codim(U 0) = dim(U ).
Furthermore, U 00 = U .
Proof. (a) Assume that
λiu∗i = 0,
i∈I
for a family (λi)i∈I (of scalars in K). Since (λi)i∈I has finite support, there is a finite subset
J of I such that λi = 0 for all i ∈ I − J, and we have
λju∗j = 0.
j∈J
Applying the linear form
λ
j∈J
j u∗
j to each uj (j ∈ J ), by Definition 4.8, since u∗
i (uj ) = 1 if
i = j and 0 otherwise, we get λj = 0 for all j ∈ J, that is λi = 0 for all i ∈ I (by definition
of J as the support). Thus, (u∗i)i∈I is linearly independent.
(b) Clearly, we have V ⊆ V 00. If V = V 00, then let (ui)i∈I∪J be a basis of V 00 such that
(ui)i∈I is a basis of V (where I ∩ J = ∅). Since V = V 00, uj ∈ V 00 for some j
0
0 ∈ J (and
thus, j0 /
∈ I). Since uj ∈ V 00, u is orthogonal to every linear form in V 0. Now, we have
0
j0
u∗j (ui) = 0 for all i ∈ I, and thus u∗ ∈ V 0. However, u∗ (uj ) = 1, contradicting the fact
0
j0
j0
0
that uj is orthogonal to every linear form in V 0. Thus, V = V 00.
0
(c) Let J = I − {1, . . . , m}. Every linear form f∗ ∈ V 0 is orthogonal to every uj, for
j ∈ J, and thus, f∗(uj) = 0, for all j ∈ J. For such a linear form f∗ ∈ V 0, let
g∗ = f ∗(u1)u∗1 + · · · + f∗(um)u∗m.
We have g∗(ui) = f ∗(ui), for every i, 1 ≤ i ≤ m. Furthermore, by definition, g∗ vanishes
on all uj, where j ∈ J. Thus, f∗ and g∗ agree on the basis (ui)i∈I of E, and so, g∗ = f∗.
This shows that (u∗1, . . . , u∗m) generates V 0, and since it is also a linearly independent family,
(u∗1, . . . , u∗m) is a basis of V 0. It is then obvious that dim(V 0) = codim(V ), and by part (b),
we have V 00 = V .
(d) Let (u∗1, . . . , u∗m) be a basis of U. Note that the map h: E → Km defined such that
h(v) = (u∗1(v), . . . , u∗m(v))
for every v ∈ E, is a linear map, and that its kernel Ker h is precisely U0. Then, by
Proposition 4.11,
E ≈ Ker (h) ⊕ Im h = U0 ⊕ Im h,
and since dim(Im h) ≤ m, we deduce that U0 is a subspace of E of finite codimension at
most m, and by (c), we have dim(U 00) = codim(U 0) ≤ m = dim(U). However, it is clear
that U ⊆ U00, which implies dim(U) ≤ dim(U00), and so dim(U00) = dim(U) = m, and we
must have U = U 00.
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CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY
Part (a) of Theorem 4.17 shows that
dim(E) ≤ dim(E∗).
When E is of finite dimension n and (u1, . . . , un) is a basis of E, by part (c), the family
(u∗1, . . . , u∗n) is a basis of the dual space E∗, called the dual basis of (u1, . . . , un).
By part (c) and (d) of theorem 4.17, the maps V → V 0 and U → U0, where V is
a subspace of finite codimension of E and U is a subspace of finite dimension of E∗, are
inverse bijections. These maps set up a duality between subspaces of finite codimension of
E, and subspaces of finite dimension of E∗.
One should be careful that this bijection does not extend to subspaces of E∗ of infinite
dimension.
When E is of infinite dimension, for every basis (ui)i∈I of E, the family (u∗i)i∈I of coor-
dinate forms is never a basis of E∗. It is linearly independent, but it is “too small” to
generate E∗. For example, if E =
(
R N), where N = {0, 1, 2, . . .}, the map f : E → R that
sums the nonzero coordinates of a vector in E is a linear form, but it is easy to see that it
cannot be expressed as a linear combination of coordinate forms. As a consequence, when
E is of infinite dimension, E and E∗ are not isomorphic.
Here is another example illustrating the power of Theorem 4.17. Let E = Mn(R), and
consider the equations asserting that the sum of the entries in every row of a matrix ∈ Mn(R)
is equal to the same number. We have n − 1 equations
n
(aij − ai+1j) = 0, 1 ≤ i ≤ n − 1,
j=1
and it is easy to see that they are linearly independent. Therefore, the space U of linear
forms in E∗ spanned by the above linear forms (equations) has dimension n − 1, and the
space U 0 of matrices sastisfying all these equations has dimension n2 − n + 1. It is not so
obvious to find a basis for this space.
When E is of finite dimension n and (u1, . . . , un) is a basis of E, we noted that the family
(u∗1, . . . , u∗n) is a basis of the dual space E∗ (called the dual basis of (u1, . . . , un)). Let us see
how the coordinates of a linear form ϕ∗ over the dual basis (u∗1, . . . , u∗n) vary under a change
of basis.
Let (u1, . . . , un) and (v1, . . . , vn) be two bases of E, and let P = (ai j) be the change of
basis matrix from (u1, . . . , un) to (v1, . . . , vn), so that
n
vj =
ai jui,
i=1
4.2. THE DUAL SPACE E∗ AND LINEAR FORMS
103
and let P −1 = (bi j) be the inverse of P , so that
n
ui =
bj ivj.
j=1
Since u∗i(uj) = δi j and v∗i(vj) = δi j, we get
n
v∗j(ui) = v∗j(
bk ivk) = bj i,
k=1
and thus
n
v∗j =
bj iu∗i,
i=1
and
n
u∗i =
ai jv∗j.
j=1
This means that the change of basis from the dual basis (u∗1, . . . , u∗n) to the dual basis
(v∗1, . . . , v∗n) is (P −1) . Since
n
n
ϕ∗ =
ϕiu∗i =
ϕiv∗i,
i=1
i=1
we get
n
ϕj =
ai jϕi,
i=1
so the new coordinates ϕj are expressed in terms of the old coordinates ϕi using the matrix
P . If we use the row vectors (ϕ1, . . . , ϕn) and (ϕ1, . . . , ϕn), we have
(ϕ1, . . . , ϕn) = (ϕ1, . . . , ϕn)P.
Comparing with the change of basis
n
vj =
ai jui,
i=1
we note that this time, the coordinates (ϕi) of the linear form ϕ∗ change in the same direction
as the change of basis. For this reason, we say that the coordinates of linear forms are
covariant . By abuse of language, it is often said that linear forms are covariant , which
explains why the term covector is also used for a linear form.
Observe that if (e1, . . . , en) is a basis of the vector space E, then, as a linear map from
E to K, every linear form f ∈ E∗ is represented by a 1 × n matrix, that is, by a row vector
(λ1, . . . , λn),
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CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY
with respect to the basis (e1, . . . , en) of E, and 1 of K, where f (ei) = λi. A vector u =
n
u
i=1
iei ∈ E is represented by a n × 1 matrix, that is, by a column vector
u
1
.
.. ,
un
and the action of f on u, namely f (u), is represented by the matrix product
u
1
λ
.
.
1
· · · λn
.
= λ1u1 + · · · + λnun.
un
On the other hand, with respect to the dual basis (e∗1, . . . , e∗n) of E∗, the linear form f is
represented by the column vector
λ
1
.
.. .
λn
Remark: In many texts using tensors, vectors are often indexed with lower indices. If so, it
is more convenient to write the coordinates of a vector x over the basis (u1, . . . , un) as (xi),
using an upper index, so that
n
x =
xiui,
i=1
and in a change of basis, we have
n
vj =
aijui
i=1
and
n
xi =
aijx j.
j=1
Dually, linear forms are indexed with upper indices. Then, it is more convenient to write the
coordinates of a covector ϕ∗ over the dual basis (u∗1, . . . , u∗n) as (ϕi), using a lower index,
so that
n
ϕ∗ =
ϕiu∗i
i=1
and in a change of basis, we have
n
u∗i =
aijv∗j
j=1
4.2. THE DUAL SPACE E∗ AND LINEAR FORMS
105
and
n
ϕj =
aijϕi.
i=1
With these conventions, the index of summation appears once in upper position and once in
lower position, and the summation sign can be safely omitted, a trick due to Einstein. For
example, we can write
ϕj = aijϕi
as an abbreviation for
n
ϕj =
aijϕi.
i=1
For another example of the use of Einstein’s notation, if the vectors (v1, . . . , vn) are linear
combinations of the vectors (u1, . . . , un), with
n
vi =
aijuj,
1 ≤ i ≤ n,
j=1
then the above equations are witten as
vi = aju
i
j ,
1 ≤ i ≤ n.
Thus, in Einstein’s notation, the n × n matrix (aij) is denoted by (aj), a (1, 1)-tensor.
i
Beware that some authors view a matrix as a mapping between coordinates, in which
case the matrix (aij) is denoted by (aij).
We will now pin down the relationship between a vector space E and its bidual E∗∗.
Proposition 4.18. Let E be a vector space. The following properties hold:
(a) The linear map evalE : E → E∗∗ defined such that
evalE(v) = evalv for all v ∈ E,
that is, evalE(v)(u∗) = u∗, v = u∗(v) for every u∗ ∈ E∗, is injective.
(b) When E is of finite dimension n, the linear map evalE : E → E∗∗ is an isomorphism
(called the canonical isomorphism).
Proof. (a) Let (ui)i∈I be a basis of E, and let v =
v
i∈I iui.
If evalE(v) = 0, then in
particular, evalE(v)(u∗i) = 0 for all u∗i, and since
evalE(v)(u∗i) = u∗i, v = vi,
we have vi = 0 for all i ∈ I, that is, v = 0, showing that evalE : E → E∗∗ is injective.
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CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY
If E is of finite dimension n, by Theorem 4.17, for every basis (u1, . . . , un), the family
(u∗1, . . . , u∗n) is a basis of the dual space E∗, and thus the family (u∗∗
1 , . . . , u∗∗
n ) is a basis of
the bidual E∗∗. This shows that dim(E) = dim(E∗∗) = n, and since by part (a), we know
that evalE : E → E∗∗ is injective, in fact, evalE : E → E∗∗ is bijective (because an injective
map carries a linearly independent family to a linearly independent family, and in a vector
space of dimension n, a linearly independent family of n vectors is a basis, see Proposition
2.8).
When a vector space E has infinite dimension, E and its bidual E∗∗ are never isomorphic.
When E is of finite dimension and (u1, . . . , un) is a basis of E, in view of the canon-
ical isomorphism evalE : E → E∗∗, the basis (u∗∗
1 , . . . , u∗∗
n ) of the bidual is identified with
(u1, . . . , un).
Proposition 4.18 can be reformulated very fruitfully in terms of pairings.
Definition 4.9. Given two vector spaces E and F over K, a pairing between E and F is a
bilinear map ϕ : E × F → K. Such a pairing is nondegenerate iff
(1) for every u ∈ E, if ϕ(u, v) = 0 for all v ∈ F , then u = 0, and
(2) for every v ∈ F , if ϕ(u, v) = 0 for all u ∈ E, then v = 0.
A pairing ϕ : E × F → K is often denoted by −, − : E × F → K. For example, the
map −, − : E∗ × E → K defined earlier is a nondegenerate pairing (use the proof of (a) in
Proposition 4.18).
Given a pairing ϕ : E × F → K, we can define two maps lϕ : E → F ∗ and rϕ : F → E∗
as follows: For every u ∈ E, we define the linear form lϕ(u) in F ∗ such that
lϕ(u)(y) = ϕ(u, y) for every y ∈ F ,
and for every v ∈ F , we define the linear form rϕ(v) in E∗ such that
rϕ(v)(x) = ϕ(x, v) for every x ∈ E.
We have the following useful proposition.
Proposition 4.19. Given two vector spaces E and F over K, for every nondegenerate
pairing ϕ : E × F → K between E and F , the maps lϕ : E → F ∗ and rϕ : F → E∗ are linear
and injective. Furthermore, if E and F have finite dimension, then this dimension is the
same and lϕ : E → F ∗ and rϕ : F → E∗ are bijections.
Proof. The maps lϕ : E → F ∗ and rϕ : F → E∗ are linear because a pairing is bilinear. If
lϕ(u) = 0 (the null form), then
lϕ(u)(v) = ϕ(u, v) = 0 for every v ∈ F ,
4.3. HYPERPLANES AND LINEAR FORMS
107
and since ϕ is nondegenerate, u = 0. Thus, lϕ : E → F ∗ is injective. Similarly, rϕ : F → E∗
is injective. When F has finite dimension n, we have seen that F and F ∗ have the same
dimension. Since lϕ : E → F ∗ is injective, we have m = dim(E) ≤ dim(F ) = n. The same
argument applies to E, and thus n = dim(F ) ≤ dim(E) = m. But then, dim(E) = dim(F ),
and lϕ : E → F ∗ and rϕ : F → E∗ are bijections.
When E has finite dimension, the nondegenerate pairing −, − : E∗ × E → K yields
another proof of the existence of a natural isomorphism between E and E∗∗. Interesting
nondegenerate pairings arise in exterior algebra. We now show the relationship between
hyperplanes and linear forms.
4.3
Hyperplanes and Linear Forms
Actually, Proposition 4.20 below follows from parts (c) and (d) of Theorem 4.17, but we feel
that it is also interesting to give a more direct proof.
Proposition 4.20. Let E be a vector space. The following properties hold:
(a) Given any nonnull linear form f ∗ ∈ E∗, its kernel H = Ker f∗ is a hyperplane.
(b) For any hyperplane H in E, there is a (nonnull) linear form f ∗ ∈ E∗ such that H =
Ker f ∗.
(c) Given any hyperplane H in E and any (nonnull) linear form f ∗ ∈ E∗ such that H =
Ker f ∗, for every linear form g∗ ∈ E∗, H = Ker g∗ iff g∗ = λf∗ for some λ = 0 in K.
Proof. (a) If f ∗ ∈ E∗ is nonnull, there is some vector v0 ∈ E such that f∗(v0) = 0. Let
H = Ker f ∗. For every v ∈ E, we have
f ∗(v)
f ∗(v)
f ∗ v −
v
f ∗(v
f ∗(v
0
= f ∗(v) −
0) = f ∗(v) − f ∗(v) = 0.
0)
f ∗(v0)
Thus,
f ∗(v)
v −
v
f ∗(v
0 = h ∈ H,
0)
and
f ∗(v)
v = h +
v
f ∗(v
0,
0)
that is, E = H + Kv0. Also, since f ∗(v0) = 0, we have v0 /
∈ H, that is, H ∩ Kv0 = 0. Thus,
E = H ⊕ Kv0, and H is a hyperplane.
(b) If H is a hyperplane, E = H ⊕ Kv0 for some v0 /
∈ H. Then, every v ∈ E can be
written in a unique way as v = h + λv0. Thus, there is a well-defined function f ∗ : E → K,
such that, f ∗(v) = λ, for every v = h + λv0. We leave as a simple exercise the verification
108
CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY
that f ∗ is a linear form. Since f ∗(v0) = 1, the linear form f ∗ is nonnull. Also, by definition,
it is clear that λ = 0 iff v ∈ H, that is, Ker f∗ = H.
(c) Let H be a hyperplane in E, and let f ∗ ∈ E∗ be any (nonnull) linear form such that
H = Ker f ∗. Clearly, if g∗ = λf ∗ for some λ = 0, then H = Ker g∗. Conversely, assume that
H = Ker g∗ for some nonnull linear form g∗. From (a), we have E = H ⊕ Kv0, for some v0
such that f ∗(v0) = 0 and g∗(v0) = 0. Then, observe that
g∗(v
g∗ −
0) f∗
f ∗(v0)
is a linear form that vanishes on H, since both f ∗ and g∗ vanish on H, but also vanishes on
Kv0. Thus, g∗ = λf ∗, with
g∗(v
λ =
0) .
f ∗(v0)
We leave as an exercise the fact that every subspace V = E of a vector space E, is the
intersection of all hyperplanes that contain V . We now consider the notion of transpose of
a linear map and of a matrix.
4.4
Transpose of a Linear Map and of a Matrix
Given a linear map f : E → F , it is possible to define a map f : F ∗ → E∗ which has some
interesting properties.
Definition 4.10. Given a linear map f : E → F , the transpose f : F ∗ → E∗ of f is the
linear map defined such that
f (v∗) = v∗ ◦ f, for every v∗ ∈ F ∗,
as shown in the diagram below:
f
E
/
f (v∗)
❇
❇
❇
❇
❇
❇
❇
❇
F
v∗
K.
Equivalently, the linear map f : F ∗ → E∗ is defined such that
v∗, f (u) = f (v∗), u ,
for all u ∈ E and all v∗ ∈ F ∗.
4.4. TRANSPOSE OF A LINEAR MAP AND OF A MATRIX
109
It is easy to verify that the following properties hold:
(f + g) = f + g
(g ◦ f) = f ◦ g
idE = idE∗.
Note the reversal of composition on the right-hand side of (g ◦ f) = f ◦ g .
The equation (g ◦ f) = f ◦ g implies the following useful proposition.
Proposition 4.21. If f : E → F is any linear map, then the following properties hold:
(1) If f is injective, then f
is surjective.
(2) If f is surjective, then f
is injective.
Proof. If f : E → F is injective, then it has a retraction r : F → E such that r ◦ f = idE,
and if f : E → F is surjective, then it has a section s: F → E such that f ◦ s = idF . Now,
if f : E → F is injective, then we have
(r ◦ f) = f ◦ r = idE∗,
which implies that f
is surjective, and if f is surjective, then we have
(f ◦ s) = s ◦ f = idF∗,
which implies that f
is injective.
We also have the following property showing the naturality of the eval map.
Proposition 4.22. For any linear map f : E → F , we have
f
◦ evalE = evalF ◦ f,
or equivalently, the following diagram commutes:
E∗∗ f
/ F ∗∗
O
O
evalE
evalF
E
/ F.
f
Proof. For every u ∈ E and every ϕ ∈ F ∗∗, we have
(f
◦ evalE)(u)(ϕ) = f
(evalE(u)), ϕ
= evalE(u), f (ϕ)
= f (ϕ), u
= ϕ, f (u)
= evalF (f (u)), ϕ
= (evalF ◦ f)(u), ϕ
= (evalF ◦ f)(u)(ϕ),
which proves that f
◦ evalE = evalF ◦ f, as claimed.
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CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY
If E and F are finite-dimensional, then evalE and then evalF are isomorphisms, so Propo-
sition 4.22 shows that if we identify E with its bidual E∗∗ and F with its bidual F ∗∗ then
(f ) = f.
As a corollary of Proposition 4.22, if dim(E) is finite, then we have
Ker (f
) = evalE(Ker (f )).
Indeed, if E is finite-dimensional, the map evalE : E → E∗∗ is an isomorphism, so every
ϕ ∈ E∗∗ is of the form ϕ = evalE(u) for some u ∈ E, the map evalF : F → F ∗∗ is injective,
and we have
f
(ϕ) = 0 iff f
(evalE(u)) = 0
iff evalF (f (u)) = 0
iff f (u) = 0
iff u ∈ Ker (f)
iff ϕ ∈ evalE(Ker (f)),
which proves that Ker (f
) = evalE(Ker (f )).
The following proposition shows the relationship between orthogonality and transposi-
tion.
Proposition 4.23. Given a linear map f : E → F , for any subspace V of E, we have
f (V )0 = (f )−1(V 0) = {w∗ ∈ F ∗ | f (w∗) ∈ V 0}.
As a consequence,
Ker f = (Im f )0
and
Ker f = (Im f )0.
Proof. We have
w∗, f (v) = f (w∗), v ,
for all v ∈ E and all w∗ ∈ F ∗, and thus, we have w∗, f(v) = 0 for every v ∈ V , i.e.
w∗ ∈ f(V )0, iff f (w∗), v = 0 for every v ∈ V , iff f (w∗) ∈ V 0, i.e. w∗ ∈ (f )−1(V 0),
proving that
f (V )0 = (f )−1(V 0).
Since we already observed that E0 = 0, letting V = E in the above identity, we obtain
that
Ker f = (Im f )0.
From the equation
w∗, f (v) = f (w∗), v ,
4.4. TRANSPOSE OF A LINEAR MAP AND OF A MATRIX
111
we deduce that v ∈ (Im f )0 iff f (w∗), v = 0 for all w∗ ∈ F ∗ iff w∗, f(v) = 0 for all
w∗ ∈ F ∗. Assume that v ∈ (Im f )0. If we pick a basis (wi)i∈I of F , then we have the linear
forms w∗i : F → K such that w∗i(wj) = δij, and since we must have w∗i, f(v) = 0 for all
i ∈ I and (wi)i∈I is a basis of F , we conclude that f(v) = 0, and thus v ∈ Ker f (this is
because w∗i, f(v) is the coefficient of f(v) associated with the basis vector wi). Conversely,
if v ∈ Ker f, then w∗, f(v) = 0 for all w∗ ∈ F ∗, so we conclude that v ∈ (Im f )0.
Therefore, v ∈ (Im f )0 iff v ∈ Ker f; that is,
Ker f = (Im f )0,
as claimed.
The following proposition gives a natural interpretation of the dual (E/U )∗ of a quotient
space E/U .
Proposition 4.24. For any subspace U of a vector space E, if p : E → E/U is the canonical
surjection onto E/U , then p is injective and
Im(p ) = U 0 = (Ker (p))0.
Therefore, p is a linear isomorphism between (E/U )∗ and U 0.
Proof. Since p is surjective, by Proposition 4.21, the map p
is injective. Obviously, U =
Ker (p). Observe that Im(p ) consists of all linear forms ψ ∈ E∗ such that ψ = ϕ ◦ p for
some ϕ ∈ (E/U)∗, and since Ker (p) = U, we have U ⊆ Ker (ψ). Conversely for any linear
form ψ ∈ E∗, if U ⊆ Ker (ψ), then ψ factors through E/U as ψ = ψ ◦ p as shown in the
following commutative diagram
p
E
/
ψ
!❈
❈
❈
❈
❈
❈
❈
❈
❈
E/U
ψ
K,
where ψ : E/U → K is given by
ψ(v) = ψ(v),
v ∈ E,
where v ∈ E/U denotes the equivalence class of v ∈ E. The map ψ does not depend on the
representative chosen in the equivalence class v, since if v = v, that is v − v = u ∈ U, then
ψ(v ) = ψ(v + u) = ψ(v) + ψ(u) = ψ(v) + 0 = ψ(v). Therefore, we have
Im(p ) = {ϕ ◦ p | ϕ ∈ (E/U)∗}
= {ψ : E → K | U ⊆ Ker (ψ)}
= U 0,
which proves our result.
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CHAPTER 4. DIRECT SUMS, THE DUAL SPACE, DUALITY
Proposition 4.24 yields another proof of part (b) of the duality theorem (theorem 4.17)
that does not involve the existence of bases