Chapter 1 Vector Spaces

1 Fields

The rational numbers Q, the real numbers R and the complex numbers C, (but notthe natural numbers N or the integers Z) with the usual operations of addition andmultiplication, provide examples of a mathematical structure called a field.

Definition 1.1. (Field) A field is a set F together with two binary operations,addition and multiplication, with the following properties.

(1) Addition: To every pair a and b in F there corresponds an element a + b of Fsuch that

(i) addition is associative: (a + b) + c = a + (b + c);

(ii) addition is commutative: a + b = b + a;

(iii) there exists an element (additive identity ) 0 of F such that a + 0 = a forevery a in F ;

(iv) for every a in F there is an element (additive inverse ) −a of F such thata + (−a) = 0.

(2) Multiplication: To every a and b in F there corresponds an element ab of Fsuch that

(i) multiplication is associative: (ab)c = a(bc);

(ii) multiplication is commutative: ab = ba;

(ii) there exists an element (multiplicative identity ) 1 of F , with 1 6= 0, suchthat a1 = a for all a in F ;

(iv) for every a in F other than 0 there is an element (multiplicative inverse )a−1 of F such that aa−1 = 1.

(3) Multiplication distributes over addition: a(b+c) = ab+ac. Note: Multiplicationtakes precedence over addition. Thus, ab + ac is read as (ab) + (ac).

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Remark (Notation) We write a − b for a + (−b); 1a

or 1/a for a−1; ab

or a/b forab−1; x2 for xx, a3 for xxx and so on.

Exercises Prove that the following hold in any field F . Below, a, b and c areelements of F .

1. Additive and multiplicative identities are unique.

2. Additive and multiplicative inverses are unique.

3. If a + b = a + c then b = c.

4. a0 = 0.

5. (−1)a = −a.

6. (−a)(−b) = ab.

7. If ab = 0 then a = 0 or b = 0.

2 Vector Space

Definition 2.1. A vector space over the field F (called the set of scalars ) is a setV (whose elements are called vectors ) together with two binary operations, vectoraddition and scalar multiplication, with the following properties.

(1) Vector addition: To every x and y in V there corresponds an element x + yof V such that

(i) vector addition is associative: (x + y) + z = x + (y + z);

(ii) vector addition is commutative: x + y = y + x;

(iii) there exists an element (additive identity or origin) 0 of V such that x+0 =x for every x in V ;

(iv) for every x in V there is an element (additive inverse) −x of V such thatx + (−x) = (−x) + x = 0. We may write x− y for x + (−y).

(2) Scalar multiplication: To every a in F and y in V there corresponds anelement ax of V such that

(i) scalar multiplication is associative: a(bx) = (ab)x;

(ii) 1x = x for every vector x.

(3) Scalar multiplication distributes over vector addition: a(x + y) = ax +ay.

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(4) Scalar multiplication distributes over addition of scalars: (a + b)x =ax + ay.

Examples Note that a field must have at least two elements (0 and 1), while a vectorspace needs only have one (the origin). An example of a vector space over a fieldF is the field F itself. A less interesting example is {0}(the singleton of 0 in F ).A less trivial example can be constructed as follows. Let C(R) denote the set ofall continuous functions f : R → R with scalar multiplication and vector additiondefined by

(1) (af)(t) = af(t)

and(2) (f + g)(t) = f(t) + g(t)

for all real t. Since the function af is continuous whenever f is continuous andthe function f + g is continuous whenever f and g are continuous, C(R) with theoperations defined in (1) and (2) is a vector space over R. The zero function 0is defined by 0(t) = 0 for all real t. The additive inverse of f is −f , defined by(−f)(t) = −f(t) for all real t. By replacing “continuous” with “differentiable”, or“Riemann integrable”, we obtain more examples of vector spaces over R. Whether 0means the scalar or the vector additive identity is resolved by context.

If we drop continuity, we still have a vector space. The set RR of all function from Rto R with the operations defined by (1) and (2) is a vector space over R.(We can replaceR with any field F .) The set RN of all functions f : N → R, with scalar multiplicationand vector addition still defined by (1) and (2), is another example of a vector space.Replace f and g with x and y and write x == (ξ0, ξ1, · · · ) and y = (η0, η1, · · · ),where ξn = x(n) and ηn = y(n) for n ∈ N .Then, (1) and (2) can be restated asax = (aξ0, aξ1, · · · ) and x + y = (ξ0 + η0, ξ1 + η1, · · · ) Further,0 = (0, 0, · · · )and−x = (−ξ0, ξ1, · · · ). We can replace R with any field F to obtain FN , the vectorspace of all infinite sequences in F .

Consider Rn for n ∈ Z, with the operations of scalar multiplication and vectoraddition still defined by (1) and (2). It is a vector space.

3 Linear Dependence and Independence

Definition 3.1. A subset S = {x1, x2, · · · , xn} of a vector space V is linearly de-pendent if there exist scalars a1, a2, · · · , an, not all zero, such that

a1x1 + a2x2 + · · ·+ anxn = 0.

Otherwise S is called linearly independent.

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Any sum of the form a1x1 + · · · + anxn is called a linear combination of thevectors in S. The collection of all linear combinations of the vectors in S (whether Sis linearly independent or not) is called the span of S and is denoted by span S. Inother words,

spanS = {x ∈ V | x = a1x1 + · · ·+ anxn, for scalars a1, · · · an}.Remark 3.1. The condition that S = {x1, x2, · · · , xn} is linearly dependent can bewritten

(∃ai)(a1x1 + · · ·+ anxn = 0 ∧ ai 6= 0).

In other words, S is linearly independent iff for all scalars a1, · · · an,

a1x1 + · · ·+ anxn = 0

impliesa1 = · · · = an = 0

Remark 3.2. If S = {x1, x2, · · · , xn} is linearly independent then for all k, 1 ≤ k ≤ n,{x1, x2, · · · , xk} is linearly independent.

Indeed, if a1x1 + · · ·+ akxk = 0 then, by choosing ak+1 = · · · = an = 0, we have

0 = a1x1 + · · ·+ akxk = a1x1 + · · ·+ akxk + ak+1xk+1 + · · ·+ anxn,

which implies a1 = · · · = an = 0, since S is linearly independent.

Examples

Remark 3.3. Note that the empty set ∅ is linearly independent and that any setcontaining the zero vector is linearly dependent. If any xk in S = {x1, · · · , xn} is alinear combination of the other vectors, then

span {x1, · · · , xn} = span {x1, · · · , xk−1, xk+1, · · · , xn}.

In fact, if xk =n∑

i=1,i6=k

bixi, then

n∑i=1

aixi =n∑

i=1,i 6=k

aixi + ak

n∑

i=1,i6=k

bixi =n∑

i=1,i6=k

(ai + akbi)xi.

If x ∈ span X, where X = {x1, x2, · · · , xn} is linearly independent, then therepresentation

x = a1x1 + · · ·+ anxn

is unique, i.e.: x uniquely determines a1, · · · , an. Indeed, if we also have x = b1x1 +· · ·+ bnxn, then from the above, we obtain (a1− b1)x1 + · · ·+ (an− bn)xn = 0, whichimplies, thanks to the linear independence of X, that ai = bi for i = 1, · · · , n.

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Theorem 3.1. A set of vectors is linearly dependent iff one of the vectors is a linearcombination of the others.

Proof. If S = {x1, · · · , xn} is linearly dependent, then there exists ak 6= 0 for some ksuch that a1x1 + · · ·+ akxk + · · ·+ anxn = 0. Therefore,

xk =n∑

i=1,i 6=k

(− ai

ak

)xi.

If, on the other hand, one of the vectors in S is a linear combination of the others,we have

xk =n∑

i=1,i6=k

aixi,

for some collection a1, · · · , ak−1, ak+1, · · · , an of scalars. Then, by choosing ak = −1we have a1x1 + · · · + anxn = 0, where not all of the ai are zero, showing that S islinearly dependent.

4 Bases and Dimension

Definition 4.1. A linearly independent subset X of a vector space V is called a basisin V if every vector in V is a linear combination of the vectors in X. A vector spaceis finite-dimensional if it has a finite basis.

ExamplesEvery vector x in R3 is of the form x = (ξ1, ξ2, ξ3). Therefore, x = (ξ1, ξ2, ξ3) =

ξ1(1, 0, 0)+ ξ2(0, 1, 0)+ ξ3(0, 0, 1), showing that X = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} spansR3. Is X linearly independent? It is, since 0 = a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1) =(a1, a2, a3) implies a1 = a2 = a3 = 0.

For a natural number n, let Pn(R) denote the set of all polynomials of the form

p(t) = a0 + a1t + · · ·+ antn

in the real variable t, where a0, · · · , an are real numbers. In other words, Pn(R) is thesubset of RR consisting of all functions p : R → R : t → p(t) of the form polynomials.It is easy to verify that Pn(R) is a vector space over R.

For k = 0, · · · , n, define xk : R → R : t → tk . Clearly, each xk is an element ofPn(R). Moreover,

p(t) = a0x0(t) + a1x1(t) + · · ·+ anxn(t)

for all real t, showing that X = {x0, · · · , xn} spans Pn(R).If p(t) = a0 + a1t + · · · + ant

n = 0 for all t then p is the zero polynomial, i.e.:a0 = · · · = an = 0. (A fact we do not prove here.) This establishes the linearindependence of X. Hence, X is a basis in Pn(R).

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Theorem 4.1. If V is a finite-dimensional vector space and Y is any linearly inde-pendent subset of V then Y ⊂ S for some basis S in V . In other words, any linearlyindependent set of vectors in a finite-dimensional vector space that is not already abasis can be augmented in such a way that the augmented set is a basis.

Proof. Suppose that V is a finite-dimensional vector space and that Y = {y1, y2, · · · , yk}is a linearly independent subset of V . If the span of Y is V ,we take S = Y and weare finished. Otherwise, we proceed inductively.

Since V is finite-dimensional, it has a finite basis, say X = {x1, · · · , xn}. DefineX1 = X \ Y = {x1,1, · · · , x1,k}, the set obtained from X by removing those vectorsthat happen to be in both sets. Then, k1 < n. Define S1 = Y ∪ X1. Clearly,span S1 = V . If S1 is linearly independent, we are finished: Take S = S1.

Otherwise, we claim that one of the vectors in X1 is a linear combination of theothers. In fact, let x1,l is the first element in X1 such that {y1, · · · , yk, x1,1, · · · , x1,l} islinearly dependent. Then there exists scalars a1, · · · , ak, b1, · · · , bl, not all zero, suchthat

a1y1 + · · ·+ akyk + b1x1,1 + · · ·+ blx1,l = 0.

Since {y1, · · · , yk, x1,1, · · · , x1,l−1} is linearly independent, we deduce that bl 6= 0.Therefore x1,l is a linear combination of the others. This proves the claim.

Remove from from X1 one of the vectors that is a linear combination of the otherelements in S1 to obtain a proper subset X2 = {x2,1, · · · , x2,k2} of X1. Then, k2 < k1.Define S2 = Y ∪X2. Nothing has been lost in the sense that the span of S2 is thespan of S1, which is V .

By repeating this process, at some point, we arrive at a set Sm = Y ∪Xm, whereXm is a nonempty proper subset of X, the span of Sm is still V and Sm is linearlyindependent. We take S = Sm.

Theorem 4.2. If a vector space V is spanned by a subset of n vectors then everysubset of V that contains n + 1 vectors is linearly dependent.

Proof. Proof Suppose the proposition is false. Then, there is a vector space V spannedby some subset X0 = {x1, · · · , xn} and there is a linearly independent subset Y ={y1, · · · , yn+1} of V . Starting with X0, we construct inductively sets X1, · · · , Xn suchthat span X0 = span X1 = · · · = span Xn = V and such that each Xi is obtainedfrom X0 by replacing i of its elements with elements of Y .

To construct X1, we proceed as follows. Since X0 spans V , y1 is in the span ofX0. Hence, there are scalars a1 = 1 and b1, · · · , bn such that

a1y1 +n∑

k=1

bkxk = 0.

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If b1 = · · · = bn = 0 then a1 = 0, a contradiction. Hence some bk 6= 0 and xk in X0 isa linear combination of the other members of the set Z0 = X0∪{y1}. We may as wellsuppose that this element xk is xn (by renaming the elements of X0, if necessary).Remove xn from Z0: Define X1 = Z0 \ {xn}. Then, span X1 = span X0. Note that

X1 = {y1} ∪ ((X0 \ {xn}) = {y1, x1, · · · , xn−1}.

Suppose that for 0 ≤ i < n a set Xi has been defined which spans V and whichconsists of all yj with 1 ≤ j ≤ i and of n − i elements of X0, say x1, · · · , xn−i, i.e.:Xi = {y1, · · · , yi, x1, · · · , xn−i}. Since Xi span s V , yi+1 is in the span of Xi. Hence,there are scalars a1, · · · , ai+1 and b1, · · · , bi−1,with ai+1 = 1, such that

ai+1yi+1 +i∑

j=1

ajyj +n−i∑

k=1

bkxk = 0.

If b1 = · · · = bn−i = 0, then, since Y is linearly independent, a1 = · · · = ai+1 = 0, acontradiction. Therefore, some xk in Xi is a linear combination of the other elementsof Zi = Xi∪{yi+1}. We may as well suppose that this element xk is xn−i (by renamingthe elements of X0, if necessary).

Define Xi+1 = Zi \ {xn−i}. Then, span Xi+1 = span Zi = V and Xi+1 has theproperties stated for Xi with i replaced by i + 1. We have Xn = {y1, · · · , yn} withyn+1 not in the span of Xn, since Y is linearly independent, a contradiction, sinceXn spans V .

Corollary 4.3. If V is a finite-dimensional vector space then any two bases have thesame number of elements.

Proof. Suppose V is a finite-dimensional vector space with bases X, with m elements,and Y , with n elements. If m > n, then Xis linearly dependent, a contradiction.Hence, m ≤ n. If n > m then Y is linearly dependent, another contradiction. Hence,n ≤ m. We conclude that m = n.

Definition 4.2. The number of elements in a basis of a finite-dimensional vectorspace V is called the dimension of V and is denoted by dim X.

Theorem 4.4. In an n-dimensional vector space any set of n linearly independentvectors is a basis.

Proof. Let V be an n-dimensional vector space. Any set X of n linearly independentvectors in V can be extended to a basis S such that X ∈ S. The number of elementsof S is n, since V is n-dimensional. Hence, X = S, since X also has n elements andX ⊂ S. Therefore, X is a basis.

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Exercises

1. Given the vectors(1,1,1) and (1,0,1) in C3 find a third vector such that the threevectors are a basis.

2. Suppose {x, y, z} is a linearly independent set. Is the set {x + y, x + z, y + z}linearly independent also? Can you give a criterior to check the independencefor general case?

3. Show that the set {x0, x1, x2} in P (R) defined by x0(t) = 1, x1(t) = t2 andx2(t) = 1 + t + t2 is linearly independent.

4. Consider the subset of R3 given by S = {(1, a, a2), (1, b, b2), (1, c, c2)}. Imposeconditions on a, b and c so that S is linearly independent.

5. Prove that C(R) is not finite-dimensional.

5 Subspaces

Definition 5.1. A nonempty subset W of a vector space V is called a subspace of Vif, for any x, y ∈ W and scalar a, one has

(i) x + y ∈ W ; and

(ii) ax ∈ W .

Theorem 5.1. Suppose W is a subset of V . Then W is a subspace of V iff, for allx, y ∈ V , and all scalars a, b,

x, y ∈ W ⇒ ax + by ∈ W.

Examples.Suppose V is an n-dimensional vector space and W is a subspace of V . Since V

cannot contain n + 1 linearly independent vectors, neither can W . Hence, no basisin W can have more that n elements. This establishes that if W has a basis then thedimension of W cannot exceed the dimension of V , but does not establish that Whas a basis. In other words, to say that W is m-dimensional, we must show that Whas a basis with m elements. Where does such a basis come from?

This issue is resolved below.

Theorem 5.2. If W is a subspace of the finite-dimensional vector space V thendim W ≤ dim V :

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Proof. Suppose that V is a vector space of dimension n and that W is a subspaceof V . We already know that a subspace of a vector space is also a vector space.Therefore W is a vector space. We have to find a basis for it with at most n elements.

If W = {0} then W is zero-dimensional. Otherwise, W contains a nonzero vectorx1. If x1 spans W then W is one-dimensional. Otherwise, W contains a vector x2 notin the span of x1 (so that x1 and x2 are linearly independent). If W = span {x1, x2}then W is two-dimensional.

We repeat this process by finding m linearly independent vectors x1, · · · , xm inW , with m ≤ n. If W = span {x1, · · · , xm} then W is m-dimensional.

This process cannot last more than n steps, for once we obtain a linearly inde-pendent subset S = {x1, · · · , xm} of W (which must be also a subset of V )if S doesnot span W then there is a vector xm+1 in W with xm+1 not in the span of S. Ifm = n then V contains n + 1 linearly independent vectors, which is a contradiction.We have shown that W contains m linearly independent vectors, m ≤ n which spanW . Hence dim W ≤ dim V

Corollary 5.3. If V is a vector space of dimension n and W is a subspace of dimen-sion m then there is a basis {x1, · · · , xm, xm+1, · · · , xn} in V such that {x1, · · · , xm}is a basis in W .

Proof. According to the proposition, there is a basis {x1, · · · , xm} in W for somem ≤ n. Extend this linearly independent subset of V to a basis in V .

Remark 5.1. 1. If X and Y are subspaces of the vector space V , is X ∪ Y a subspaceof V ?

2. It is not clear whether the intersection X ∩ Y of two subspaces is a subspace.This is clarified below.

6 Intersection and Sum of Subspaces

Theorem 6.1. An intersection of subspaces of a vector space V is also a subspace ofV .

Proof. Let V be a vector space. Let {Xi | i ∈ I} be an arbitrary nonempty collectionof subspaces of V , where I is a index set. Let

X =⋂i∈I

Xi.

Every Xi is a vector space. Hence, 0 ∈ Xi for all i ∈ I. Therefore, 0 ∈ X. Thisshows that the intersection X is not empty.

Suppose that x and y are in X. Then x and y also belong to Xi for all i ∈ I.Therefore, for any pair of scalars a and b, the vector ax + by also belong to every Xi,

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since each is a vector space. It follows that ax + by belongs to X. We conclude thatX is a subspace of V .

Definition 6.1. If X and Y are subspaces of the vector space V , define the vectorspace

X + Y = {x + y ∈ V | x ∈ X and y ∈ Y }.In other words, X + Y = span (X ∪ Y ).

Definition 6.2. Two subspaces X and Y of a vector space V are called disjoint ifX

⋂Y = {0}. If X and Y are disjoint and X +Y = V then Y is called a complement

of X.

Remark 6.1. (a) Note that in the above definition X and Y are not disjoint sets(X

⋂Y 6= ∅)

(b) Do not confuse the relative complement of a set with the complement of asubspace relative to the whole space. A subspace of a vector space can have manycomplements. For example, consider the vector space R2 and the subspace X of allvectors of the form (ξ1, 0) (the horizontal axis). Other than the horizontal axis, everyline Y containing the origin is a complement of X and R2 = X + Y . To be specific,for real a let Ya denote the subspace consisting of all vectors of the form (ξ1, aξ1).Then, Y0 = X. Suppose that a 6= 0. Then, Ya is a complement of X.

ExampleIn C3, the vector space of all triples (ξ1, ξ2, ξ3) of complex numbers, defineX to be the subspace of all complex triples of the form (ξ1, ξ2, 0),

Y to be the subspace of all complex triples of the form (0, ξ2, 0) andZ to be the subspace of all complex triples of the form (0, 0, ξ3). Then,

1. Y ∩ Z = X ∩ Z = {0}.2. X ∩ Y is the subspace of V consisting of all triples of the form (0, ξ2, 0).3. Y +Z is the subspace of V consisting of all complex triples of the form (0, ξ2, ξ3).4. X + Y is the subspace of V consisting of all triples of the form (ξ1, ξ2, 0).5. X + Z = X + Y + Z = V .6. X ∩ (Y + Z) is the subspace of V consisting of all triples of the form (0, ξ2, 0).7. (X ∩ Y ) + (X ∩ Z) is the subspace of V consisting of all triples of the form

(0, ξ2, 0).

Theorem 6.2. (Dimension Formula) If X and Y are subspaces of the finite-dimensional vector space V then

dim X + dim Y = dim(X + Y ) + dim(X ∩ Y ).

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Proof. Assume the dimensions of X and Y are n1 and n2 and the dimension of X ∩Yis m. Choose a basis

{α1, α2, · · · , αm}of X ∩ Y . By Theorem ??, it can be augmented to form a basis

{α1, α2, · · · , αm, β1, β2, · · · , βn1−m}

of X. It can also be augmented to form a basis

{α1, α2, · · · , αm, γ1, γ2, · · · , γn2−m}

of Y .We claim that

{α1, α2, · · · , αm, β1, β2, · · · , βn1−m, γ1, γ2, · · · , γn2−m}

is a basis of X + Y . If the claim is true then the dimension of X + Y is n1 + n2 −mand the theorem is proved.

We prove the the claim. Since

X = span {α1, α2, · · · , αm, β1, β2, · · · , βn1−m}

andY = span {α1, α2, · · · , αm, γ1, γ2, · · · , γn2−m},

it is obvious that

X + Y = span {α1, α2, · · · , αm, β1, β2, · · · , βn1−m,

γ1, γ2, · · · , γn2−m}.

We only need to prove that

{α1, α2, · · · , αm, β1, β2, · · · , βn1−m, γ1, γ2, · · · , γn2−m}

is linearly independent. Assume that

k1α1 + · · ·+ kmαm + p1β1 + ·+ pn1−mβn1−m

+ q1γ1 + · · ·+ qn1−mγn2−m = 0

Let

α = k1α1 + · · ·+ kmαm + p1β1 + ·+ pn1−mβn1−m

= −q1γ1 − · · · − qn1−mγn2−m.

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From the first equality we know that α ∈ X and from the second equality we deducethat α ∈ Y . Therefore α ∈ X ∩ Y . Therefore, α can be expressed as a linearcombination of {α1, α2, · · · , αm}, that is

α = l1α1 + l2α2 + · · ·+ lmαm.

Then,l1α1 + l2α2 + · · ·+ lmαm + q1γ1 + · · ·+ qn1−mγn2−m = 0.

Since {α1, α2, · · · , αm, γ1, γ2, · · · , γn2−m} is linearly independent, l1 = l2 = · · · = lm =0 and q1 = q2 = · · ·+ qn2−m = 0. Therefore, α = 0. The first equality above becomes

k1α1 + · · ·+ kmαm + p1β1 + · · ·+ φn1−mβn1−m = α = 0.

It conclude that

k1 = k2 = · · · = km = p1 = · · · = pn1−m = 0,

since {α1, α2, · · · , αm, β1, β2, · · · , βn1−m} is linearly independent. Therefore

{α1, α2, · · · , αm, β1, β2, · · · , βn1−m, γ1, γ2, · · · , γn2−m}

is linearly independent. This completes the proof.

Corollary 6.3. Let V1 and V2 be subspaces of a n-dimensional vector space. If thesum of the dimensions of V1 and V2 is greater than the n, then V1 and V2 contains acommon nonzero vector

Proof. It leaves to you as a homework.

Exercises1. Find a vector space V and subspaces X, Y and Z such that

X ∩ (Y + Z) 6= (X ∩ Y ) + (X ∩ Z).

2. Show that if X and Y are two-dimensional subspaces of a three-dimensionalspace then X and Y are not disjoint.

3. Prove that if X is an m-dimensional subspace of an n-dimensional vector spacethen every complement of X has dimension n−m.

4. A function f in RR is called even if f(−t) = f(t) for all real t and is called oddif f(−t) = −f(t) for all real t. Prove that the even functions and the odd functionsare disjoint subspaces of RR .

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7 Direct Sums

An important case of sum of subspaces is the direct sum.

Definition 7.1. Let V1 and V2 be subspaces of a vector space V . The sum V1 + V2 iscalled a direct sum if, for any vector α in V1 + V2, there is a unique decomposition

α = α1 + α2,

where α1 ∈ V1 and α2 ∈ V2. The direct sum is denoted by V1

⊕V2.

Theorem 7.1. The sum V1 + V2 is the direct sum if and only if that

α1 + α2 = 0, αi ∈ Vi (i = 1, 2)

implies that α1 = α2 = 0.

Proof. The condition of the theorem states that the decomposition of the zero vectoris unique. It is necessary obviously. Let us prove the sufficiency.

Suppose that α ∈ V1 + V2 has two decompositions:

α = α1 + α2 = β1 + β2,

where αi, βi ∈ Vi, for i = 1, 2. Then (α1−β1)+ (α2−β2) = 0. Note that αi−βi ∈ Vi

for i = 1, 2. By the assumption of the theorem, we deduce that αi − βi = 0, i.e.,αi = βi for i = 1, 2. That is the decomposition is unique.

Corollary 7.2. Sum V1 + V2 is the direct sum if and only if V1

⋂V2 = {0}.

Proof. Sufficiency. Suppose that α1 + α2 = 0 and αi ∈ Vi for i = 1, 2. Then α1 =−α2 ∈ V1

⋂V2. By the assumption, we deduce that α1 = α2 = 0 That is the sum is

the direct sum.Necessity. For any vector α ∈ V1

⋂V2, we write 0 = α + (−α) where α ∈ V1

and −α ∈ V2. Since the sum V1 + V2 is the direct sum, we deduce that α = 0, i.e.,V1

⋂V2 = {0}.

Theorem 7.3. Let V1 and V2 be subspaces of a vector space V and W = V1 + V2.Then W = V1

⊕V2 if and only if

dim(V1 + V2) = dim V1 + dim V2.

Proof. It is a direct consequence of Dimension Formula.

13

Theorem 7.4. Let U be a subspace of a n-dimensional space W . There exists asubspace of W such that W = U

⊕V . In other words, every subspace of a finite-

dimensional vector space has a complement.

Proof. Let W be a vector space of dimension n and let U be a subspace of dimensionm. Then, there is a basis X = {x1, · · · , xm} in U . By a previous proposition, thereis a basis Z = {x1, · · · , xm, y1, · · · , yn−m} in W . Define

V = span {y1, · · · , yn−m}.

Then, the dimension of V is n−m since {y1, · · · , yn} is linearly independent.We need to show that W = U

⊕V . We show that for every z in W there is

exactly one pair of vectors x in U and y in V such that z = x + y.Indeed, since Z is a basis in W , for every z in W there are unique scalars a1, · · · , am

and b1, · · · , bn−m such that z = a1x1 + · · ·+ amxm + b1y1 + · · ·+ bn−myn−m Therefore,since U ∩ V = {0}, there are uniquely defined vectors x = a1x1 + · · · + amxm andy = b1y1 + · · ·+ bn−myn−m such that z = x + y.

The concept of direct sums can be extended to many subspaces.

Definition 7.2. Let V1, V2, · · · , Vs be subspaces of a vector space V . The sum V1+V2+· · ·+ Vs is called the direct sum of V1, V2, · · · , Vs if, for any vector α in V1, V2, · · · , Vs,there is a unique decomposition

α = α1 + α2 + · · ·+ αs,

where αi ∈ Vi for i = 1, 2, · · · s. The direct sum is denoted by V1

⊕V2

⊕ · · ·⊕ Vs.

Similarly, we have,

Theorem 7.5. Let V1, V2, · · · , Vs be subspaces of a vector space V . The the followingstatements are equivalent

(i) W =∑

Vi is the direct sum.

(ii) Decomposition of the zero vector is unique.

(iii) Vi

⋂ ∑j 6=i

Vj = {0} for i = 1, 2, · · · , s.

(iv) dim W = Σ dim Vi.

14

8 Direct Sums

We begin our presentation of direct sums by considering two examples.Example. Let F be a field. Define U = Fm, V = F n and W = Fm+n. If

x = (x1, · · · , xn) is a vector in U and y = (y1, · · · , ym) is a vector in V , the vectorz = (x1, · · · , xn, y1, · · · , ym) is a vector in W . We might write z = (x, y) and think ofthis construction as a method for obtaining longer vectors from shorter ones. Let ustemporarily give an unofficial definition of the symbol

⊕, by proposing to define this

method of obtaining W from U and V as the result of the operation W = U⊕

V .Each vector x in U can be identified with (x, 0) in W = U

⊕V and each vector y

in V can be identified with (0, y) in W = U⊕

V . Can this construction be extendedto a more general scheme that can be applied to every pair of vector spaces? Weanswer this question below. Before we do so, it is worthwhile to consider anotherexample.

5.2 Example Suppose we start with W = Fm+n. Define U to be the subspace ofFm+n consisting of all vectors x of the form x = (x1, · · · , xm, 0, · · · , 0). Define V to bethe subspace of Fm+n consisting of all vectors y of the form y = (0, · · · , 0, y1, · · · , yn).Then, U and V are disjoint subspaces of W and W = U + V , i.e.: U and V arecomplements of each other. Since U is isomorphic to Fm and V is isomorphic to F n,we can identify U with Fm and V with F n. But, if we do this, we can no longer writeW = U + V , since neither the vectors in U nor the vectors in V are necessarily alsovectors in W . (We can, however, write W = U

⊕V , consistently with the previous

example.)If we agree that this distinction is only technical and not substantive, we can

simply ignore it. Indeed, this is the course of action we shall take in what follows.

Remark 8.1. The operation U⊕

V described in the first example above is an exampleof an “external direct sum” of two vector spaces (over the same field), while theoperation U

⊕V described in the second example above is an example of an “interior

direct sum” of two disjoint subspaces of a vector space. Strictly speaking, they aredifferent operations.

Definition 8.1. Definition Let U and V be vector spaces over the same field. Wedefine their direct sum U

⊕V to be the collection of all ordered pairs (x, y) with x in

U and y in V . We make U⊕

V into a vector space by defining scalar multiplicationby a(x, y) = (ax, ay) and vector addition by (x, y) + (x′, y′) = (x + x′, y + y′).

The zero vector in U⊕

V is (0, 0). Note that• U is isomorphic to {(x, y) ∈ U

⊕V | y = 0},a subspace of U

⊕V , and

• V is isomorphic to {(x, y) ∈ U⊕

V | x = 0}, a subspace of U⊕

V .So, although neither U nor V is a subspace of U

⊕V , each is isomorphic to a subspace

of U⊕

V . With the proper identifications, by abuse of language, we speak of U and

15

V as being subspaces of U⊕

V . Note that U ∩ V = {0} (the singleton of the zerovector in U

⊕V ).

Now, let’s turn things around. If U and V are subspaces of a vector space W ,when can we say that W = U

⊕V ? We answer this question below.

Theorem 8.1. Let U and V be subspaces of a vector space W . The following threeconditions are equivalent.

(1) W = U⊕

V .

(2) U and V are complements of each other. That is, W = U +V and U ∩V = {0}.(3) For every vector z in W there is exactly one pair of vectors x in U and y in V

such that z = x + y.

Proof. We prove that (1) implies (2). Every element z of W can be written z =(x, y) = (x, 0) + (0, y). Hence, Z = U + V . If W = U

⊕V then z = (x, y) ∈ U and

z = (x, y) ∈ V imply y = 0 and x = 0, i.e., z = 0. Hence, U ∩ V = {0}.We prove that (2) implies (3). If W = U + V then every element z of W is of the

form z = x + y, with x in U and y in V . If z = x′ + y′ also, with x′ in U and y′ in V ,then x− x′ = y − y′. Since x− x′ is in U , y − y′ is in V , and U ∩ V = {0}, it followsthat x− x′ = 0 and y − y′ = 0, showing that the representation z = x + y is unique.

We prove that (3) implies (1). Suppose that for every z in W the representationz = x + y, with x in U and y in V is unique. Since we have agreed to identify everyx in U with (x, 0) in U

⊕V and to identify every y in V with (0, y) in U

⊕V , z can

be identified uniquely with (x, y), since (x, y) = (x, 0) + (0, y) = x + y = z. With thisidentification, the elements z of W are precisely the elements (x, y) of U

⊕V .

Theorem 8.2. If U and V are vector spaces over the same field then

dim U⊕

V = dim U + dim V.

Proof. It is a direct consequence of Dimension Formula.

Theorem 8.3. Let U be a subspace of a n-dimensional space W . There exists asubspace of W such that W = U

⊕V . In other words, every subspace of a finite-

dimensional vector space has a complement.

Proof. Let W be a vector space of dimension n and let U be a subspace of dimensionm. Then, there is a basis X = {x1, · · · , xm} in U . By a previous proposition, thereis a basis Z = {x1, · · · , xm, y1, · · · , yn−m} in W . Define

V = span {y1, · · · , yn−m}.

16

Then, the dimension of V is n−m since {y1, · · · , yn} is linearly independent.We need to show that W = U

⊕V . We show that for every z in W there is

exactly one pair of vectors x in U and y in V such that z = x + y.Indeed, since Z is a basis in W , for every z in W there are unique scalars a1, · · · , am

and b1, · · · , bn−m such that z = a1x1 + · · ·+ amxm + b1y1 + · · ·+ bn−myn−m Therefore,since U ∩ V = {0}, there are uniquely defined vectors x = a1x1 + · · · + amxm andy = b1y1 + · · ·+ bn−myn−m such that z = x + y.

9 Isomorphisms

A bijection between two sets is an isomorphism of sets, i.e.: two isomorphic sets canat most differ in the naming of their elements. A bijection between vector spaces isnot enough to consider the two vector spaces as differing at most by a renaming oftheir elements, the bijection must preserve the vector space structure. We elaborate

Definition 9.1. If V and W are vector spaces over the same field F ,a mappingT : V → W is called a linear transformation if

(1) T (x + y) = T (x) + T (y) for all x and y in V ; and

(2) T (ax) = aT (x) for all x in V all a in F .

Condition (1) and (2) is equivalent to the following• T (ax + by) = aT (x) + bT (y) for all x and y in V and all a and b in F .Note that in the left-handside of (1)the expression ax + by refers to scalar multi-

plication and vector addition in V , while in the right-hand side of (1) the expressionaT (x)+ bT (y) refers to scalar multiplication and vector addition in W . If, in additionto being linear, T is also bijective, it is called a linear isomorphism, or a vectorspace isomorphism between V and W , in which case the two spaces are said to beisomorphic.

Note that the inverse T−1 of a linear isomorphism T is linear. Indeed, the bijectionT has a bijective inverse T−1. To see that it is linear, evaluate T−1(ax′ + by′). IfT (x) = x′ and T (y) = y′ then T (ax + by) = aT (x) + bT (y) = ax′ + by′, which implies

T−1(ax′ + by′) = T−1(T (ax + by)) = ax + by = aT−1(x′) + bT−1(y′).

Remark 9.1. While an isomorphism of sets is any bijection between the two sets, anisomorphism of vector spaces is a linear bijection between the two vector spaces.

Remark 9.2. Suppose V , W and Z are vector spaces over the same field. If T1 : V →W and T2 : W → Z are linear isomorphisms then T2 ◦ T1 : V → Z is also a linearisomorphism.

17

Indeed, for all x and y in V and all scalars a and b we have

(T2 ◦ T1)(ax + by) =T2(T1(ax + by))

= T2(aT1(x) + bT1(y))

= aT2(T1(x)) + bT2(T1(y))

= a(T2T1)(x) + b(T2T1)(y))

Let us, but just temporarily, write V ∼ W to indicate that the vector spaces V andW (over the same field) are isomorphic. Then,

(i) V ∼ V (take T : V → V : x → x),

(ii) V ∼ W ⇒ W ∼ V (given T : V → W ,we have T−1 : W → V ), and

(iii) (V ∼ W ) and (W ∼ Z) ⇒ V ∼ Z (given T1 : V → W and T2 : W → Z,we haveT2 ◦ T1 : V → Z).

It is an equivalence relation over the set of all vector spaces.

Theorem 9.1. Every n-dimensional vector space V over the field F is isomorphic toF n.

Proof. Let V be an n-dimensional vector space over the field F and let

X = {x1, · · · , xn}be a basis in V . Define T : V → F n as follows. Every x in V has a unique represen-tation of the form

x = a1x1 + · · ·+ anxn.

DefineT (x) = (a1, · · · , an).

It is called the coordinate of x under the basis X.The mapping T is surjective, since for every (a1, · · · , an) in F n there is some x in

V such that T (x) = (a1, · · · , an). (In fact, x = a1x1 + · · ·+ anxn.)The mapping T is also injective, since the representation x = a1x1 + · · ·+ anxn is

unique. (Hence, T (x) 6= T (y) implies x 6= y.)Further, T is also linear. Indeed, if x = a1x1 + · · · + anxn and y = b1x1 + · · · +

bnxn,then

T (ax + by) = T (a(a1x1 + · · ·+ anxn) + b(b1x1 + · · ·+ bnxn))

= T ((aa1 + bb1)x1 + · · ·+ (aan + bbn)xn)

= ((aa1 + bb1), · · · , (aan + bbn))

= (aa1, · · · , aan) + (bb1, · · · , bbn)

= a(a1, · · · , an) + b(b1, · · · , bn)

= aT (x) + bT (y).

18

Theorem 9.2. Two finite dimensional spaces over a field F are isomorphic if andonly if the two spaces have the same dimension.

Exercises1. Prove that Pn(R) is isomorphic to Rn+1.2. Consider C as a vector space over R by defining scalar multiplication as the

usual multiplication of a real number by a complex number and by defining vectoraddition as the usual addition of complex numbers. Prove that (the vector space) Cis isomorphic to R2.

3. Prove that if X and Y are subspaces of a vector space V then X ∪ Y is asubspace of V iff X ⊂ Y or Y ⊂ X.

4. For any set V the power set P (V ) of V is the set of all subsets of V . If V is avector space, let S(V ) denote the set(a subsetof P (V )) of all subspaces of V . For afinite-dimensional vector space V define

σ : P (V ) → S(V ) : S → span S.

(a) Prove that σ(σ(S)) = σ(S) for every S in P (V ).(b) Prove that (S1 ⊂ S2) ⇒ σ(S1) ⊂ σ(S2) for all S1 and S2 in P (V ). (c) Prove

that σ(S1 ∪ S2) = σ(S1) + σ(S2) for all S1 and S2 in P (V ).

10 Linear Functionals

Definition 10.1. Let V be a vector space over the field F . A linear functional on Vis a function λ : V → F such that for all vectors x and y and for all scalars a and b

λ(ax + by) = aλ(x) + bλ(y).

The kernel of the linear functional λ is the set ker λ = {x ∈ V | λ(x) = 0}.A hyperplane is a subspace that has dimension one less than the whole space.

ExamplesEx. 1 On Cn, the vector space of all n-tuples x = (ξ1, · · · , ξn) of complex numbers,

the following are examples of linear functionals

(a) λ(x) = 0,

(b) λ(x) = ξ1,

1. λ(x) = a1ξ1 + · · ·+ anξn, for scalars a1, · · · , an.

19

Ex. 2 On C(R), the vector space of all continuous functions f : R → R, thefollowing are examples of linear functionals

(a) λ(f) = 0,

(b) λ(f) = f(0),

(c) λ(f) =b∫

a

α(t)f(t)dt, for any bounded interval (a, b) and any continuous a func-

tion α : [a, b] → R.

Remark 10.1. If λ1 and λ2 are linear functionals on a vector space V over a field F ,for any pair α and β of scalars, define λ : V → F by

λ(x) = αλ1(x) + λ2(x)

for all x in V . Then λ is a linear functional, denoted by λ = αλ1 + βλ2. Indeed,

λ(ax + by) = αλ1(ax + by) + βλ2(ax + by)

= αaλ1(x) + αbλ1(y) + βaλ2(x) + βbλ2(y)

= a(αλ1(x) + βλ2(x)) + b(αλ1(y) + βλ2(y))

= aλ(x) + bλ(y)

for all vectors x and y and for all scalars a and b. This suggests the definitions givenbelow.

Definition 10.2. Given a vector space V over the field F , denote by V ′ the collectionof all linear functionals on V . We make V ′ into a vector space over F by defining thefollowing.

(1) The zero element of V ′ is the linear functional 0, defined by 0(x) = 0 for all xin V .

(2) Scalar multiplication: For λ in V ′ and scalar a define aλ in V ′ by

(aλ)(x) = aλ(x)

for all x in V .

(3) Vector addition: For λ1 and λ2 in V ′ define λ1 + λ2 in V ′ by

(λ1 + λ2)(x) = λ1(x) + λ2(x)

for all x in V .

20

The vector space V ′ is called the dual space of V . We write 〈x, λ〉 for λ(x).

Remark 10.2. The reason for calling V ′ the dual space of V and for introducing thenotation 〈x, λ〉 becomes transparent when we observe that

〈ax1 + bx2, λ〉 = a〈x1, λ〉+ b〈x2, λ〉 (10.1)

and〈x, aλ1 + bλ2〉 = a〈x, λ1〉+ b〈x, λ2〉. (10.2)

In other words, we have a function 〈·, ·〉 : V × V ′ → F that is linear in the firstargument, i.e.: (10.1) holds, and is also linear in the second argument, i.e.: (10.2)holds. The function 〈·, ·〉 is called a bilinear functional. Later we will see an obviousreason to adopt this notation.

11 Dual Bases

Theorem 11.1. Let V be a vector space with basis {x1, · · · , xn}. To every set{a1, · · · , an} of scalars there corresponds a unique λ in V ′ such that

〈xi, λ〉 = ai (11.1)

for i = 1, · · · , n.

Proof. Suppose that V is a vector space over the field F and that {x1, · · · , xn} is abasis in V . Let {a1, · · · , an} be a given set of scalars.

We show existence. Define λ : V → F as follows. Every vector x in V has theunique representation x = b1x1 + · · ·+ bnxn. Define

〈x, λ〉 = b1a1 + · · ·+ bnan.

Then 〈xi, λ〉 = ai for i = 1, · · · , n, i.e.: λ satisfies (11.1).We show that λ is linear. Suppose that y is also a vector in V . Then, y has the

unique representation y = c1x1 + · · · + cnxn. Hence, for scalars a and b, the vectorax + by has the unique representation

ax + by = a(b1x1 + · · ·+ bnxn) + b(c1x1 + · · ·+ cnxn)

= (ab1 + bc1)x1 + · · ·+ (abn + bcn)xn.

Therefore,

〈ax + by, λ〉 = (ab1 + bc1)a1 + · · ·+ (abn + bcn)an

= a(b1a1 + · · ·+ bnan) + b(c1a1 + · · ·+ cnan)

= a〈x, λ〉+ b〈y, λ〉,

21

establishing the linearity of λ.We show uniqueness. Let λ1 and λ2 be linear functionals on V that satisfy (11.1).

Every x in V has the unique representation of the form x = b1x1 + · · ·+ bnxn. Hence,thanks to (11.1),

〈x, λ1〉 − 〈x, λ2〉 = (b1〈x1, λ1〉+ · · ·+ bn〈xn, λ1〉)− (b1〈x1, λ2〉+ · · ·+ bn〈xn, λ2〉)= (b1a1 + · · ·+ bnan)− (b1a1 + · · ·+ bnan)

= 0

Therefore, (11.1) implies 〈x, λ1〉 = 〈x, λ2〉 for every vector x in V , i.e.: λ1 = λ2.

Theorem 11.2. (Dual Basis) Let V be an n-dimensional vector space. To everybasis X = {x1, · · · , xn} in V there corresponds exactly one basis X ′ = {λ1, · · · , λn}in V ′ such that 〈xi, λj〉 = 0 for i 6= j and 〈xi, λi〉 = 1.

Proof. Thanks to the previous proposition, for each j, 1 ≤ j ≤ n,there is a unique λj

in V ′ such that 〈xi, λj〉 = 0 for i 6= j and 〈xi, λi〉 = 1. Define X ′ = {λ1, · · · , λn}. Weprove that X ′ spans V ′. Given any λ in V ′, define ai = 〈xi, λ〉 for i = 1, · · · , n. Then,since every x in V is of the form x = b1x1 + · · ·+ bnxn, we have

〈x, λ〉 = 〈b1x1 + · · ·+ bnxn, λ〉= b1〈x1, λ〉+ · · ·+ bn〈xn, λ〉= b1a1 + · · ·+ bnan

and, for i = 1, · · · , n,

〈x, λi〉 = 〈b1x1 + · · ·+ bnxn, λi〉= b1〈x1, λi〉+ · · ·+ bn〈xn, λi〉= bi.

Hence,

〈x, λ〉 = a1〈x, λ1〉+ · · ·+ an〈x, λn〉= 〈x, a1λ1 + · · ·+ anλn〉,

which holds for every x in V . Therefore, λ = a1λ1 + · · · + anλn. This shows that{λ1, · · · , λn} spans V ′.

We prove that X ′ is linearly independent. Suppose that

a1λ1 + · · ·+ anλn = 0. (11.2)

22

Then, for every x in V

〈x, a1λ1 + · · ·+ anλn〉 = a1〈x, λ1〉+ · · ·+ an〈x, λn〉.Hence, for x = xi the above yields

0 = a1〈xi, λ1〉+ · · ·+ an〈xi, λn〉 = ai〈xi, λi〉 = ai.

The above holds for i = 1, · · · , n. Hence, (11.2) implies a1 = · · · = an = 0. Thisshows that {λ1, · · · , λn} is linearly independent.

Corollary 11.3. The dimension of a finite-dimensional vector space and the dimen-sion of its dual space are the same.

Proof. X and X ′ have the same number of elements.

Theorem 11.4. Let V be a finite-dimensional vector space. For every nonzero x inV there exists a linear functional λ on V such that 〈x, λ〉 6= 0.

Proof. Let V be an n-dimensional vector space and let x be any vector in V . There isa basis X = {x1, · · · , xn} in V and a dual basis X ′ = {λ1, · · · , λn} in V ′. The vectorx has the unique representation x = a1x1 + · · · + anxn and we have 〈x, λi〉 = ai fori = 1, · · · , n. Therefore, with N = {1, · · · , n},

(∀λ ∈ V ′)(〈x, λ〉 = 0) ⇒ (∀i ∈ N)(〈x, λi〉 = 0) ⇒ (∀i ∈ N)(ai = 0) ⇒ x = 0.

Then, by contraposition,

x 6= 0 ⇒ (∃i ∈ N)(ai 6= 0) ⇒ (∃i ∈ N)(〈x, λi〉 6= 0) ⇒ (∃λ ∈ V ′)(〈x, λ〉 6= 0).

Corollary 11.5. If V is a finite-dimensional vector space then for any two distinctvectors x and y in V the exists a linear functional λ such that 〈x, λ〉 6= 〈y, λ〉.Proof. Let V be a finite-dimensional vector space and let x and y be elements of V .If x 6= y then x−y 6= 0 and there exists a λ in V ′ such that 〈x−y, λ〉 6= 0). Therefore,〈x, λ〉 6= 〈y, λ〉.

Exercises

1. Define a nonzero λ ∈ (R3)′ such that 〈x1, λ〉 = 〈x2, λ〉 = 0 for the two vectorsx1 = (1,−1, 1) and x2 = (−1, 1, 1).

2. Define λ in (Cn)′ by 〈x, λ〉 = ξ0 + · · ·+ ξn for all vectors x = (ξ0, · · · , ξn) in Cn.Find a basis for the subspace X = {x ∈ Cn| 〈x, λ〉 = 0}.

3. Consider the basis X = {(1, 0, 0), (1, 1, 0), (0, 1, 1)} in C3. What is the dualbasis X ′?

23

12 The Second Dual

V The dual V ′ of a vector space V is also a vector space. Therefore, V ′ has a dualspace (V ′)′, which we shall simply denote by V ′′ . The vector space V ′′ is called thesecond dual of V . So far, the only thing we know about the second dual is that, sincea finite-dimensional vector space and its dual have the same dimension, is that if Vis finite-dimensional then dim

dim V ′′ = dim V ′ = dim V.

Now, V , V ′ and V ′′ are vector spaces over the same field and, if V is finite dimensional,all three spaces have the same dimension. Hence, they are isomorphic. However, thisdoes not reveal a much deeper connection between the elements of V and those of V ′′,a correspondence we shall call “ natural isomorphism” between a finite-dimensionalvector space and its second dual.

Let V be a finite-dimensional vector space over the field F . For each x in V defineµx = 〈x, ·〉, i.e.: define

µx : V ′ → F : λ → 〈x, λ〉.Then, for any two elements λ1 and λ2 of V ′ and scalars a and b,

µx(aλ1 + bλ2) = 〈x, aλ1 + bλ2〉 = a〈x, λ1〉+ b〈x, λ2〉 = aµx(λ1) + bµx(λ2).

We have defined a linear mapping from V ′ to F . Therefore, 〈x, ·〉 is an element of V ′′.In other words, 〈x, ·〉 is a linear functional on V ′ for every x in V . Is every

functional µ on V ′ of the form µ = 〈x, ·〉 for some x in V ? We answer this questionbelow.

Theorem 12.1. If V is a finite-dimensional vector space then Φ : V → V ′′ : x →〈x, ·〉 is a vector space isomorphism (the natural isomorphism ) between V and itssecond dual V ′′. In other words, to every µ in V ′′ there corresponds exactly one x inV such that µ(λ) = 〈x, λ〉 for all λ in V ′.

Proof. Suppose V is a finite-dimensional vector space. Then, for every x and y in Vand for all scalars a and b,

Φ(ax + by)(λ) = 〈ax + by, λ〉 = a〈x, λ〉+ b〈y, λ〉 = aΦ(x)(λ) + bΦ(y)(λ)

for all λ in V ′, i.e.: Φ(ax + by) = aΦ(x) + bΦ(y). This shows that Φ is linear.We show that Φ is injective. Suppose that x and y are in V . Then Φ(x)(λ) = 〈x, λ〉

and Φ(y)(λ) = 〈y, λ〉 for all λ in V ′. Hence,

Φ(x) = Φ(y) ⇒ (∀λ ∈ V ′)(Φ(x)(λ) = Φ(y)(λ))

⇒ (∀λ ∈ V ′)(Φ(x− y)(λ) = 0))

⇒ x = y.

24

We show that Φ is surjective. Let W denote the range of Φ, i.e.: the subset ofV ′′ consisting of all elements of the form 〈x, ·〉 for some x in V . Since 〈0, ·〉 is in Wand, for all scalars a and b, 〈ax + by, ·〉 is also in W whenever 〈x, ·〉 and 〈y, ·〉 are inW , it follows that W is a subspace of V ′′. By the injectivity of Φ, this function isan isomorphism between V and W . Hence, the dimension of W is the dimension ofV , which is also the dimension of V ′′. Hence, W is a subspace of V ′′ with the samedimension as V ′′. It follows that W = V ′′, i.e.: that every µ in V ′′ is of the formµ = 〈x, ·〉 for some x in V .

Remark 12.1. Remark A finite-dimensional vector space that is isomorphic to its sec-ond dual is called reflexive. We have shown that every finite-dimensional vector spaceis reflexive. Since every finite-dimensional vector space and its dual have the samedimension, the two are isomorphic. Why has this isomorphism not been talked aboutand, instead, we elaborated on the isomorphism between a finite-dimensional vectorspace and its second dual? The reason is that there is always a natural isomorphismbetween a finite-dimensional vector space and its second dual, while a correspond-ing “natural isomorphism” does not always exist between a finite-dimensional vectorspace and its (first) dual. Later we shall consider the proper setting where a “naturalisomorphism” can be constructed between some vector spaces and their dual spaces.It is commonplace to blur the difference between mathematical structures that aredifferent, but isomorphic in some natural way. Mathematics is full of such abusesof language. Accordingly, we identify any finite-dimensional vector space V with itssecond dual space. In other words, we blur the difference between x in V and 〈x, ·〉 inV ′′. To put it even more specifically, every x in V “is a linear functional” on V ′ viaλ → 〈x, λ〉 and every functional µ on V ′′ (which corresponds a unique x in V suchthat: µ(λ) = 〈x, λ〉 for all λ in V ) “is a vector” x in V . An obvious consequenceof this identification is that any basis in a finite-dimensional vector space V is also abasis in V ′′.

If the last statement above seems to be too abstract, consider this. Every µ in V ′′

is a linear functional on V ′, i.e.:

µ(λ) = 〈x, λ〉, (12.1)

for all λ in V ′, where x in V is uniquely determined by µ. If X = {x1, · · · , xn} is abasis in V and x = a1x1 + · · ·+ anxn, then, thanks to (12.1), by linearity,

µ(λ) = 〈a1x1 + · · ·+ anxn, λ〉 = a1〈x1, λ〉+ · · ·+ an〈xn, λ〉,for all λ in V ′. Therefore, for the given µ

µ = a1〈x1, ·〉+ · · ·+ an〈xn, ·〉,showing that {〈x1, ·〉, · · · , 〈xn, ·〉}, which we identify with {x1, · · · , xn}, is a basis inV ′′.

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13 Annihilators

Definition 13.1. Let V be a vector space. Given any subset S of V the annihilatorS0 of S is defined by

S0 = {λ ∈ V ′| (∀x ∈ S)(〈x, λ〉 = 0)}.

In other words, the annihilator S0 of S is the collection of all linear functionals on Vthat vanish at every x in S.

Remark 13.1. If V is a vector space, then {0}0 = V ′ and V 0 = {0}, the singleton ofthe zero functional on V . If V is finite-dimensional and S contains a nonzero vectorx then there is a linear functional λ on V such that 〈x, λ〉 6= 0. Hence, S0 cannotcontain all linear functionals on V , showing that S0 must be a proper subset of V ′.

Remark 13.2. S0 is always a vector space (a subspace of V ′, of course), even whenS is not a subspace of V . Indeed, if 〈x, λ1〉 = 0 and 〈x, λ2〉 = 0 for all x in S, then0 = a〈x, λ1〉+ b〈x, λ2〉 = 〈x, aλ1 + bλ2〉 for all x in S, showing that

λ1 ∈ S0 and λ2 ∈ S0 ⇒ aλ1 + bλ2 ∈ S0

for all scalars a and b.

Theorem 13.1. Suppose V is a finite-dimensional vector space. If X is a subspaceof V then dim X0 = dim V − dim X.

Proof Suppose V is an n-dimensional vector space and X is a subspace of V . LetY = {y1, · · · , yn} be a basis in V such that {y1, · · · , ym} spans X. Consider thedual basis Y ′ = {λ1, · · · , λn} in V ′ and let Z denote the subspace of V ′ spanned by{λm+1, · · · , λn}. We have dim V = n, dim X = m and dim Z = n − m. We provethat Z = X0 to establish the proposition.

We prove that Z ⊂ X0. Every x in X is of the form x = a1y1 + · · ·+amym. Hence,for every j, m + 1 ≤ j ≤ n, and for all x in X we have

〈x, λj〉 = 〈a1y1 + · · ·+ amym, λj〉= a1〈y1, λj〉+ · · ·+ am〈ym, λj〉= 0,

since 〈yi, λj〉 = 0 for 1 ≤ i ≤ m and m + 1 ≤ j ≤ n (i.e.: i 6= j). We have shown thatλj ∈ X0 for j = m + 1, · · · , n. Hence, any linear combination of λm+1, · · · , λn alsobelongs to X0, i.e.: Z = span {λm+1, · · · , λn} ⊂ X0.

We prove that X0 ⊂ Z. Every λ in X0 is of the form λ = b1λ1 + · · · + bnλn (asis every element of V ′). Further, λ in X0 implies that λ vanishes on every element of

26

X and, in particular, on the elements y1, · · · , ym. Hence, for i = 1, · · · ,m,

0 = 〈yi, λ〉= 〈yi, b1λ1 + · · ·+ bnλn〉= b1〈yi, λ1〉+ · · ·+ bn〈yi, λn〉= bi,

showing that

λ = b1λ1 + · · ·+ bmλm + bm+1λm+1 + · · ·+ bnλn = bm+1λm+1 + · · ·+ bnλn,

i.e.: λ ∈ span {λm+1, · · · , λn} = Z. We proved that λ ∈ X0 implies λ ∈ Z, i.e.:X0 ⊂ Z.

Remark 13.3. Since we identify every finite-dimensional vector space with its seconddual, does it follow that the annihilator of an annihilator of a subspace can be iden-tified with that subspace? Below, we show that the answer is affirmative. First, letus agree to write S00 for (S0)0.

Theorem 13.2. Let V be a finite-dimensional vector space. If X is any subspace ofV then X00 = X.

Proof. Let V be an n-dimensional vector space and let X be a subspace of V . Then,by identifying V with its second dual,

X00 = {x ∈ V | (∀λ ∈ X0)(〈x, λ〉 = 0)}.Recall the definition of

X0 := {λ ∈ V ′| (∀x ∈ X)(〈x, λ〉 = 0).

Hence, (∀x ∈ X)(∀λ ∈ X0)(〈x, λ〉 = 0), which implies

x ∈ X ⇒ (∀λ ∈ X0)(〈x, λ〉 = 0),

which says x ∈ X ⇒ x ∈ X00, i.e.: X ⊂ X00.Can X be a proper subset of X00? Let’s count dimensions. We have

dim X0 = dim V − dim X,

dim X00 = dim V ′ − dim X0,

dim V = dim V ′.

Hence, dim X = dim V − dim X0 = dim V ′ − dim X0 = dim X00. The only subspaceof X00 with the same dimension as X00 is X00. Hence, X00 = X.

27

Theorem 13.3. Let U and V be subspaces of a vector space W . If W = U⊕

V thenthe following hold.

(1) The annihilator U0 of U and the annihilator V 0 of V are disjoint, i.e.: U0∩V 0 ={0}.

(2) W ′ = U0⊕

V 0.

(3) The dual space V ′ of V is isomorphic to the annihilator U0 of U .

(4) The dual space U ′ of U is isomorphic to the annihilator V 0 of V .

Proof. We prove (1). Recall that

λ ∈ U0 ⇔ (∀x ∈ U)(〈x, λ〉 = 0)

andλ ∈ V ′ ⇔ (∀y ∈ V )(〈y, λ〉 = 0).

Suppose that λ ∈ U0 ∩ V 0. Then,〈x, λ〉 = 0 for all x in U and 〈y, λ〉 = 0 for all y inV . Hence, since every z in W has the unique representation z = x + y, with x in Uand y in V , we have

〈z, λ〉 = 〈x + y, λ〉 = 〈x, λ〉+ 〈y, λ〉 = 0.

We have shown that λ ∈ U0 ∩ V 0 ⇒ λ = 0,i.e.: U∩V 0 = {0}.We prove (2). Corresponding to every λ in W ′ we define the linear functionals λ1

and λ2 on W as follows. Every z in W has the unique representation z = x + y, withx in U and y in V . Define λ1 by 〈z, λ1〉 = 〈y, λ〉 and define λ2 by 〈z, λ2〉 = 〈x, λ〉. Weclaim that λ1 ∈ U0 and λ2 ∈ V 0. Indeed (keeping in mind the unique representationz = x + y),

z ∈ U ⇒ 〈z, λ1〉 = 〈0, λ〉 = 0 ⇒ λ1 ∈ U0,

andz ∈ V ⇒ 〈z, λ2〉 = 〈0, λ〉 = 0 ⇒ λ2 ∈ V 0.

On the other hand, for every z in W

〈z, λ〉 = 〈x + y, λ〉 = 〈x, λ〉+ 〈y, λ〉 = 〈z, λ1〉+ 〈z, λ2〉 = 〈z, λ1 + λ2〉,

i.e.: λ = λ1 + λ2, showing that W ′ = U0 + V 0. Hence, since U0 ∩ V 0 = {0},W ′ = U0

⊕V 0.

We prove (3). We need a linear isomorphism between V ′ and U0. In the mappingW ′ → U0 : λ → λ1 defined above, λ1 is defined by

〈z, λ1〉 = 〈y, λ〉 (13.1)

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(Keep in mind the unique representation z = x + y.) If we consider all z of the formz = y (all z in V ), then the rule given by (13.1) reduces to 〈y, λ1〉 = 〈y, λ〉 and definesa bijection between V ′ and U0.

To establish that this bijection is linear, suppose that µ is a functional on V (sothat µ1 in U0 is defined by 〈y, µ1〉 = 〈y, µ〉). Then, for all scalars a and b,

〈y, aλ + bµ〉 = a〈y, λ〉+ b〈y, µ〉 = a〈y, λ1〉+ b〈y, µ1〉,showing that aλ + bµ → aλ1 + bµ1. We have obtained the desired linear isomorphismbetween U ′ and V 0.

The proof of (4) is carried out by making obvious modifications to the proof of(3).

Remark 13.4. The above proof uses neither coordinates nor finite-dimensional argu-ments. The vector spaces need not be finite-dimensional.

Exercises

1. Prove that S00 = span S for any subset S of a finite-dimensional vector space.

2. Suppose that S1 and S2 are subsets of a vector space V . Prove that if S1 ⊂ S2

then S02 ⊂ S0

1 .

3. Suppose X and Y are subspaces of a finite-dimensional vector space.

(a) Prove that (X ∩ Y )0 = X0 + Y 0.

(b) Prove that (X + Y )0 = X0 ∩ Y 0.

4. Show that the cancellation law X⊕

Y = X⊕

Z ⇒ Y = Z.

does not always apply to the direct sum of subspaces. In other words, givean example of a vector space V with subspaces X, Y and Z such that V =X

⊕Y = X

⊕Z and Y 6= Z.

5. Suppose U , V and W are vector spaces.

(a) Does U⊕

(V⊕

W ) = (U⊕

V )⊕

W hold? If not, is there a linear iso-morphism between the two sides of the equation?

(b) Does U⊕

V = V⊕

U hold? If not, is there a linear isomorphism betweenthe two sides of the equation?

6. The subspaces X, Y and Z of the vector space V are said to be independent ifthe sum of any two is disjoint from the third. Further, define X + Y + Z in theobvious way:

X + Y + Z = {x + y + z ∈ V | x ∈ X, y ∈ Y and z ∈ Z}.

29

(a) Suppose that x, y and z are vectors in V . Further, suppose that X =span{x}, Y = span{y} and Z = span{z}. Prove that {x, y, z} is linearlyindependent iff X, Y and Z are independent.

(b) Prove that if V = X⊕

(Y⊕

Z) then V = X + Y + Z and the subspacesare independent. Do the same for V = (X

⊕Y )

⊕Z.

(c) If V = X + Y + Z and the subspaces X, Y , Z, are independent, does itfollow that i. V = X

⊕(Y

⊕Z)? ii. V = (X

⊕Y )

⊕Z?

14 Quotient Sets

Definition 14.1. A binary relation ' on a set S is a subset of S×S. We write x ' yfor (x, y) ∈'. The binary relation ' is called an equivalence relation if it is

(1) reflexive: x ' x for all x in S,

(2) symmetric: x ' y ⇒ y ' x for all x and y in S,

(3) transitive: x ' y and y ' z ⇒ x ' z for all x, y and z in S.

If ' is an equivalence relation on S we define the function φ' : S → P (S) byφ'(x) = {y ∈ S| x ' y}, where P (S) denotes the power set of S, the set of all subsetsof S. The range of φ ' is called the quotient set of S modulo ' and is denoted byS/ '. We define the projection p' : S → S/ ': x → φ ' (x). Clearly, p' is surjectiveand S/ ' does not contain the empty set, since x ∈ p ' (x) for every x in S (x ' xforevery x in S). The set p ' (x) is called the equivalence class of x. In other words,S/ '= {p ' (x) ∈ P (S)| x ∈ S} is the set of all equivalence classes defined by ' onS.

A partition P of a set S is any collection of disjoint nonempty subsets of S whoseunion is S.

Theorem 14.1. If ' is an equivalence relation on the set S then the elements ofS/ ' are disjoint.

Proof If z ∈ p'(x) and z ∈ p'(y) then x ' z and y ' z. Hence, thanks tosymmetry, x ' z and z ' y. Therefore, thanks to transitivity, x ' y, which impliesp'(x) = p'(y). Hence, p'(x) 6= p ' (y) implies p'(x) ∩ p'(y) = ∅.Corollary 14.2. If ' is an equivalence relation on the set S then S/ ' is a partitionof S.

Proof We already know that S/ ' is a collection of nonempty disjoint subsets ofS. It remains to be shown that S is the union of the elements in S/ '. Indeed, everyx in S is an element of p'(x). Hence, S ⊂ ∪x∈Sp'(x). On the other hand, p'(x) is asubset of S. Hence, ∪x∈Sp'(x) ⊂ S.

30

Theorem 14.3. Every partition of a set defines an equivalence relation on that set.

Proof Let P be a partition of S. Then, since the union of the elements of P equalsS, every x in S belongs to some X in P . Moreover, since the elements of P aredisjoint, this X is unique.

Define the binary relation ' on S by

x ' y ⇔ (∃X inP )(y ∈ X ∧ x ∈ X).

Then, x ' x for every x in S. Further, the symmetry of the definition implies thatif x ' y then y ' x. Finally, if x ' y and y ' z then x and y belong to some X inP ,and y and z belong to some Y in P . Since z belongs to exactly one set in P , itfollows that X = Y and, hence, that x ' z.

Theorem 14.4. If ' is an equivalence relation on the set S then to every functionf : S → X such that x ' y ⇒ f(x) = f(y) there corresponds exactly one functiong : S/ '→ X such that f = g ◦ p ' , i.e.: such that the diagram below commutes.

Sp'−→ S/ '↘f

yg

X

Moreover, if f is surjective and f(x) = f(y)) ⇒ x ' y then g is bijective.

Proof Suppose ' is an equivalence relation on the set S and f : S → X is suchthat x ' y ⇒ f(x) = f(y). Then f is constant on the elements of each equivalenceclass p'(x), i.e.: if f(x) = c then f(y) = c for all y ∈ p'(x). Define g : S/ '→ X byg(p'(x)) = f(x). Then,

(g ◦ p')(x) = g(p'(x)) = f(x).

If f is surjective then so is g. Suppose f(x) = f(y) ⇒ x ' y. Then g(p'(x)) =g(p'(y)) implies f(x) = f(y), which implies x ' y, i.e.: p'(x) = p'(y), showing thatg is injective.

15 Quotient Spaces

Definition 15.1. (Cosets) Let U be a subspace of a vector space V and let v be avector in V . Define the coset (also called affine subspace) v + U of U by

v + U = {v + x ∈ V | x ∈ U}.We say that S is a coset of U if S = v + U for some vector v.

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Remark 15.1. Note that U is a coset of U , since U = 0+U . Note that v +U = u+Udoes not imply v = u, i.e.: it is possible that v + U = u + U and v 6= u, as shownin Figure 2 for the case V = R2. In R2, if U is a line containing the origin and v isany vector then v + U is a line parallel to U . The collection of all cosets of U is thecollection of all lines parallel to U (including, of course, U).

Definition 15.2. Define X + Y = {x + y ∈ V ‖x ∈ X ∧ y ∈ Y } for any two cosets Xand Y of any subspace U of the vector space V .

Theorem 15.1. Let U be a subspace of a vector space V . If X and Y are cosets ofU so is X + Y .

Proof Suppose U is a subspace of a vector space V . If X and Y are cosets of Uthen X = u + U and Y = v + U for some vectors u and v in V .

We show that X + Y ⊂ (u + v) + U . Each vector x1 in X is of the form u + z1

for some z1 in U and each vector x2 in Y is of the form x2 = v + z2 for some z2 in U .Therefore, each vector x in X + Y is of the form x = (u + v) + (z1 + z2). Since U is avector space, z1 + z2 is in U . Hence, x = (u + v) + z for some z in U , i.e.: x belongsto the coset (u + v) + U of U .

We show that (u + v) + U ⊂ X + Y . Every vector x in (u + v) + U is of the formx = (u+ v)+ z for some z in U and, therefore, of the form x = (u+ z)+ (v +0), withz and 0 both in U , i.e.: u + z is in u + U (which is X)and v + 0 is in v + U (which isY ). Hence, x is in X + Y .

Remark 15.2. The above proposition establishes that

(u + U) + (v + U) = (u + v) + U = (u + v) + (U + U)

for any two cosets u+U and v +U of U .(Note that U +U = U , since U is a subspace.

Corollary 15.2. Coset addition is associative and commutative.

Proof Commutativity is an immediate consequence of the definition of coset ad-dition. We show associativity. Consider any three cosets X, Y and Z of U .ThenX = u + U , Y = v + U and Z = w + U for some vectors u, v and w in V . Then,since vector addition is associative and U = U + U = (U + U) + U = U + (U + U),

(X + Y ) + Z = ((u + U) + (v + U)) + (w + U)

= ((u + v) + (U + U)) + (w + U)

= ((u + v) + w)) + ((U + U) + U)

= (u + (v + w) + (U + (U + U))

= (u + U) + ((v + U)) + (w + U)

= X + (Y + Z)

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Definition 15.3. Let U be a subspace of a vector space V . We make the set of allcosets of U into a vector space as follows.

· Coset addition is already defined and is associative and commutative.

· Recall that U = 0+U is a coset of U . For any coset X of U we have X+U = X,i.e.: U is the origin.

· Scalar multiplication is defined as follows. For a scalar a, a 6= 0, define aX asthe set of all vectors ax with x in X. Define 0X = U .

· Define −X as set of all vectors −x with x in X. Then X + (−X) = U .

With above definitions, the set of all cosets of U is a vector space, the quotient spaceof V modulo U , and is denoted by V/U (read V modulo U).

Theorem 15.3. Suppose U and V are complementary subspaces of a vector spaceW , i.e.: U + V = W and U ∩ V = {0}.T he mapping

Φ : V → W/U : v → v + U

is a vector space isomorphism.

Proof We prove that Φ is injective. If Φ(v1) = Φ(v2) then v1 + U = v2 + U . Then,v1 is an element of v2 +U . Hence, v1 = v2 +x for some x in U . Therefore, v1−v2 = x.It follows that v1−v2 = 0, since v1−v2 belongs to V , x belongs to U and U∩V = {0}.

We prove that Φ is surjective. Choose any coset v +U of U . Recall that v = x+ ywith x in U and y in V , since U + V = W . Hence, since U + U = U and x + U = U ,

v + U = (x + y) + U = (x + U) + (y + U) = U + (y + U) = y + U = Φ(y).

Therefore, for every coset v + U of U there is some y in V such that Φ(y) = v + U .We prove that Φ is linear. For all scalars a and b and all v1 and v2 in V

Φ(av1 + bv2) = (av1 + bv2) + U

= (av1 + U) + (bv2 + U)

= (av1 + aU) + (bv2 + bU)

= a(v1 + U) + b(v2 + U)

= aΦ(v1) + bΦ(v2)

since U + U = U = aU = bU .

Corollary 15.4. If U is a subspace of the finite-dimensional vector space W thendim(W/U) = dim W − dim U

33

Proof Suppose W is a finite-dimensional vector space and U is a subspace of W .There is a subspace V of W such that W = U

⊕V and dim W = dim U + dim V .

Thanks to the theorem, V is isomorphic to W/U . Hence, dim(W/U) = dim V =dim W − dim U .

Exercises1. Suppose U is a subspace of the vector space V . For any two vectors u and v in

V write u ' v (read “u and v are congruent modulo U”) if u− v is a vector in U .

(a) Prove that ' is an equivalence relation on V .

(b) Prove that if u1 ' v1 and u2 ' v2 then au1 + bu2 ' av2 + bv2 for all scalars aand b.

c) Prove that V/ '= V/U .

Hint: To visualize the geometric meaning of ' for the case V = R2, consider thesituation shown in Figure 3. Given a subspace U , the vector u defines the coset X.The vector v also defines the same coset. Therefore, the vector u − v is parallel to U ,i.e.: is an element of U . More generally, when u and v are not necessarily in the cosetX shown in the figure, u− v ∈ U is another way of saying that u + U and v + U arethe same coset (u + U = v + U).

2. Suppose U is a subspace of a vector space V over the field F . Define f :(V/U)′ → F V : λ → µ, where µ(x) = 〈x + U, λ〉 for all x in V .

(a) Prove that f(λ) is a linear functional on V for every λ in (V/U)′.

(b) Prove that f is a linear isomorphism between (V/U)′ and U0.

3. Suppose U is a subspace of the vector space V . Define

f : V ′ → U ′ : λ → µ,

where 〈x, µ〉 = 〈x, λ〉.(a) Prove that if λ1 + U0 = λ2 + U0 then f(λ1) = f(λ2).

(b) Find a linear isomorphism g between V ′/U0 and U ′. Hint: Fill in the questionmark in the diagram below, then obtain g.

34