groups

Let us first recall that a binary operation • on a nonempty set X is a map from

X× X into X, but we write x • y instead of •(x, y) for any x, y ∈ X (Section A.2.1)

Dhilqlwlrq Let X be any nonempty set, and + a binary operation on X Thedoubleton (X, +) is called a group if the following four properties are satisfied

(i) (Associativity) (x + y) + z = x + (y + z) for all x, y, z ∈ X;

(ii) (Existence of an identity element) There exists an element 0 ∈ X such that

0+ x = x = x + 0for all x ∈ X;

(iii) (Existence of inverse elements) For each x ∈ X, there exists an element −x ∈ Xsuch that x +−x = 0 = −x + x

If, in addition, we have

(iv) (Commutativity) x + y = y + x for all x, y ∈ X,

then (X, +) is said to be an Abelian (or commutative) group

Notation For any group (X, +), and any nonempty subsets A and B of X, we let

In this book we will work exclusively (and often implicitly) with Abelian groups.2

In fact, even Abelian groups do not provide sufficiently rich algebraic structure forour purposes, so we will shortly introduce more discipline into the picture But weshould first consider some examples of Abelian groups to make things a bit moreconcrete

E{dpsoh 1 [1] (Z, +), (Q, +), (R, +), (Rn, +) and (R\{0}, ·) are Abelian groupswhere + and · are the usual addition and multiplication operations (Note In(R\{0}, ·) the number 1 plays the role of the identity element.) On the other hand,(R, ·) is not a group because it does not satisfy the requirement (iii) (What would

be the inverse of 0 in (R, ·)?) Similarly, (Z, ·) and (Q, ·) are not groups

[2] (R[0,1], +) is an Abelian group where + is defined pointwise (i.e., f + g ∈ R[0,1]

is defined by (f + g)(t) := f (t) + g(t))

[3] Let X be any nonempty set and X the class of all bijective self-maps on X.Then (X , ◦) is a group, but it is not Abelian unless |X| ≤ 2

2 “Abelian” in the term Abelian group honors the name of Niels Abel (1802-1829) who was able

to make lasting contributions to group theory in his very short life span.

[4] Let X := {x ∈ R2 : x21+ x22 = 1} and define x + y := (x1y1− x2y2, x1y2+ x2y1)for any x, y ∈ X It is easily checked that x + y ∈ X for each x, y ∈ X, so this well-defines + as a binary operation on X In fact, (X, +) is an Abelian group (Verify!Hint The identity element here is (1, 0).)

[5] Let (X, +) be any group The identity element 0 of this group is unique For,

if y ∈ X is another candidate for the identity element, then 0+y = 0, but this implies

0= 0 + y = y.Similarly, the inverse of any given element x is unique Indeed, if y is

an inverse of x, then

y = y + 0 = y + (x +−x) = (y + x) + −x = 0 + −x = −x

In particular, the inverse of the identity element is itself

[6] (Cancellation Laws) The usual cancellation laws apply to any Abelian group(X, +) For instance, x + y = z + x iff y = z, −(−x) = x and −(x + y) = −x + −y(Section A.2.1)

These examples should convince you that we are on the right track The notion

of Abelian group is a useful generalization of (R, +) It allows us to “add” members

of arbitrary sets in concert with the intuition supplied by R

Exercise 1 LetX be a nonempty set, and defineA B := (A\B) ∪ (B\A) for any

A, B ⊆ X.Show that (2X, ) is an Abelian group.

Exercise 2.H Let(X, +) be a group, and define the binary operation +on 2X

Exercise 4.H Let (X, +) be a group If∅ = Y ⊆ X and (Y, +) is a group (where

we use the restriction of +toY × Y, of course), then(Y, +) is called a subgroup

of(X, +)

(a) For any nonempty subsetY of X, show that(Y, +) is a subgroup of (X, +) iff

x +−y ∈ Y for all x, y∈ Y

(b) Give an example to show that if(Y, +)and(Z, +)are subgroups of(X, +),then

(Y ∪ Z, +)need not be a subgroup of (X, +)

(c) Prove that if(Y, +),(Z, +)and (Y ∪ Z, +)are subgroups of(X, +),then either

Y ⊆ Z orZ ⊆ Y

Exercise 5 Let (X, +) and (Y,⊕) be two groups A function f ∈ YX is said to

be a homomorphism from (X, +) into(Y,⊕)iff (x + x ) = f (x)⊕ f(x )for all

x, x ∈ X.If there exists a bijective such map, then we say that these two groups are homomorphic.

(a) Show that(R, +) and (R\{0}, ·)are homomorphic.

(b) Show that if f is a homomorphism from (X, +) into (Y,⊕), then f (0) is the identity element of(Y,⊕),and (f (X),⊕)is a subgroup of (Y,⊕).

(c) Show that if (X, +) and (Y,⊕)are homomorphic, then (X, +) is Abelian iff so

is(Y,⊕)

1.2 Linear Spaces: Definition and Examples

So far so good, but we are only half way through our abstraction process We wish

to have a generalization of the linear structure of Rn in a way that would allow

us to algebraically represent some basic geometrical objects For instance, the linesegment between the vectors (1, 2) and (2, 1) in R2 can be described algebraically as{λ(1, 2) + (1 − λ)(2, 1) : 0 ≤ λ ≤ 1}, thanks to the linear structure of R2.By contrast,the structure of an Abelian group falls short of letting us represent even such anelementary geometric object This is because it is meaningful to “multiply” a vector

in a Euclidean space with a real number (often called a scalar in this context), whilethere is no room for doing this within an arbitrary Abelian group

The next step is then to enrich the structure of Abelian groups by defining ascalar multiplication operation on them Once this is done, we will be able to describealgebraically a “line segment” in a very general sense This description will indeedcorrespond to the usual geometric notion of line segment in the case of R2, andmoreover, it will let us define and study a general notion of convexity for sets Asyou have probably already guessed, the abstract model that we are after is noneother than that of linear space.3 Chapter G will demonstrate that this model indeedprovides ample room for powerful geometric analysis

Dhilqlwlrq Let X be a nonempty set The list (X, +, ·) is called a linear (orvector) space if (X, +) is an Abelian group, and if · is a mapping that assigns toeach (λ, x) ∈ R × X an element λ · x of X (which we denote simply as λx) such that,for all α, λ ∈ R and x, y ∈ X, we have

(v) (Associativity) α(λx) = (αλ)x;

(vi) (Distributivity) (α + λ)x = αx + λx and λ(x + y) = λx + λy;

(vii) (The unit rule) 1x = x

In a linear space (X, +, ·), the mappings + and · are called addition and scalarmultiplication operations on X, respectively When the context makes the nature

of these operations clear, we may refer to X itself as a linear space The identity

3 We owe the first modern definition of linear space to the 1888 work of Giuseppe Peano While initially ignored by the profession, the original treatment of Peano was amazingly modern I recall Peter Lax telling me once that he would not be able to tell from reading certain parts of Peano’s work that it was not written instead in 1988.

element 0 is called the origin (or zero), and any member of X is referred to as avector If x ∈ X\{0}, then we say that x is a nonzero vector in X.

Notation Let (X, +, ·) be a linear space, and A, B ⊆ X and λ ∈ R Then,

A + B :={x + y : (x, y) ∈ A × B} and λA :={λx : x ∈ A}

For simplicity, we write A + y for A + {y}, and similarly, y + A := {y} + A

E{dpsoh 2.[1]The most trivial example of a linear space is a singleton set where theunique member of the set is designated as the origin Naturally, this space is denoted

as {0} Any linear space that contains more than one vector is called nontrivial

[2] Let n ∈ N A very important example of a linear space is, of course, ourbeloved Rn (Remark A.1) On the other hand, endowed with the usual addition andscalar multiplication operations, Rn

++is not a linear space since it does not contain theorigin This is of course not the only problem After all, Rn+is not a linear space either(under the usual operations), for it does not contain the inverse of any nonzero vector

Is [−1, 1]n

a linear space? How about {x ∈ Rn : x1 = 0}? {x ∈ Rn : Sn

xi = 1}?{x ∈ Rn:Sn

xi = 0}?

[3] In this book, we always think of the sum of two real functions f and g defined

on a nonempty set T as the real function f + g ∈ RT with (f + g)(t) := f (t) + g(t).Similarly, the product of λ ∈ R and f ∈ RT

is the function λf ∈ RT defined by(λf )(t) := λf (t) (Recall Section A.4.1.) In particular, we consider the real sequencespace R∞ (which is none other than RN) and the function spaces RT and B(T ) (forany nonempty set T ) as linear spaces under these operations Similarly, p (for any

1≤ p ≤ ∞), along with the function spaces CB(T ) and C(T ) (for any metric space

T), are linear spaces under these operations (why?), and when we talk about thesespaces, it is these operations that we have in mind The same goes as well for otherfunction spaces, such as P(T ) or the space of all polynomials on T of degree m ∈ Z+(for any nonempty subset T of R)

[4]Since the negative of an increasing function is decreasing, the set of all ing real functions on R (or on any compact interval [a, b] with a < b) is not a linearspace under the usual operations Less trivially, the set of all monotonic self-maps

increas-on R is not a linear space either To see this, observe that the self-map f defined increas-on

R by f(t) := sin t + 2t is an increasing function (For, f (t) = cos t + 2 ≥ −1 + 2 ≥ 0for all t ∈ R.) But the self-map g defined on R by g(t) := sin t − 2t is a decreasingfunction, and yet f + g is obviously not monotonic For another example, we notethat the set of all semicontinuous self-maps on R does not form a linear space underthe usual operations (Why?)

It will become clear shortly that linear spaces provide an ideal structure for aproper investigation of convex sets For the moment, however, all you need to do is

to recognize that a linear space is an algebraic infrastructure relative to which theaddition and scalar multiplication operations behave in concert with intuition, that

is, the way the corresponding operations behave on a Euclidean space To derive thispoint home, let us derive some preliminary facts about these operations that we shalllater invoke in a routine manner

Fix a linear space (X, +, ·) First of all, we have λ0 = 0 for any λ ∈ R Indeed,λ0 + λ0 = λ(0 + 0) = λ0 = 0 + λ0 so that, by Example 1.[6], λ0 = 0 A similarreasoning gives us that 0x = 0 for any x ∈ X More generally, for any x = 0,

(Here −(λx) is the inverse of λx, while (−λ)x is the “product” of −λ and x; so theclaim is not trivial.) Indeed, 0 = 0x = (λ − λ)x = λx + (−λ)x for any (λ, x) ∈ R × X,and if we add −(λx) to both sides of 0 = λx + (−λ)x, we find −(λx) = (−λ)x.Thanks to (2), there is no difference between (−1)x and −x, and hence between

x + (−1)y and x + −y In what follows, we shall write x − y for either of the latterexpressions since we are now confident that there is no room for confusion

Exercise 6 Define the binary operations+1 and+2 onR2 byx+1y := (x1+y1, x2+

y2)andx +2y := (x1+ y1, 0) Is (R2, +i,·),where·maps each (λ, x)∈ R × X to

(λx1, λx2)∈ R2,a linear space, i = 1, 2? What if· maps each(λ, x)to (λx1, x2)?

What if ·maps each (λ, x)to(λx1, 0)?

Exercise 7 Let (X, +,·) be a linear space, and x, y ∈ X Show that ({αx + λy :

α, λ∈ R},+, ·) is a linear space.

Henceforth, we use the notation X instead of (X, +, ·) for a linear space, butyou should always keep in mind that what makes a linear space “linear” is the twooperations defined on it Two different types of addition and scalar multiplicationoperations on a given set may well endow this set with different linear structures, andhence yield two very different linear spaces

1.3 Linear Subspaces, Affine Manifolds and Hyperplanes

One method of obtaining other linear spaces from a given linear space X is to considerthose subsets of X which are themselves linear spaces under the inherited operations

Dhilqlwlrq Let X be a linear space and ∅ = Y ⊆ X If Y is a linear space withthe same operations of addition and scalar multiplication as with X, then it is called

a linear subspace of X.4 If, further, Y = X, then Y is called a proper linearsubspaceof X

The following exercise provides an alternative (and of course equivalent) definition

of the notion of linear subspace We will use this alternative formulation freely inwhat follows

Exercise 8 Let X be a linear space and ∅ = Y ⊆ X Show that Y is a linear subspace ofX iff λx + y∈ Y for eachλ∈ R and x, y ∈ Y

E{dpsoh 3 [1] [0, 1] is not a linear subspace of R whereas {x ∈ R2 : x1 + x2 = 0}

is a proper linear subspace of R2

[2] For any n ∈ N, Rn×n is a linear space under the usual matrix operations ofaddition and scalar multiplication (Remark A.1) The set of all symmetric n × nmatrices, that is, {[aij]n×n : aij = aji for each i, j} is a linear subspace of this space

[3] For any n ∈ N and linear f : Rn

→ R (Section D.5.3), {x ∈ Rn : f (x) = 0} is

a linear subspace of Rn Would this conclusion be true if all we knew was that f isadditive?

[4]For any m ∈ N, the set of constant functions on [0, 1] is a proper linear subspace

of the set of all polynomials on [0, 1] of degree at most m The latter set is a properlinear subspace of P[0, 1] which is a proper linear subspace of C[0, 1] which is itself aproper linear subspace of B[0, 1] Finally, B[0, 1] is a proper linear subspace of R[0,1]

Exercise 9 (a) Is 1 a linear subspace of ∞?

(b) Is the setc of all convergent real sequences a linear subspace of ∞? Of 1?

(c) Let c0 be the set of all real sequences all but finitely many terms of which are zero Is c0 a linear space (under the usual (pointwise) operations) Is it a linear subspace ofc?

(d ) Is{(xm)∈ c0 : xi = 1for somei}a linear subspace ofc0? OfR∞?

4 Put more precisely, if the addition and scalar multiplication operations on X are + and ·, respectively, then by a linear subspace Y of X, we mean the linear space (Y, ⊕, ), where Y is a nonempty subset of X, and ⊕ : Y 2 → Y is the restriction of + to Y 2 and : R × Y → Y is the restriction of · to R × Y.

Exercise 10 Show that the intersection of any collection of linear subspaces of a linear space is itself a linear subspace of that space.

Exercise 11 If Z is a linear subspace of Y and Y a linear subspace of the linear spaceX, isZ necessarily a linear subspace of X?

Exercise 12.H LetY and Z be linear subspaces of a linear space X.Prove:

(a) Y + Z is a linear subspace of X;

(b) IfY ∪ Z is a linear subspace ofX, then eitherY ⊆ Z orZ ⊆ Y

Clearly, {x ∈ R2 : x1+ x2 = 1} is not a linear subspace of R2 (This set doesnot even contain the origin of R2 And yes, this is crucial!) On the other hand,geometrically speaking, this set is very “similar” to the linear subspace {x ∈ R2 :

x1 + x2 = 0} Indeed, the latter is nothing but a parallel shift (translation) of theformer set It is thus not surprising that such sets play an important role in geometricapplications of linear algebra They certainly deserve a name

Dhilqlwlrq A subset S of a linear space X is said to be an affine manifold of

X if S = Z + x∗ for some linear subspace Z of X and some vector x∗ ∈ X.5 If Z

is a ⊇-maximal proper linear subspace of X, S is then called a hyperplane in X.(Equivalently, a hyperplane is a ⊇-maximal proper affine manifold.)

As for examples, note that there is no hyperplane in the trivial space {0} Sincethe only proper linear subspace of R is {0}, any one-point set in R and R itself are theonly affine manifolds in R So, all hyperplanes in R are singleton sets In R2,any one-point set, any line (with no endpoints) and the entire R2 are the only affine manifolds

A hyperplane in this space is necessarily of the form {x ∈ R2 : a1x1+ a2x2 = b} forsome real numbers a1, a2 with at least one of them being nonzero, and some realnumber b Finally, all hyperplanes are of the form of (infinitely extending) planes in

R3

A good way of thinking intuitively about the notion of affinity in linear analysis

is this:

affinity = linearity + translation

Since we think of 0 as the origin of the linear space X — this is a geometric pretation; don’t forget that the definition of 0 is purely algebraic — it makes sense toview a linear subspace of X as untranslated (relative to the origin of the space), for alinear subspace “passes through” 0 The following simple but important observationthus gives support to our informal equation above: An affine manifold S of a linearspace X is a linear subspace of X iff 0 ∈ S (Proof If S = Z + x∗ for some linearsubspace Z of X and x∗ ∈ X, then 0 ∈ S implies −x∗ ∈ Z (Yes?) Since Z is a linearspace, we then have x∗ ∈ Z, and hence Z = Z + x∗ = S.) An immediate corollary of

inter-5 Reminder Z + x∗:= {z + x∗: z ∈ Z}.

this is: If S is an affine manifold of X, then S − x is a linear subspace of X for any

x∈ S Moreover, this subspace is determined independently of x, because if S is anaffine manifold, then

We leave the proof of this assertion as an easy exercise.6

The following result, a counterpart of Exercise 8, provides a useful characterization

of affine manifolds

Proposition 1 Let X be a linear space and ∅ = S ⊆ X Then S is an affine manifold

of X if, and only if,

λx + (1− λ)y ∈ S for any x, y ∈ S and λ ∈ R (4)

Proof If S = Z + x∗ for some linear subspace Z of X and x∗ ∈ X, then for any

x, y∈ S there exist zx and zy in Z such that x = zx+ x∗ and y = zy + x∗ It followsthat

λx + (1− λ)y = (λzx+ (1− λ)zy) + x∗ ∈ Z + x∗for any x, y ∈ S and λ ∈ R Conversely, assume that S satisfies (4) Pick any x∗ ∈ Sand define Z := S − x∗.Then S = Z + x∗,so we will be done if we can show that Z is

a linear subspace of X Thanks to Exercise 8, all we need to do is, then, to establishthat Z is closed under scalar multiplication and vector addition To prove the formerfact, notice that if z ∈ Z then z = x − x∗ for some x ∈ S, so, by (4),

λz = λx− λx∗ = (λx + (1− λ)x∗)− x∗ ∈ S − x∗ = Zfor any λ ∈ R To prove the latter fact, take any z, w ∈ Z, and note that z = x − x∗and w = y − x∗ for some x, y ∈ S By (4), 2x − x∗ ∈ S and 2y − x∗ ∈ S Therefore,applying (4) again,

z + w = 12(2x− x∗) + 12(2y− x∗)− x∗ ∈ S − x∗ = Z,and the proof is complete

The geometric nature of affine manifolds and hyperplanes makes them able tools for convex analysis, and we will indeed use them extensively in what follows.For the time being, however, we leave their discussion at this primitive stage, andinstead press on reviewing the fundamental concepts of linear algebra We will revisitthese notions at various points in the subsequent development

indispens-6 Just to fix the intuition, you might want to verify everything I said in this paragraph in the special case of the affine manifolds {x ∈ R 2 : x1+ x2 = 1}, {x ∈ R 3 : x1+ x2 = 1} and {x ∈

R 3 : x1+ x2+ x3= 1} (In particular, draw the pictures of these manifolds, and see how they are obtained as translations of specific linear subspaces.)

1.4 Span and Affine Hull of a Set

Let X be a linear space and (xm)∈ X∞.We define

mSi=1

xi := x1+· · · + xm and

mSi=k

x∈Tx for the sum of all members of T ) As in the case of sums of real numbers, wewill write Sm

xi for Sm

i=1xi in the text

Dhilqlwlrq For any m ∈ N, by a linear combination of the vectors x1, , xm

in a linear space X, we mean a vector Sm

λixi is referred to as an affine combination of the vectors

x1, , xm If, on the other hand, λi ≥ 0 for each i, then Sm

λixi is called a positivelinear combination of x1, , xm Finally, if λi ≥ 0 for each i and Sm

λi = 1,then Sm

λixi is called a convex combination of x1, , xm Equivalently, a linear(affine (convex)) combination of the elements of a nonempty finite subset T of X isS

λixi : m ∈ N and (xi, λi)∈ S × R, i = 1, , m

,

or equivalently,

span(S) =

Sx∈Tλ(x)x : T ∈ P(S) and λ ∈ RT

,

where P(S) is the class of all nonempty finite subsets of S By convention, we letspan(∅) = {0}

Given a linear space X, span({0}) = {0} and span(X) = X while span({x}) ={λx : λ ∈ R} The span of a set in a linear space is always a linear space (yes?), and it

is thus a linear subspace of the mother space What is more, span(S) is the smallest(i.e ⊇-minimum) linear subspace of the mother space that contains S Especiallywhen S is finite, this linear subspace has a very concrete description in that everyvector in it can be expressed as linear combination of all the vectors in S (Why?)

Exercise 13 Let X be a linear space and ∅ = S ⊆ X Show that span(S) ⊆ Y

for any linear subspaceY of X with S ⊆ Y Conclude thatspan(S) is the smallest linear subspace ofX that containsS

Dhilqlwlrq The set of all affine combinations of finitely many members of a empty subset S of a linear space X is called the affine hull of S (in X), and isdenoted by aff (S) That is, for any ∅ = S ⊆ X,

non-aff (S) :=

Sx∈T

λ(x)x : T ∈ P(S) and λ ∈ RT with S

x∈Tλ(x) = 1

Propo-These observations help clarify the nature of the tight connection between thenotions of span and affine hull of a set Put precisely, for any nonempty subset S of

a linear space X, we have

aff (S) = span(S − x) + x for any x ∈ S (5)Indeed, for any x ∈ X, span(S−x)+x is an affine manifold of X that contains S, thus,being the smallest such affine manifold, aff (S) must be contained in span(S − x) + x.Conversely, if x ∈ S, then aff (S) − x is a linear subspace of X that contains S − x(why?), and hence, being the smallest such linear subspace of X, span(S − x) must

be contained in aff (S) − x, that is, aff (S) ⊇ span(S − x) + x

1.5 Linear and Affine Independence

Dhilqlwlrq Let X be a linear space A subset S of X is called linearly dent in X if it either equals {0} or at least one of the vectors in S can be expressed

depen-as a linear combination of finitely many vectors in S\{x} For any m ∈ N, any tinct vectors x1, , xm ∈ X are called linearly dependent if {x1, , xm} is linearlydependent in X

dis-Dhilqlwlrq Let X be a linear space A subset of X is called linearly independent

in X if no finite subset of it is linearly dependent in X For any m ∈ N, the vectors

7 What is the affine hull of {x ∈ R 2 : d2(x, 0) = 1} in R 2 Of {x ∈ R 2 : x1+ x2 = 1}? Of {0, (1, 0), (0, 1)}?

x1, , xm ∈ X are called linearly independent if {x1, , xm} is linearly independent

in X

That is, a nonempty subset S of X is linearly independent in X iff, for everynonempty finite subset T of S and λ ∈ RT,

Sx∈Tλ(x)x = 0 implies λ = 0

(Why?) In particular, for any m ∈ N, the vectors x1, , xm ∈ X are linearly pendent iff, for every (λ1, , λm)∈ Rm,

inde-mSi=1

λixi = 0 implies λ1 =· · · = λm = 0

You have probably seen these definitions before, but let’s reflect on them a bitmore anyway Note first that ∅ and any singleton set other than {0} is a linearlyindependent set in any linear space Moreover, it follows from these definitions thatany set S in a linear space X with 0 ∈ S is linearly dependent in X (Why?) Forany distinct x, y ∈ X\{0}, the set {x, y} is linearly dependent in X iff y ∈ span({x})

iff span({x}) = span({y}) Hence, one says that two nonzero vectors x and y arelinearly dependent iff they both lie on a line through the origin More generally, asubset S of X\{0} is linearly dependent in X iff there exists an x ∈ S such that x ∈span(S\{x})

A fundamental principle of linear algebra is that there cannot be more than mlinearly independent vectors in a linear space spanned by m vectors Many importantfindings concerning the structure of linear spaces follow from this observation

Proposition 2 Let X be a linear space, and A,B ⊆ X If B is linearly independent

in X and B ⊆ span(A), then |B| ≤ |A|

Proof Assume that B is linearly independent in X and B ⊆ span(A) If A =

∅, then the latter hypothesis implies that either B = ∅ or B = {0} By linearindependence of B, then, B = ∅, so the claim follows Similarly, if |A| = ∞, there isnothing to prove So let m := |A| ∈ N, and enumerate A as {x1, , xm} To derive acontradiction, let us assume that |B| > m Take any y1

∈ B We have y1 = 0,because

B is linearly independent Since y1

∈ span(A), therefore, it can be written as a linearcombination of x1, , xm,with at least one coefficient of the linear combination beingnonzero This implies that at least one member of A, say x1 (relabeling if necessary),can be written as a linear combination of y1, x2, , xm (Why?) Thus,

span(A) = span({y1, x2, , xm})

(Why?) But repeating the same reasoning successively (and keeping in mind that B

is linearly independent), we would find vectors y2, , ym

∈ B withspan(A) = span({y1, , ym})

(How, exactly?) This means that any y ∈ B\{y1, , ym} (and there is such a ybecause |B| > m) can be written as a linear combination of y1, , ym, contradictingthe linear independence of B.

Recall that, for any n ∈ N and i ∈ {1, , n}, the ith unit vector in Rn— denoted

by ei — is the n-vector whose ith term equals 1 and whose all other terms equal 0.Obviously, e1, , en

are linearly independent in Rn.It then follows from Proposition

3 that there can be at most n linearly independent vectors in Rn, or, equivalently,any S ⊆ Rn

with |S| > n is linearly dependent in Rn

Exercise 14 (a) Iff, g ∈ C(R)are defined byf (t) := et andg(t) := tet,show that

Dhilqlwlrq Let X be a linear space A finite subset T of X is called affinelyindependent in X if {z − x : z ∈ T \{x}} is linearly independent in X for any

x∈ T An arbitrary nonempty subset S of X is affinely independent in X if everyfinite subset of S is affinely independent in X S is called affinely dependent in X

if it is not affinely independent in X

Since, at most n vectors can be linearly independent in Rn, n = 1, 2, , it followsthat there can be at most n + 1 affinely independent vectors in Rn It is also useful

8 In view of (3), this fact is hardly surprising But still, please provide a proof for it.

to note that a nonempty subset S of a linear space X is affinely independent in X

iff, for every nonempty finite subset T of S and α ∈ RT,

Sx∈T

α(x)x = 0 and S

x∈Tα(x) = 0 imply α = 0

(Prove!9)

Recall that we have invoked Carathéodory’s Theorem in Section E.5.3 when ing Kakutani’s Fixed Point Theorem, but, shamelessly, omitted its proof Equippedwith the notion of affine independence, we are now ready to settle that score

prov-E{dpsoh 4 (Proof of Carathéodory’s Theorem) Let us first recall what y’s Theorem is about Fix any n ∈ N and recall that, for any subset A of Rn, co(A)corresponds to the set of all convex combinations of finitely many members of A.Carathéodory’s Theorem states the following:

Carathéodor-Given any nonempty S ⊆ Rn

, for every x ∈ co(S) there exists a Tx ⊆ S suchthat x ∈ co(Tx) and |Tx| ≤ n + 1

Here goes the proof Take any x ∈ co(S) By definition, there exists a finite T ⊆ Ssuch that x ∈ co(T ) Let Tx be a ⊇-minimal such T (Thanks to finiteness, such a Txmust exist Right?) Then there exists a λ : Tx → R++ such that

x = Sy∈T x

and C := {y ∈ Tx : λ(y)− θα(y) = 0}

Clearly, θ > 0 and C = ∅ Moreover, by definition of θ, λ(y)−θα(y) ≥ 0 for all y ∈ Tx

x = Sy∈T x

λ(y)y− θ S

y∈T x

α(y)y = S

y∈T x \C(λ(y)− θα(y)) y,

so it follows that x ∈ co(Tx\C) Since C = ∅, this contradicts the ⊇-minimality of

Tx, and the proof is complete

Actually, this argument delivers a bit more than what is claimed by y’s Theorem It shows that what matters is not the underlying linear space Rn, butrather the affine hull of S Put more precisely, what we have established above is this:For every x ∈ co(S) there exists a Tx ⊆ S such that x ∈ co(Tx) and |Tx| ≤ m(S) +1,where m(S) is the maximum number of affinely independent vectors in S.10

Carathéodor-∗ Exercise 17.H (Radon’s Lemma) Let n ∈ N Given any nonempty S ⊆ Rn with

|S| ≥ n+2,there exist two disjoint setsAandBinSsuch thatco(A)∩ co(B) = ∅.11

Radon’s Lemma makes it quite easy to prove Helly’s Intersection Theorem, one of the most famous results of convex analysis.

∗ Exercise 18.H Letn∈ N.Prove:

(a) (Helly’s Intersection Theorem) If S is a finite class of convex subsets ofRn such that|S| ≥ n + 1, andW

T = ∅ for anyT ⊆ S with|T | = n + 1,thenW

1.6 Bases and Dimension

Since span(S) is the linear space that consists of all finite linear combinations of thevectors in S, all there is to know about this space is contained in S Thus if we wish

to think of a linear space as a span of a certain set, then it makes sense to choose thatset as small as possible This would allow us to economically “represent” the linearspace that we are interested in, thereby letting us view S as some sort of “nucleus”

of the space that it spans This prompts the introduction of the following importantconcept

Dhilqlwlrq A basis for a linear space X is a ⊇-minimal subset of X that spans X.That is, S is a basis for X if, and only if,

(i) X = span(S); and

(ii) If X = span(T ), then T ⊂ S is false

10 There are various refinements of Carathéodory’s Theorem that posit further structure on S and provide better bounds for the number of elements of S that is needed to express any given x ∈ co(S)

as a convex combination For instance, one can show that if S ⊆ R n is connected, then every x ∈ co(S) can be expressed as a convex combination of at most n members of S This is (a special case of) the Fenchel-Bunt Theorem.

11 Check first that the claim is true in R 2 with a set S that consists of 4 vectors, but it is false with a set of cardinality 3!

Dhilqlwlrq If a linear space X has a finite basis, then it is said to be finite mensional, and its dimension, dim(X), is defined as the cardinality of any one ofits bases If X does not have a finite basis, then it is called infinite dimensional,

di-in which case we write dim(X) = ∞

It may seem like there is an ambiguity in the definition of “the” dimension of afinite dimensional linear space X What if two bases for X have different number

of elements? In fact, there is no ambiguity, for the cardinality of any two bases for

a linear space must indeed coincide The following characterization of a basis for alinear space will help settle this matter

Proposition 3 A subset S of a linear space X is a basis for X if, and only if, S islinearly independent and X = span(S)

Exercise 19 Prove Proposition 3.

Combining Propositions 2 and 3, we obtain the following fact which removes thepotential ambiguity that surrounds the notion of dimension

Corollary 1.Any two bases of a finite dimensional linear space have the same number

of elements

It is important to note that a nontrival linear space admits numerous bases Infact, in an m-dimensional linear space, m ∈ N, any linearly independent set of vectors

of cardinality m is itself a basis For, if {x1, , xm

} is a basis for a linear space X, and{y1, , ym} is an arbitrary linearly independent set in X, then the argument outlined

in the proof of Proposition 2 shows that

span({x1, , xm}) = span({y1, , ym})

so, by Proposition 3, {y1, , ym

} is a basis for X

E{dpsoh 5 [1]∅ is a basis for the trivial space {0} and hence dim({0}) = 0

[2]For any n ∈ N, a basis for Rnis the set of all unit vectors S = {ei : i = 1, , n}where ei is the real n-vector the ith component of which equals 1 and the rest of thecomponents of which are 0 It is easy to check that S is linearly independent in Rn,and, since x =Sn

xiei for all x ∈ Rn, we have span(S) = Rn Thus, by Proposition

3, S is a basis for Rn

— it is called the standard basis for Rn We thus have thefollowing comforting fact: dim(Rn) = n

As noted above, any linearly independent subset S of Rn with n elements serves

as a basis for Rn For instance, {iei : i = 1, , n} is a basis for Rn which is distinctfrom the standard basis

[3]Fix m ∈ N, and let X denote the set of all polynomials on R which are of degree

at most m If fi ∈ R[0,1] is defined by fi(t) := ti

for each i = 0, , m, then {f0, , fm}

is a basis for X (Note Linear independence of {f0, , fm} follows from the fact thatany polynomial of degree k ∈ N has at most k roots.) We have dim(X) = m + 1

[4] Let e1 := (1, 0, 0, ), e2 := (0, 1, 0, ), etc It is easily checked that {ei : i =

1, 2, } is linearly independent in p, 1≤ p ≤ ∞

For any given 1 ≤ p ≤ ∞, what is the dimension of p? Suppose dim( p) = m forsome m ∈ N Then, as we noted above, any linearly independent subset of p withcardinality m, say {e1, , em}, would qualify as a basis for p But, of course, this isimpossible; em+1 cannot be written as a linear combination of e1, , em.Conclusion:For any 1 ≤ p ≤ ∞, dim( p) =∞

[5] Let T be any nonempty set Is the linear space B(T ) finite dimensional? Thediscussion of [2] and [4] is easily adapted here to yield the complete answer If T isfinite, then {1{t} : t∈ T } is a basis for B(T ), where 1{t} is the indicator function of{t} in T (Why?) Conversely, if |T | = ∞, then {1{t} : t ∈ T } is linearly independent

in B(T ), and hence B(T ) is an infinite dimensional linear space Conclusion: Forany nonempty set T, dim(B(T )) = |T | 12

Remark 1 The dimension of an affine manifold S in a linear space X, denoted bydim(S), is defined as the dimension of the linear subspace of X that is “parallel” to

S.Put more precisely,

\{∅}, we define dim(S) := dim(aff (S)) This is, of course, a generalization

of the previous definitions Indeed, by (6) and (5), for any x ∈ S,

dim(S) = dim(aff (S)) = dim(aff (S) − x) = dim(span(S − x))

For instance, dim({x}) = 0 and dim({λx + (1 − λ)y : 0 ≤ λ ≤ 1}) = 1 for any

x, y ∈ Rn, n = 1, 2, Less trivially, dim(O) = n for any open subset O of Rn.(Prove!)

The following simple uniqueness result will prove useful on many occasions

Corollary 2 Let S be a basis for a linear space X Any nonzero vector x ∈ X can

be expressed as a linear combination of finitely many members of S with nonzero

12 We will sharpen this observation in Chapter J It turns out that there does not exist even a countably infinite basis for B(T ) when T is infinite.

coefficients in only one way.13 If X is finite dimensional, then every vector in X can

be uniquely written as a linear combination of all vectors in S

Proof We only need to prove the first assertion Take any x ∈ X\{0} Sincespan(S) = X, there exists a finite subset A of S and a map λ : A → R\{0} such that

x =S

y∈Aλ(y)y.Now suppose that B is another finite subset of S and α : B → R\{0}

is a map such that x =S

z∈Bα(z)z ThenS

with the convention that any of the three terms on the left hand side equals 0 if it is

a sum over the empty set Since, by Proposition 3, S is linearly independent in X, so

is (A\B) ∪ (B\A) ∪ (A ∩ B) It follows that λ(y) = 0 for all y ∈ A\B, α(z) = 0 forall z ∈ B\A, and λ(w) = α(w) for all w ∈ A ∩ B Since neither λ nor α ever takesvalue zero, this can happen only if A = B and λ = α

Exercise 20 Show that{f0, f1, }is a basis forP[0, 1],wherefi ∈ P[0, 1]is defined

by fi(t) := ti, i = 0, 1, Conclude that dim(P[0, 1]) = ∞

Exercise 21 Show that ifY is a proper linear subspace of a finite dimensional linear spaceX, thenY is also finite dimensional anddim Y < dim X

Exercise 22 Show that every linearly independent set in a finite dimensional linear space can be extended to a basis for the entire space.

Exercise 23 H (Products of Linear Spaces) For any linear spacesX andY,we define the following addition and scalar multiplication operations onX× Y:

(x, y) + (x , y ) := (x + x , y + y ) and λ(x, y) := (λx, λy)

for all x, x ∈ X, y, y ∈ Y, and λ ∈ R.14 Under these operations X × Y becomes

a linear space — it is called the linear product (or direct sum) ofX andY (This space is often denoted asX⊕Y in linear algebra, but we shall stick with the notation

X× Y instead.) Show that

dim(X × Y ) = dim(X) + dim(Y )

13 If x = 0, we obviously need to allow for zero coefficients, hence in this case the claim needs to

be altered slightly (because 0 = 0x = 0x + 0y for any x, y ∈ S) This trivial problem doesn’t arise in the finite dimensional case, because we can then use all the basis elements in the associated linear expressions.

14 Of course, we use the addition operation on X when computing x + x , and that on Y when writing y + y , even though we do not use a notation that makes this explicit A similar remark applies to λx and λy as well.

It is a triviality that every finite dimensional linear space has a basis In fact, byExercise 22, we can pick an arbitrary vector x in such a space, and then pick anothervector y such that {x, y} is linearly independent (provided that the dimension of thespace exceeds 1, of course), and then pick another, and so on, secure in the knowledgethat we will arrive at a basis in finitely many steps But what if the space at hand

is not finite dimensional? In that case it is not at all obvious if this procedure wouldactually get us anywhere Well, it turns out that it would work out just fine, if one isready to invoke the Axiom of Choice (Section A.1.7) Our final task in this section is

to demonstrate how a basis can be obtained for any linear space by using this axiom.But be warned that “how” is actually the wrong word here, for our argument willhardly be constructive We shall rather use the Axiom of Choice to establish that onecan always find a basis for a linear space (exploiting this property as usual “to go fromfinite to infinite”), but you will notice that the argument will be silent about how toachieve this in actuality At any rate, the following result is a beautiful illustration

of the power of the Axiom of Choice, or of its equivalent formulation, Zorn’s Lemma

Theorem 1 Every linear space has a basis

Proof Let X be a linear space, and pick any linearly independent set Y in X

We would like to show that Y can be extended to a basis for X Towards this end,denote by Y the set of all linearly independent sets in X that contain Y Since Y ∈ Y,

we have Y = ∅ Moreover, (Y, ⊇) is obviously a poset Let (Z, ⊇) be a loset suchthat Z ⊆ Y We claim that V

Z ∈ Y, that is, it is a linearly independent set in

X To see this, pick arbitrary z1, , zm inV

Z, and consider any linear combination

Sm

λizi that equals 0 By definition, each zi belongs to some linearly independentset in X, say Zi Since Z is linearly ordered by ⊇, it is without loss of generality tolet Z1 ⊆ · · · ⊆ Zm But then all zis must belong to Zm, and since Zm is a linearlyindependent set, it follows that λ1 = · · · = λm = 0 This proves our claim, andlets us conclude that any loset in (Y, ⊇) has an ⊇-upper bound in Y But then, byZorn’s Lemma, (Y, ⊇) must possess a ⊇-maximal element, say S This set is obviouslylinearly independent in X Moreover, we have X = span(S), because if there existed

an x ∈ X\span(S), then S ∪ {x} would be a linearly independent set which wouldcontradict the ⊇-maximality of the set S in Y Thus S is a basis for X

Please note that this proof yields something even stronger than Theorem 1:

Any linearly independent subset Y of a linear space X can be enlarged to

• For any m ∈ N, s1, , sm ∈ S and q1, , qm ∈ Q, we have Sm

(b) Let S be a Hamel basis for R Then, for every nonzero x ∈ R, there exist

a unique mx ∈ N, a unique subset {s1(x), , sm x(x)} of S, and a unique subset

{q1(x), , qm x(x)} ofQ\{0}such thatx =Sm xqi(x)si(x) Define the self-map f

onRbyf (x) :=Sm x

qi(x)f (si(x)),wheref is defined arbitrarily onS,except that

f (s) = 0 and f (s ) = 1 for some s, s ∈ S (We have|S| ≥ 2, right?) Show that

f (x + y) = f (x) + f (y)for allx, y ∈ R,but there is noα∈ Rsuch thatf (x) = αx

for allx∈ R.Conclusion: Cauchy’s Functional Equation admits nonlinear solutions (Compare with Lemma D.2).

∗ Exercise 25.16 Prove: If A and B are bases for a linear space,then A ∼card B

2.1 Definitions and Examples

Since in a linear space we can talk about linear combinations of vectors, those tions that map one linear space into another in a way that preserves linear combina-tions are of obvious importance Such maps are said to be linear and are formallydefined as follows

func-Dhilqlwlrq Let X and Y be two linear spaces A function L : X → Y is called alinear operator(or a linear transformation) if

L(αx + x ) = αL(x) + L(x ) for all x, x ∈ X and α ∈ R, (7)

αiL

xifor all m ∈ N and (xi, αi)∈ X × R, i = 1, , m

The set L−1(0) is called the null space (or the kernel) of L, and is denoted bynull(L), that is,

null(L) := {x ∈ X : L(x) = 0}

16 This exercise, which generalizes Corollary 1, presumes familiarity with the cardinality theory sketched in Subsection B.3.1.

A real-valued linear operator is referred to as a linear functional on X.17

Notation The set of all linear operators from a linear space X into a linear space

Y is denoted as L(X, Y ) So, L is a linear functional on X iff L ∈ L(X, R)

Warning While, as is customary, we don’t adopt a notation that makes this explicit,the + operation on the left of the equation in (7) is not the same as the + operation

on the right of this equation Indeed, the former one is the addition operation on Xwhile the latter is that on Y (The same comment applies to the · operation as well

— we use the scalar multiplication operation on X when writing αx, and that on Ywhen writing αL(x).)

E{dpsoh 6 For any m, n ∈ N, A ∈ Rm×n and x ∈ Rn, we define

Ax :=

#nSj=1

a1jxj, ,

nSj=1

amjxj

$,

where A := [aij]m×n (Section A.1.6.) Then L : Rn

→ Rm defined by L(x) := Ax is alinear operator More importantly: All linear operators from Rn to Rm arise in thisway To see this, take any L ∈ L(Rn

, Rm), and define Li : Rn

→ R as the functionthat maps each x ∈ Rn to the ith component of L(x), i = 1, , m Since L is a linearoperator, each Li is a linear functional on Rn, and hence, for any x ∈ Rn, we have

Li(x) = Li

#nSj=1

xjej

$

=nSj=1

Li(ej)xj

for each i = 1, , m (Here {e1, , en

} is the standard basis for Rn (Example 5.[2]).)Therefore, letting A := [Li(ej)]m×n, we find that L(x) = Ax for all x ∈ Rn

In particular, L : Rn → R is a linear functional on Rn iff there exist real numbers

α1, , αn such that L(x) = α1x1+· · · + αnxn for all x ∈ Rn.Put slightly differently,and using the standard inner product notation, we can say that L ∈ L(Rn

, R) iffthere is an n-vector ω such that L(x) = ωx for all x ∈ Rn.18

E{dpsoh 7 [1] Recall that C1[0, 1] is the linear space of all continuously tiable functions on [0, 1], and define D : C1[0, 1]→ C[0, 1] by D(f) := f Then D is

differen-17 Quiz What is the connection between the notion of a linear operator and that of a phism (Exercise 5)?

homomor-18 This is not the whole story Suppose L was a linear functional defined on a linear subspace Y

of R n What would it look like then? If you think about it for a moment, you will notice that the answer is not obvious After all, Y need not have some (or any) of the unit vectors of R n , and this complicates things But, with some effort, one can still show that, even in this case, we can find an

ω ∈ R n such that L(x) = ωx for all x ∈ Y (This vector will, however, not be uniquely determined

by L, unless dim(Y ) = n.) In the following section I will prove something even stronger than this (Example 9).

a linear operator — it is called the differentiation operator on C1[0, 1] It is easilyseen that null(D) equals the set of all constant real maps on [0, 1].

[2] Let L1, L2 ∈ RC[0,1] be defined by

L1(f ) := f (0) and L2(f ) :=

] 1 0

f (t)dt

Both L1and L2are linear functionals on C[0, 1] (Linearity of L2follows from ExerciseA.57.) Clearly, null(L1) ={f ∈ C[0, 1] : f(0) = 0} On the other hand, null(L2) isthe class of all continuous functions on [0, 1] whose integrals vanish In particular, byExample A.61, null(L2)∩ R[0,1]+ ={0}

E{dpsoh 8 In the following examples X and Y stand for arbitrary linear spaces

[1]Any L ∈ L(X, Y ) maps the origin of X to that of Y, that is, 0 ∈ null(L) For,L(0) = L(0 + 0) = L(0) + L(0) = 2L(0) so that L(0) = 0 (Here, of course, 0 onthe left hand side is the origin of X, and 0 on the right hand side is the origin ofY.) Therefore, there is only one constant linear operator from X into Y ; the one thatmaps the entire X to the origin of Y If Y = R, then this functional is equal to zeroeverywhere, and it is called the zero functional on X

[2] Suppose L ∈ L(X, Y ) is injective Then, since L(0) = 0, we must havenull(L) = {0} Interestingly, the converse is also true Indeed, if null(L) = {0}, then,for any x, x ∈ X, L(x) = L(x ) implies L(x − x ) = 0 by linearity of L, so x − x = 0.Conclusion: An L ∈ L(X, Y ) is injective iff null(L) = {0}

[3] For any L ∈ L(X, Y ), null(L) is a linear subspace of X, and L(X) is a linearsubspace of Y (Proofs?)

Exercise 26.H Let Y be the set of all polynomials on R of degree at most2 Define the mapL : R2×2→ Y by L([aij]2×2)(t) := a21t + (a11+ a12)t2 for all t∈ R

(a) Show thatL∈ L(R2, Y )

(b) Compute null(L) and find a basis for it.

(c) ComputeL(R2×2) and find a basis for it.

Exercise 27 Let X denote the set of all polynomials onR Define the self-maps D

and L on X by D(f ) := f and L(f )(t) := tf (t) for all real t, respectively Show thatD and Lare linear operators and computeD◦ L − L ◦ D

Exercise 28.H LetK andLbe two linear functionals on a linear spaceX,and define the functionKL∈ RX byKL(x) := K(x)L(x).Prove or disprove: KL is a linear functional onX iff it is the zero functional onX

Exercise 29 Let X be the linear space of all polynomials onR of degree at most 1 Define the linear functionalsK and L on X by K(f ) := f (0) and L(f ) := f (1),

respectively Show that {K, L} is a basis for L(X, R) How would you express

L1, L2 ∈ L(X, R)with

L1(f ) :=

] 1 0

f (t)dt and L2(f ) := f ,

as linear combinations of K and L?

The following is a very important result — it is a version of the Fundamental Theorem

of Linear Algebra.

Exercise 30.H Prove: Given any two linear spaces X and Y,

dim(null(L)) + dim(L(X)) = dim(X) for any L∈ L(X, Y )

Exercise 31.H (Linear Correspondences) Let X and Y be two linear spaces, and

Γ : X ⇒ Y a correspondence We say thatΓ is a linear correspondence if

αΓ(x)⊆ Γ(αx) and Γ(x) + Γ(x )⊆ Γ(x + x ) (8)

for all x, x ∈ X and α∈ R

(a) Show that the following statements are equivalent for a linear correspondence

Γ : X ⇒ Y: (i)Γis single-valued; (ii)Γ(0) ={0};(iii) we can replace each⊆with

=in (8).

(b) Show that the following are equivalent for anyΓ : X ⇒ Y: (i) Γ is linear; (ii)

Gr(Γ)is a linear subspace ofX× Y (Exercise 23); (iii)αΓ(x) + Γ(x ) = Γ(αx + x )

for all x, x ∈ X and α∈ R\{0}

(c) Show that if Γ : X ⇒ Y is linear, thenΓ(0) and Γ(X)are linear subspaces of

Y,and Γ(x) = y + Γ(0)for any y∈ Γ(x)

(d ) Show that ifL∈ L(X, Y ),thenL−1 is a linear correspondence from Y intoX

Also prove that, for any linear subspace Z of Y, the correspondence Γ : X ⇒ Y

2.2 Linear and Affine Functions

In Example 6 we have derived a general characterization of all linear functionalsdefined on a Euclidean space While it is extremely useful, this result does notreadily identify the structure of linear functions defined on an arbitrary subset of agiven Euclidean space We need a slight twist in the tale for this

Dhilqlwlrq Let S be a nonempty subset of a linear space X, and denote by P(S)the class of all nonempty finite subsets of S A real map ϕ ∈ RS is called linear if

ϕ

Sx∈Aλ(x)x



x∈Aλ(x)ϕ(x)

for any A ∈ P(S) and λ ∈ RA such thatS

x∈Aλ(x)x∈ S.19

E{dpsoh 9 Let n ∈ N and ∅ = S ⊆ Rn Under the shadow of Example 6, we havethe following characterization: ϕ ∈ RS is linear iff there exist real numbers α1, , αnsuch that L(x) =Sn

L(x) := S

z∈A

λx(z)ϕ(z)

L extends ϕ, because, for any x ∈ S, we have x =S

z∈Aλx(z)z,so, by linearity of ϕ,

L(x) = S

z∈A

λx(z)ϕ(z) = ϕ

Sz∈A



x∈Aλ(x)ϕ(x)

for any A ∈ P(S) and λ ∈ RA such thatS

x∈Aλ(x) = 1 and S

x∈Aλ(x)x∈ S

19 This definition is duly consistent with that of a linear functional When S is a linear subspace

of X, f ∈ L(S, R) iff f ∈ R S is linear.

Recall that we have motivated affinity as the combination of linearity and lation in the previous section Precisely the same idea applies here A map ϕ on alinear space X is affine iff its graph is an affine manifold on X × R, that is, it is atranslation of a linear subspace of X × R (Exercise 23).20 It follows that ϕ is affine

trans-iff ϕ − ϕ(0) is a linear functional on X It is important to note that this fact remainstrue even if the domain of the affine map is not a linear space

Lemma 1 Let T be a subset of a linear space with 0 ∈ T Then ϕ ∈ RT is affine if,and only if, ϕ − ϕ(0) is a linear real map on T

Proof If ϕ ∈ RT

is affine, and A ∈ P(T ) and λ ∈ RA satisfy S

x∈Aλ(x)x ∈ T,then



1− Sx∈Aλ(x)

0



− ϕ(0)

x∈Aλ(x)ϕ (x) +



1− Sx∈Aλ(x)

ϕ(0)− ϕ(0)

x∈Aλ(x)(ϕ (x)− ϕ(0)),that is, ϕ − ϕ(0) is linear Conversely, if ϕ − ϕ(0) ∈ RT

is linear, and A ∈ P(T ) and

λ∈ RA satisfy S

x∈Aλ(x) = 1 andS

x∈Aλ(x)x∈ T, thenϕ

Sx∈Aλ(x)x



=

ϕ

Sx∈Aλ(x)x



− ϕ(0)

+ ϕ(0)

=

Sx∈Aλ(x)(ϕ (x)− ϕ(0))

+ ϕ(0)

x∈Aλ(x)ϕ (x)−

Sx∈Aλ(x)

ϕ(0) + ϕ(0)

x∈Aλ(x)ϕ (x) ,

that is, ϕ is affine

Here is a useful application of this result which generalizes the characterization oflinear functions that we obtained in Example 9

Proposition 4 Let n ∈ N and ∅ = S ⊆ Rn Then ϕ ∈ RS is affine if, and only if,there exist real numbers α1, , αn, β such that ϕ(x) =Sn

αixi+ β for all x ∈ S.Proof We only need to prove the “only if” part of the assertion Let ϕ ∈ RS

be an affine map Pick any x∗ ∈ S, let T := S − x∗, and define ψ ∈ RT by ψ(y) :=

20 No, this is not entirely obvious But all will become clear momentarily.