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Chapter 4

Direct Sums, The Dual Space, Duality

4.1

Sums, Direct Sums, Direct Products

Before considering linear forms and hyperplanes, we define the notion of direct sum and

prove some simple propositions. There is a subtle point, which is that if we attempt to

define the direct sum E

F of two vector spaces using the cartesian product E × F , we

don’t quite get the right notion because elements of E × F are ordered pairs, but we want

F = F

E. Thus, we want to think of the elements of E

F as unordrered pairs of

elements. It is possible to do so by considering the direct sum of a family (Ei)i∈{1,2}, and

more generally of a family (Ei)i∈I. For simplicity, we begin by considering the case where

I = {1, 2}.

Definition 4.1. Given a family (Ei)i∈{1,2} of two vector spaces, we define the (external)

direct sum E1

E2 (or coproduct) of the family (Ei)i∈{1,2} as the set

E2 = {{ 1, u , 2, v } | u ∈ E1, v ∈ E2},

with addition

{ 1, u1 , 2, v1 } + { 1, u2 , 2, v2 } = { 1, u1 + u2 , 2, v1 + v2 },

and scalar multiplication

λ{ 1, u , 2, v } = { 1, λu , 2, λv }.

We define the injections in1 : E1 → E1

E2 and in2 : E2 → E1

E2 as the linear maps

defined such that,

in1(u) = { 1, u , 2, 0 },

and

in2(v) = { 1, 0 , 2, v }.

CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY

Note that

E1 = {{ 2, v , 1, u } | v ∈ E2, u ∈ E1} = E1

E2.

Thus, every member { 1, u , 2, v } of E1

E2 can be viewed as an unordered pair consisting

of the two vectors u and v, tagged with the index 1 and 2, respectively.

Remark: In fact, E1

E2 is just the product

i∈{1,2}

i of the family (Ei)i∈{1,2}.

This is not to be confused with the cartesian product E1 × E2. The vector space E1 × E2

is the set of all ordered pairs u, v , where u ∈ E1, and v ∈ E2, with addition and

multiplication by a scalar defined such that

u1, v1 + u2, v2 = u1 + u2, v1 + v2 ,

λ u, v = λu, λv .

There is a bijection between

i∈{1,2}

i and E1 × E2, but as we just saw, elements of

i∈{1,2}

i are certain sets.

The product E1 × · · · × En of any number of vector spaces

can also be defined. We will do this shortly.

The following property holds.

Proposition 4.1. Given any two vector spaces, E1 and E2, the set E1

E2 is a vector

space. For every pair of linear maps, f : E1 → G and g : E2 → G, there is a unique linear

map, f + g : E1

E2 → G, such that (f + g) ◦ in1 = f and (f + g) ◦ in2 = g, as in the

following diagram:

in1

f +g

7 G

♥

♥ g

♥

Proof. Define

(f + g)({ 1, u , 2, v }) = f(u) + g(v),

for every u ∈ E1 and v ∈ E2. It is immediately verified that f + g is the unique linear map

with the required properties.

We already noted that E1

E2 is in bijection with E1 × E2. If we define the projections

π1 : E1

E2 → E1 and π2 : E1

E2 → E2, such that

π1({ 1, u , 2, v }) = u,

and

π2({ 1, u , 2, v }) = v,

we have the following proposition.

4.1. SUMS, DIRECT SUMS, DIRECT PRODUCTS

Proposition 4.2. Given any two vector spaces, E1 and E2, for every pair of linear maps,

f : D → E1 and g : D → E2, there is a unique linear map, f × g : D → E1

E2, such that

π1 ◦ (f × g) = f and π2 ◦ (f × g) = g, as in the following diagram:

7 E1

♥

f ×g

Proof. Define

(f × g)(w) = { 1, f(w) , 2, g(w) },

for every w ∈ D. It is immediately verified that f × g is the unique linear map with the

required properties.

Remark: It is a peculiarity of linear algebra that direct sums and products of finite families

are isomorphic. However, this is no longer true for products and sums of infinite families.

When U, V are subspaces of a vector space E, letting i1 : U → E and i2 : V → E be the

inclusion maps, if U

V is isomomorphic to E under the map i1 + i2 given by Proposition

4.1, we say that E is a direct sum of U and V , and we write E = U

V (with a slight abuse

of notation, since E and U

V are only isomorphic). It is also convenient to define the sum

U1 + · · · + Up and the internal direct sum U1 ⊕ · · · ⊕ Up of any number of subspaces of E.

Definition 4.2. Given p ≥ 2 vector spaces E1, . . . , Ep, the product F = E1 × · · · × Ep can

be made into a vector space by defining addition and scalar multiplication as follows:

(u1, . . . , up) + (v1, . . . , vp) = (u1 + v1, . . . , up + vp)

λ(u1, . . . , up) = (λu1, . . . , λup),

for all ui, vi ∈ Ei and all λ ∈ K. With the above addition and multiplication, the vector

space F = E1 × · · · × Ep is called the direct product of the vector spaces E1, . . . , Ep.

As a special case, when E1 = · · · = Ep = K, we find again the vector space F = Kp.

The projection maps pri : E1 × · · · × Ep → Ei given by

pri(u1, . . . , up) = ui

are clearly linear. Similarly, the maps ini : Ei → E1 × · · · × Ep given by

ini(ui) = (0, . . . , 0, ui, 0, . . . , 0)

CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY

are injective and linear. If dim(Ei) = ni and if (ei1, . . . , ein ) is a basis of Ei for i = 1, . . . , p,

then it is easy to see that the n1 + · · · + np vectors

(e11, 0, . . . , 0),

. . . ,

(e1n , 0, . . . , 0),

(0, . . . , 0, ei1, 0, . . . , 0), . . . , (0, . . . , 0, ein , 0, . . . , 0),

(0, . . . , 0, ep1),

. . . ,

(0, . . . , 0, epn )

form a basis of E1 × · · · × Ep, and so

dim(E1 × · · · × Ep) = dim(E1) + · · · + dim(Ep).

Let us now consider a vector space E and p subspaces U1, . . . , Up of E. We have a map

a : U1 × · · · × Up → E

given by

a(u1, . . . , up) = u1 + · · · + up,

with ui ∈ Ui for i = 1, . . . , p. It is clear that this map is linear, and so its image is a subspace

of E denoted by

U1 + · · · + Up

and called the sum of the subspaces U1, . . . , Up. It is immediately verified that U1 + · · · + Up

is the smallest subspace of E containing U1, . . . , Up.

If the map a is injective, then Ker a = 0, which means that if ui ∈ Ui for i = 1, . . . , p and

u1 + · · · + up = 0

then u1 = · · · = up = 0. In this case, every u ∈ U1 + · · · + Up has a unique expression as a

sum

u = u1 + · · · + up,

with ui ∈ Ui, for i = 1, . . . , p. It is also clear that for any p nonzero vectors ui ∈ Ui, u1, . . . , up

are linearly independent.

Definition 4.3. For any vector space E and any p ≥ 2 subspaces U1, . . . , Up of E, if the

map a defined above is injective, then the sum U1 + · · · + Up is called a direct sum and it is

denoted by

U1 ⊕ · · · ⊕ Up.

The space E is the direct sum of the subspaces Ui if

E = U1 ⊕ · · · ⊕ Up.

4.1. SUMS, DIRECT SUMS, DIRECT PRODUCTS

Observe that when the map a is injective, then it is a linear isomorphism between

U1 × · · · × Up and U1 ⊕ · · · ⊕ Up. The difference is that U1 × · · · × Up is defined even if the

spaces Ui are not assumed to be subspaces of some common space.

Now, if p = 2, it is easy to determine the kernel of the map a : U1 × U2 → E. We have

a(u1, u2) = u1 + u2 = 0 iff u1 = −u2, u1 ∈ U1, u2 ∈ U2,

which implies that

Ker a = {(u, −u) | u ∈ U1 ∩ U2}.

Now, U1 ∩ U2 is a subspace of E and the linear map u → (u, −u) is clearly an isomorphism,

so Ker a is isomorphic to U1 ∩ U2. As a result, we get the following result:

Proposition 4.3. Given any vector space E and any two subspaces U1 and U2, the sum

U1 + U2 is a direct sum iff U1 ∩ U2 = (0).

An interesting illustration of the notion of direct sum is the decomposition of a square

matrix into its symmetric part and its skew-symmetric part. Recall that an n × n matrix

A ∈ Mn is symmetric if A = A, skew -symmetric if A = −A. It is clear that

S(n) = {A ∈ Mn | A = A} and Skew(n) = {A ∈ Mn | A = −A}

are subspaces of Mn, and that S(n) ∩ Skew(n) = (0). Observe that for any matrix A ∈ Mn,

the matrix H(A) = (A + A )/2 is symmetric and the matrix S(A) = (A − A )/2 is skew-

symmetric. Since

A + A

A − A

A = H(A) + S(A) =

we see that Mn = S(n) + Skew(n), and since S(n) ∩ Skew(n) = (0), we have the direct sum

Mn = S(n) ⊕ Skew(n).

Remark: The vector space Skew(n) of skew-symmetric matrices is also denoted by so(n).

It is the Lie algebra of the group SO(n).

Proposition 4.3 can be generalized to any p ≥ 2 subspaces at the expense of notation.

The proof of the following proposition is left as an exercise.

Proposition 4.4. Given any vector space E and any p ≥ 2 subspaces U1, . . . , Up, the fol-

lowing properties are equivalent:

(1) The sum U1 + · · · + Up is a direct sum.

(2) We have

Ui ∩

= (0),

i = 1, . . . , p.

j=1,j=i

CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY

(3) We have

i−1

Ui ∩

= (0),

i = 2, . . . , p.

j=1

Because of the isomorphism

U1 × · · · × Up ≈ U1 ⊕ · · · ⊕ Up,

we have

Proposition 4.5. If E is any vector space, for any (finite-dimensional) subspaces U1, . . .,

Up of E, we have

dim(U1 ⊕ · · · ⊕ Up) = dim(U1) + · · · + dim(Up).

If E is a direct sum

E = U1 ⊕ · · · ⊕ Up,

since every u ∈ E can be written in a unique way as

u = u1 + · · · + up

for some ui ∈ Ui for i = 1 . . . , p, we can define the maps πi : E → Ui, called projections, by

πi(u) = πi(u1 + · · · + up) = ui.

It is easy to check that these maps are linear and satisfy the following properties:

if i = j

j ◦ πi =

if i = j,

π1 + · · · + πp = idE.

For example, in the case of the direct sum

Mn = S(n) ⊕ Skew(n),

the projection onto S(n) is given by

A + A

π1(A) = H(A) =

and the projection onto Skew(n) is given by

A − A

π2(A) = S(A) =

Clearly, H(A)+S(A) = A, H(H(A)) = H(A), S(S(A)) = S(A), and H(S(A)) = S(H(A)) =

A function f such that f ◦ f = f is said to be idempotent. Thus, the projections πi are

idempotent. Conversely, the following proposition can be shown:

4.1. SUMS, DIRECT SUMS, DIRECT PRODUCTS

Proposition 4.6. Let E be a vector space. For any p ≥ 2 linear maps fi : E → E, if

if i = j

j ◦ fi =

if i = j,

f1 + · · · + fp = idE,

then if we let Ui = fi(E), we have a direct sum

E = U1 ⊕ · · · ⊕ Up.

We also have the following proposition characterizing idempotent linear maps whose proof

is also left as an exercise.

Proposition 4.7. For every vector space E, if f : E → E is an idempotent linear map, i.e.,

f ◦ f = f, then we have a direct sum

E = Ker f ⊕ Im f,

so that f is the projection onto its image Im f .

We now give the definition of a direct sum for any arbitrary nonempty index set I. First,

let us recall the notion of the product of a family (Ei)i∈I. Given a family of sets (Ei)i∈I, its

product

i∈I

i, is the set of all functions f : I →

i∈I

i, such that, f (i) ∈ Ei, for every

i ∈ I. It is one of the many versions of the axiom of choice, that, if Ei = ∅ for every i ∈ I,

then

i∈I

i = ∅. A member f ∈

i∈I

i, is often denoted as (fi)i∈I . For every i ∈ I , we

have the projection πi :

i∈I

i → Ei, defined such that, πi((fi)i∈I ) = fi. We now define

direct sums.

Definition 4.4. Let I be any nonempty set, and let (Ei)i∈I be a family of vector spaces.

The (external) direct sum

i∈I

i (or coproduct ) of the family (Ei)i∈I is defined as follows:

i∈I

i consists of all f ∈

i∈I

i, which have finite support, and addition and multi-

plication by a scalar are defined as follows:

(fi)i∈I + (gi)i∈I = (fi + gi)i∈I,

λ(fi)i∈I = (λfi)i∈I.

We also have injection maps ini : Ei →

i∈I

i, defined such that, ini(x) = (fi)i∈I , where

fi = x, and fj = 0, for all j ∈ (I − {i}).

The following proposition is an obvious generalization of Proposition 4.1.

CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY

Proposition 4.8. Let I be any nonempty set, let (Ei)i∈I be a family of vector spaces, and

let G be any vector space. The direct sum

i∈I

i is a vector space, and for every family

(hi)i∈I of linear maps hi : Ei → G, there is a unique linear map

hi :

Ei → G,

i∈I

such that, (

i∈I

i) ◦ ini = hi, for every i ∈ I .

Remark: When Ei = E, for all i ∈ I, we denote

i∈I

i by E(I).

In particular, when

Ei = K, for all i ∈ I, we find the vector space K(I) of Definition 2.13.

We also have the following basic proposition about injective or surjective linear maps.

Proposition 4.9. Let E and F be vector spaces, and let f : E → F be a linear map. If

f : E → F is injective, then there is a surjective linear map r : F → E called a retraction,

such that r ◦ f = idE. If f : E → F is surjective, then there is an injective linear map

s : F → E called a section, such that f ◦ s = idF .

Proof. Let (ui)i∈I be a basis of E. Since f : E → F is an injective linear map, by Proposition

2.13, (f (ui))i∈I is linearly independent in F . By Theorem 2.7, there is a basis (vj)j∈J of F ,

where I ⊆ J, and where vi = f(ui), for all i ∈ I. By Proposition 2.13, a linear map r : F → E

can be defined such that r(vi) = ui, for all i ∈ I, and r(vj) = w for all j ∈ (J − I), where w

is any given vector in E, say w = 0. Since r(f (ui)) = ui for all i ∈ I, by Proposition 2.13,

we have r ◦ f = idE.

Now, assume that f : E → F is surjective. Let (vj)j∈J be a basis of F . Since f : E → F

is surjective, for every vj ∈ F , there is some uj ∈ E such that f(uj) = vj. Since (vj)j∈J is a

basis of F , by Proposition 2.13, there is a unique linear map s : F → E such that s(vj) = uj.

Also, since f (s(vj)) = vj, by Proposition 2.13 (again), we must have f ◦ s = idF .

The converse of Proposition 4.9 is obvious. We now have the following fundamental

Proposition.

Proposition 4.10. Let E, F and G, be three vector spaces, f : E → F an injective linear

map, g : F → G a surjective linear map, and assume that Im f = Ker g. Then, the following

properties hold. (a) For any section s : G → F of g, we have F = Ker g ⊕ Im s, and the

linear map f + s : E ⊕ G → F is an isomorphism.1

(b) For any retraction r : F → E of f, we have F = Im f ⊕ Ker r.2

E o

F o

1The existence of a section s : G → F of g follows from Proposition 4.9.

2The existence of a retraction r : F → E of f follows from Proposition 4.9.

4.1. SUMS, DIRECT SUMS, DIRECT PRODUCTS

Proof. (a) Since s : G → F is a section of g, we have g ◦ s = idG, and for every u ∈ F ,

g(u − s(g(u))) = g(u) − g(s(g(u))) = g(u) − g(u) = 0.

Thus, u − s(g(u)) ∈ Ker g, and we have F = Ker g + Im s. On the other hand, if u ∈

Ker g ∩ Im s, then u = s(v) for some v ∈ G because u ∈ Im s, g(u) = 0 because u ∈ Ker g,

and so,

g(u) = g(s(v)) = v = 0,

because g ◦ s = idG, which shows that u = s(v) = 0. Thus, F = Ker g ⊕ Im s, and since by

assumption, Im f = Ker g, we have F = Im f ⊕ Im s. But then, since f and s are injective,

f + s : E ⊕ G → F is an isomorphism. The proof of (b) is very similar.

Note that we can choose a retraction r : F → E so that Ker r = Im s, since

F = Ker g ⊕ Im s = Im f ⊕ Im s and f is injective so we can set r ≡ 0 on Im s.

Given a sequence of linear maps E −→ F −→ G, when Im f = Ker g, we say that the

sequence E −→ F −→ G is exact at F . If in addition to being exact at F , f is injective

and g is surjective, we say that we have a short exact sequence, and this is denoted as

0 −→ E −→ F −→ G −→ 0.

The property of a short exact sequence given by Proposition 4.10 is often described by saying

that 0 −→ E −→ F −→ G −→ 0 is a (short) split exact sequence.

As a corollary of Proposition 4.10, we have the following result.

Theorem 4.11. Let E and F be vector spaces, and let f : E → F be a linear map. Then,

E is isomorphic to Ker f ⊕ Im f, and thus,

dim(E) = dim(Ker f ) + dim(Im f ) = dim(Ker f ) + rk(f ).

Proof. Consider

Ker f

−→ E −→ Im f,

where Ker f

−→ E is the inclusion map, and E −→ Im f is the surjection associated

with E −→ F . Then, we apply Proposition 4.10 to any section Im f

−→ E of f to

get an isomorphism between E and Ker f ⊕ Im f, and Proposition 4.5, to get dim(E) =

dim(Ker f ) + dim(Im f ).

Remark: The dimension dim(Ker f ) of the kernel of a linear map f is often called the

nullity of f .

We now derive some important results using Theorem 4.11.

CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY

Proposition 4.12. Given a vector space E, if U and V are any two subspaces of E, then

dim(U ) + dim(V ) = dim(U + V ) + dim(U ∩ V ),

an equation known as Grassmann’s relation.

Proof. Recall that U + V is the image of the linear map

a : U × V → E

given by

a(u, v) = u + v,

and that we proved earlier that the kernel Ker a of a is isomorphic to U ∩ V . By Theorem

4.11,

dim(U × V ) = dim(Ker a) + dim(Im a),

but dim(U × V ) = dim(U) + dim(V ), dim(Ker a) = dim(U ∩ V ), and Im a = U + V , so the

Grassmann relation holds.

The Grassmann relation can be very useful to figure out whether two subspace have a

nontrivial intersection in spaces of dimension > 3. For example, it is easy to see that in

R ,

there are subspaces U and V with dim(U ) = 3 and dim(V ) = 2 such that U ∩ V = 0; for

example, let U be generated by the vectors (1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), and V be

generated by the vectors (0, 0, 0, 1, 0) and (0, 0, 0, 0, 1). However, we claim that if dim(U ) = 3

and dim(V ) = 3, then dim(U ∩ V ) ≥ 1. Indeed, by the Grassmann relation, we have

dim(U ) + dim(V ) = dim(U + V ) + dim(U ∩ V ),

namely

3 + 3 = 6 = dim(U + V ) + dim(U ∩ V ),

and since U + V is a subspace of

R , dim(U + V ) ≤ 5, which implies

6 ≤ 5 + dim(U ∩ V ),

that is 1 ≤ dim(U ∩ V ).

As another consequence of Proposition 4.12, if U and V are two hyperplanes in a vector

space of dimension n, so that dim(U ) = n − 1 and dim(V ) = n − 1, the reader should show

that

dim(U ∩ V ) ≥ n − 2,

and so, if U = V , then

dim(U ∩ V ) = n − 2.

Here is a characterization of direct sums that follows directly from Theorem 4.11.

4.1. SUMS, DIRECT SUMS, DIRECT PRODUCTS

Proposition 4.13. If U1, . . . , Up are any subspaces of a finite dimensional vector space E,

then

dim(U1 + · · · + Up) ≤ dim(U1) + · · · + dim(Up),

and

dim(U1 + · · · + Up) = dim(U1) + · · · + dim(Up)

iff the Uis form a direct sum U1 ⊕ · · · ⊕ Up.

Proof. If we apply Theorem 4.11 to the linear map

a : U1 × · · · × Up → U1 + · · · + Up

given by a(u1, . . . , up) = u1 + · · · + up, we get

dim(U1 + · · · + Up) = dim(U1 × · · · × Up) − dim(Ker a)

= dim(U1) + · · · + dim(Up) − dim(Ker a),

so the inequality follows. Since a is injective iff Ker a = (0), the Uis form a direct sum iff

the second equation holds.

Another important corollary of Theorem 4.11 is the following result:

Proposition 4.14. Let E and F be two vector spaces with the same finite dimension

dim(E) = dim(F ) = n. For every linear map f : E ?