Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations
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triangle
inequality
narmed
linear
space
natural
distance
in
a
normed
linear
space
parallelogram
law
1.2 INNER
PROOUCT
31
1. The norm of the zero vector is zero:
11011
= 0.
2.
lIall
2::
0, and
lIall
= 0 if and only
if
la) = 10).
3. lIaall
= lailiall tor any? complex o.
4. lIa+bll
::0
lIall
+
IIbli.
This property is called the
triangle
ineqnality.
A vector space on which a
nonn
is defined is called a
normed
linear
space.
One can
introduce
the idea of the
"distance"
between two vectors in a nonned
linear space. The distance between la) and
Ib)----<!enoted
by
d(a,
b)-is
simply
the norm
of
their difference: d(a, b) es lIa - bll.
It
can be readily shown that this
has all the properties one expects of the distance (or metric) function introduced
in Chapter
O.
However, one does not need a
nonned
space to define distance. For
exannple,as explainedin Chapter0, one can define the distancebetweentwo points
on the surface
of
a sphere, but the addition of two points on a
sphere-a
necessary
operation for vector space
structure-is
not defined. Thus the points on a sphere
forma metricspace,butnota vectorspace.
Inner product spaces are automatically nonned spaces, but the converse is not,
in general, true: There are nonned spaces, i.e., spaces satisfying properties
1-4
above that cannot be promoted to inner product spaces. However, if the norm
satisfies the paraUelogram law,
(1.7)
then one can define
and show that it is indeed an inner product. In fact, we have (see [Frie 82, pp.
203-204] for a proof) the foUowing theorem.
1.2.8.
Theorem.
A normed linear space is an innerproduct space ifand only if
the norm satisfies the parallelogram law.
Now consider any N-dimensionalvector space
V.Choose a basis Ilain!:! in
V, and for any vector
la)
whose components are la;
1!:1
in this basis, define
N
lIall
2
==
~)ail2.
i=l
The reader may check that this defines a norm, and that the norm satisfies the
parallelogrann law. From Theorem 1.2.8 we have the foUowing:
1.2.9. Theorem. Everyfinite-dimensional
vector
space
can be turnedinto an inner
product space.
32 1.
VECTORS
ANO
TRANSFORMATIONS
<C"
has
many
different
distance
functions
1.2.10. Example. Let the
space
be
en.
The
natural
inner
product
of
en
givesriseto a
norm, which,forthe
vector
la} = (cq , aZ
I
•••
, an) is
n
lIall=~=
~)a;l2.
i=l
This
norm
yields
the
following
distance
between
la) and
Ib)
= Uh.lh.....
fJn):
n
d(a,b)
= Iia-bil
=)(a-bla-b)=
~)ai
-fJiI
2
•
i=l
Onecan
define
nthernorms,
such
aslIa
IIJ
sa
LI
~l
lai I,
which
hasallthereqnired
properties
of anonn,andleadstothe
distance
n
dl (a. b) = lIa- bill =
~)aj
-
fJ;I·
i=1
Another
norm
defined
on
en
is givenby
lI
ali
p
=
(~Ia;lp
)
lip
where
p isa
positive
integer.
It
is
proved
in
higher
mathematical
analysis
that
II
.lip hasall
the
properties
ofanonn.(The
nontrivial
part
oftheproofistoverify
the
triangle
inequality.)
The
associated
distance
is
("
)t
lP
dp(a,
b) = lIa- blip =
\;~
laj
-
fJ;lP
The
other
twonoons
introduced
above
arespecialcases,forp = 2 andp = 1.
III
1.3 Linear Transformations
Wehave made progress in enriching vector spaces with structures such as norms
andinner products.However,this enrichment,althoughimportant,will be of little
value
if
it is imprisoned in a single vector space. We would like to give vector
spacepropertiesfreedomof movement, so they can go from one spaceto another.
The vehicle that carries these properties is a linear transformation which
is the
subject of this section. However,first it is instructive to review the concept of a
mapping (discussed in Chapter 0) by considering some examples relevant. to the
present
discussion.
1.3.1.
Example. The
following
area
few
familiar
examples
of
mappings.
9Thefirst
property
followsfromthisby lettinga =
O.
1.3 LINEAR TRANSFORMATIONS 33
1. Let f :
IR
->-1R
be givenby
f(x)
= x
2
•
2. Let g :
1R
2
->-
IR
be givenby
g(x,
y) = x
2
+ y2 - 4.
3. Let
F :
1R
2
->-
iC
begivenby F(x, y) = U(x, y) +
iV(x,
y), where U :
1R
2
->-
IR
and V :
1R
2
->-
lR.
4. Let T :
IR
->-
1R
2
be givenby
T(t) = (t +3, 2t - 5).
5.
Motion ofapointparticlein space can be consideredas a mapping M : [a, b] --+
JR
3
,
where [a, b] is an intervalof the real line. For each t E [a, b], we define
M(t)
=
(x(t), y(t),
z(t»,
where x(t), y(t), and z(t) are real-valned functions of t.
If
we
identify
t withtime,whichis
assumed
to havea
value
in the
interval
[a, b],
then
M(t)
describes
the
path
of the
particle
as a
function
of time,anda
and
b arethe
beginning
and
theendof the
motion,
respectively.
II
Let
us consider an arbitrary mapping F : V
--->
W from a vector space V
to another vector space W.
It
is assumed that the two vector spaces are over the
same scalars, say
Co
Consider la)
and
Ib) in V
and
Ix)
and
Iy) in W such that
F(la)
= Ix) and F(lb) = Iy).
In
general, F does
not
preserve the vector space
structnre.
That
is, the image
of
a linear combination
of
vectors is
not
the same as
thelinearcombination of theimages:
F(a
la) +
{J
Ib)
i'
aF(lx)
+
{JF(ly).
This is the case for all the mappings
of
Example 1.3.1 except the fourth item.
Thereare many applications in whichthe preservation
of
the vectorspace structure
(preservation
of
the linearcombination) is desired.
1.3.2. Definition. A linear transformation from the complex vector space V to
the complex vector space
W is a mapping T : V
--->
W such that
linear
transformation,
linear
operator,
endomorphism
T(a
la) +{JIb)) = aTria)) +{JT(lb))
Via)
, Ib) E V
and
a,
{J
E C.
A linear transformation T : V
--->
Vis called an
endomorphism
of
V or a linear
operatoron V. The action
of
a linear transformation on a vectoriswritten without
the parentheses: T(la)) sa T la).
The
same definition applies to real vector spaces. Note that the definition
demands that both vector spaces have the same set
of
scalars: The same scalars
multiply vectors in
V on the LHS and those in W on the RHS. An immediate
consequence
of
the definition is the following:
1.3.3. Box. Twa linear transformations T : V
--->
Wand
U : V
--->
Ware
equal
if
and
only
ifT
la,) = U lail
for
all la,) in same basis
afV.
Thus, a
linear transformation is uniquely determined by its action an same basis
of
its domain space.
34 1.
VECTORS
AND
TRANSFORMATIONS
linear
functional
qv, W)isa
vector
space
dual
vector
space
v·
derivative
operator
integration
operator
integration
isa
linear
functional
on
the
space
of
continuous
functions
The
equality in this box is simply the set-theoretic equality
of
maps discussed
in Chapter O.Au importaut example
of
a linear trausformation occurs when the
second vector space,
W, happens to be the set
of
scalars, C or
JR,
in which case
the linear trausformation is called a
linear
functional.
The
set
of
lineartrausformations from Vto Wis denotedby
r:.
(V, W), aud this
set happens to be a vectorspace.
The
zero transformation, 0, is definedto take every
vectorin
V to the zero vector
of
W.
The
sum
of
two lineartrausformations Taud U
is the lineartrausformationT
+U,
whose action on a vector la) E V is definedto be
(T+U) la) es T la) + U la). Similarly, define
aT
by
(aD
la) es
arT
tal)
=
aT
la).
The set
of
endomorphisms
of
Vis denoted by
r:.
(V) rather thau t:(V,
V).
The
set
oflinear
functionals
r:.
(V,
iC)--or
t:(V,
JR)
if
Vis a real vector space---
is denoted by
V* aud is called the dual
space
of
V.
1.3.4. Example. Thefollowing aresome
examples
of
linear
operators
in
various
vector
spaces.Theproofsof
linearity
are
simpleinall cases andareleftasexercises forthe
reader.
I.
Let {Jal} , la2) ,
...
, lam}} be an arbitraryfiniteset of vectorsin V, and {I), 12,
...
,
1
m
}
an
arbitrary
set of linear functionals on V.Let
m
A'"
L
lak)lk
E
qV)
k~l
be definedby A [x) =
L:~llak)
Ik(lx» =
L:~1
Ik(lx» lak}. Then A is a linear
operator
onV.
2.
Let"
be a permutation(shnffiing) of the integers
It,
2,
...
, n}.
If
[x} =
(~1,
~2,
...
,
fJn)
isa
vector
in
en,
wecanwrite
An Ix} =
(~n(l)'
~n(2),
...
,
~n(n).
ThenAn is a
linear
operator.
3. Forany Ix) E
:P'[t],
withx(t)
=
L:Z~o
cxktk,write Iy} = D
[x},
whereIy) isdefined
as
yet) =
Lk=l
kaktk-l.
Then
Dis a
linear
operator,
thederivativeoperator.
4. For every [z} E
:P'[t],
with
x(t)
=
L:Z~o
cxktk, write Iy) = S Ix}, where Iy) E
:P'[t]
is definedas yet) =
L:Z~O[CXk/(k
+
l)]t
H I.
Then S is a linearoperator, the
integrationoperator.
5. DefinetheoperatorInl : eO(a, b)
-+
IRby
inl(f)
= l
b
f(t)
dt.
Thenintis a
linear
functional
onthevectorspaceeO(a,
b)~
6. Let en(a, b) be the set ofreat-valued functionsdefinedin the interval [a, b] whose
first n derivatives exist and are continoous.For any
If)
E en(a, b) define lu} =
G
If},
with u(t) =
g(t)f(t)
and get) a fixedfunctionin en(a, b). Then Gis linear.
In
particular.
the
operation
of multiplying by t, whose
operator
is
denoted
by T,is
~~
.
1.3
LINEAR
TRANSFORMATIONS
35
An immediate consequence
of
Definition 1.3.2 is that the image
of
the zero
vector in V is the zero vector in W. This is not true for a generalmapping, but it is
necessarily true for a linearmapping. As the zero vector
of
V is mapped onto the
zero vector
of
W, othervectors
of
V may also be dragged along. In fact, we have
the following theorem.
1.3.5.
Theorem.
The set
of
vectors in V that are mapped onto the zero vector
of
W under the linear transformation T : V --+ W form a subspace
of
V called the
Tremel,
or
null
space,
ofT
and denoted by
kerT.
kernel
ofa
linear
transformation
Proof
The
proofis left as an exercise".
D
nullity
The
dimension
of
ker T is also called the
nullity
of
V.
The
proof
of
the following is also left as an exercise.
rank
ofa
linear
transformation
1.3.6.
Theorem.
The range T(V)
of
a linear transformation T
subspace
ofW.
The dimension
ofT(V)
is called the
rank
ofT.
V--+Wisa
1.3.7.
Theorem.
A linear transformation is
I-I
(injective) iffits kernel is zero.
Proof
The
"only
if"
partis trivial. For the
"if"
part, suppose T lat) = T la2);then
linearity
of
T implies that T(lat) - la2)) =
O.
Since
kerT
= 0, we must have
lat) = la2). D
Suppose we start with a basis
of
ker T and add enough linearly independent
vectorsto it
toget a basis for V.Withoutloss
of
generality, let us assumethatthe first
n vectors in this basis form a basis
ofker
T. So let B = (Ial) , la2),
...
, IaN)}be a
basis for V and
B' = (Iat) , la2),
...
, Ian)}be a basis for kerT. Here N =dim V
andn
= dim
kerT.
II is straightforward to show that (Tlan+t),
...
, T IaN)}is a
basis for T(V). We therefore have the following result.
dimension
theorem
1.3.8.
Theorem.
Let T : V --+ W be a linear transformation.
Then!O
dim V = dim ker T +dim T(V)
Thistheorem is calledthe
dimension
theorem.
One
of
its consequencesis that
an injective
endomorphism
is
automatically
surjective,
and
vice versa:
1.3.9.
Proposition.
An endomorphism
of
a finite-dimensional vector space is bi-
jective zfit is either injective or surjective.
The
dimension theorem is obviously valid only for finite-dimensional vec-
tor spaces.
In particnlar, neither swjectivity nor injectivity implies bijectivity for
infinite-dimensional vector spaces.
lORecall that the dimension of a vector space depends on the scalars used in that space. Although we are dealing with two
different vector spaces here, since they are both over the same set
of
scalars (complex or real), no confusion in the concept
of
dimension arises.
36 1.
VECTORS
AND
TRANSFORMATIONS
1.3.10. Example. Letus try to findthekernel
ofT
: ]R4
-+
]R3 givenby
2xl
+x2
+x3
-X4
=0,
XI
+x2
+2X3
+2X4
= 0,
XI - X3 - 3X4 = 0.
The
"solution"
tothese
equations
is Xl = x3 +3X4 andX2 =
-3X3
- 5X4.
Thus,
tobe in
kerT,a vectorinr mustbeof theform
where
X3 andX4
are
arbitrary
real
numbers.
It follows
that
kerTconsists of
vectors
that
can
be
written
aslinearcombinations of the two linearly independent vectors(1,
-3,
1,0) and
(3,
-5,0,
I). Therefore, dim
kerT
= 2. Theorem 1.3.8then saysthat dimT(V) = 2; that
is, the
range
ofTis
two-dimensional.
Thisbecomesclearwhenonenotes
that
and
therefore
T(x!. X2. X3,
X4),
an
arbitrary
vectorinthe
range
ofT, is a
linear
combination
of only twolinearlyindependentvectors,
(I,
0,
I)
and (0,
I,
-I).
III
isomorphism
and
automorphism
In many cases, two vector spaces
may
"look" different, while in reality they
are very
much
the same. For example, the set
of
complex numbers C is a two-
dimensional vector space
over the reals, as is
1R
2.
Although we call the vectors
of
these two spaces by differentnames, they have very similarproperties. Thisnotion
of
"similarity" is made precise in the following definition.
1.3.11. Definition. A vector space V is said to be isomorphic to another vector
space Wifthere exists a bijective linear mapping T ;
'17--->
W.Then Tis called an
isomorphism.
II
A bijective linear map
of
Vonto itselfis calledan automorphism
ofV.
The set
of
automarphisms
of
Vis denoted by
GL(V).
For
all practicalpurposes, two isomorphic vector spaces are differentmanifes-
tations
of the
"same"
vector
space.
Inthe
example
discussed
above,
the
correspon-
dence
T;
C
--->
1R
2
,
with T(x +
iy)
= (x,
y),
establishes an isomorphism between
the two vectorspaces.
It
should be emphasized that only as vector spaces are C
and
1R
2
isomorphic.
If
we go beyond the vector space structures, the two sets are
qnite different. For example, C has a natural multiplication for its elements, but
1R
2
does not. The following theorem gives a working criterionfor isomorphism.
llThe word"isomorphism," as we shall see, is used in
conjunction
with many
algebraic
structures.
To
distinguish
them,
qualifiers
needto be used.In the
present
context,
we speakof linear isomorphism.We shalluse
qualifiers
when
necessary.
However,
thecontextusuallymakesthe
meaning
of
isomorphism
clear.
only
two
N-dimensional
vector
spaces
1.3 LINEAR
TRANSFORMATIONS
37
1.3.12. Theorem. A linear surjective map T : V
--+
W is an isomorphism ifand
only
ifits nullity is zero.
Proof
The "only
if"
part is obvious. Toprove the
"if"
part, assume that the nnllity
is zero. Then by Theorem 1.3.7, T is
I-I.
Since it is already surjective, Tmust be
bijective. D
1.3.13. Theorem. An isomorphism T : V
--+
W carries linearly independent sets
of
vectors onto linearly independent sets
of
vectors.
Proof
Assume that {Iai)
}I~I
is a set of linearly independentvectors in V.To show
that{T
lai))I=1
islinearly independentin W, assume that there exist eq , az,
...
,am
such that
2:1~1
a;T lai) =
10).
Then the linearity
of
T and Theorem 1.3.12 give
T(2:7:"1
a; lai)) =
10),
or 2:;'::1 a; laj) =
10),
and the linear independence of
the
laj) implies that a, = °for all i. Therefore, {Tlaj
))I~I
must be linearly
independent. D
The following theorem showsthat finite-dimensionalvector spaces are severely
limited in number:
1.3.14. Theorem. Twofinite-dimensionalvector spaces areisomorphic ifand only
ifthey have the same dimension.
Proof
Let
Bv
=
{lai)}~1
be a basis for V and
Bw
=
{lbi))~1
a basis for W.
Define the linear transformation
Tlai) = Ibi), i = 1,2,
...
, N. The rest of the
proofinvolves showing that T is an isomorphism. We leave this as an exercise for
the reader. D
A consequence of Theorem 1.3.14 is that all N-dimensional vector spaces
over
JR
are isomorphic to
JRN
and all complex N-dimensional vector spaces are
isomorphic to
eN.
So, for all practicalpurposes, we have only two N-dimensional
vectorspaces,
IRN
and
eN.
1.3.1 Moreon LinearFunctionals
An example
of
isomorphism is that between a vector space and its dual space,
which we discuss now. Consider an N-dimensional vector space with a basis
B =
{Ial),
laz),
...
, ION)). For any given set of N scalars,
{ai,
az,
...
,
aN},
define the linear functional fa by fa laj) = at. Whenfa acts on any arbitrary vector
Ib) =
2:~dJ;
laj) in V, the result is
(1.9)
This expressionsuggests that
Ib)can berepresentedas acolumnvectorwith entries
th,
fJz,
...
,fJN and fa as a row vector with entries
ai,
az,···,
aN.
Then fa Ib) is
38 1.
VECTORS
ANO
TRANSFORMATIONS
Every
setofN
scalars
defines
a
linear
functional.
merely the matrix product'< of the row vector (on the left) and the column vector
(on the right).
fa is uniquely detennioed by the set
{aI,
a2,
...
,aN
j.
In
other words, corre-
sponding to every set of
N scalars there exists a unique linear functional. This
leads us to a particular set
of
functionals,
fl,
f2, , fN corresponding, respec-
tively, to the sets of scalars {I, 0, 0,
...
,
OJ,
{O,
1,0,
,
OJ,
.,.,
(O,
0, 0,
...
,
I],
This means that
fl
101)
= 1
f21a2) = 1
or that
and
and
and
fllaj}
=0
f2laj}
=0
for j
i'
1,
for
j
i'
2,
for
j
i'
N,
(1.10)
where
8ij is the Kronecker delta.
The functionals
of
Equation (1.10) form a basis
of
the dual space V*. To
show this, consider an arbitrary 9 E V*, which is uniquely determined by its
action on the vectors in a basis B =
lIal)
,
la2}
,
...
,
ION}
j. Let 9 lai} = Yi E C.
Then we claim that 9
=
L:;:"I
Yifi.
In
fact, consider an arbitrary vector
10)
in
V with components
(aI,
a2,
...
,
aN)
with respect to B. Then, on the one hand,
9
10)
=
g(L:;:"1
a; lai) =
L:;:"I
aig
lai} =
L:;:"!
aiYi. On the other hand,
(~Yifi)
la} =
(~Yifi)
(~aj
la
j))
N N N N N
=
LYi
Lajfi
laj) =
LYi
LaAj
=
LYiai.
;=1
j=l
;=1
j=l
i=l
Since the actions
of
9 and
L:;:"I
Yifi yield equal results for arbitrary 10),we con-
clude that 9 =
L:;:"I
Yifi, i.e., {fi
j;:"1
span V'. Thus, we have the following result.
1.3.15.
Theorem.
If
V is an N-dimensional vector space with a basis B =
lIal} ,
102)
,
...
,
ION}
j, then there is a corresponding unique basis B* = {f;};:"!
in V*with the property that fi
10
j} = 8ij.
By this theorem the dual space
of
an N-dinnensional vector space is also
N-
dual
basis
dinnensional, and thus isomorphic to it. The basis B* is called the dual basis of
B. A corollary to Theorem 1.3.15 is that to every vector in V there corresponds a
12Matrices
will be
taken
upin
Chapter
3.Here,we assumeonlya
nodding
familiarity
with
elementary
matrix
operations.
1.3 liNEAR
TRANSFORMATIONS
39
unique linear functional in V*. This can be seen by noting that every vector la) is
uniquely determinedby its components
(ai,
a2,
...
,aN)
in a basis B. The unique
linear functional
fa corresponding to [a), also called the dual of la), is simply
Lf::l
ajfj,
withIi E
B*.
annihilator
ofa
vector
and
a
subspace
1.3.16. Definition. An annihilator
of
la) E V is a linear junctional f E V* such
that
f la) =
O.
Let W be a subspace
ofV.
The set
of
linear functionals in V* that
annihilate all vectors in
W is denoted by
WO.
The reader may check that
WO
is a subspace
of
V*. Moreover,
if
we extend
a basis
{Iai))f~t
of W to a basis B = (Iaj)
1[:,1
of
V, then we can show that the
functionals
{fj
17=k+l'
chosen from the basis B* = (fj 17=t
dnalto
B,
span
WO.
I!
then follows that
dim V
=
dimW+dimWO.
(1.11)
dual,
or
pull
back,
of
a
linear
transformation
We shall have occasions to use annihilators later on when we discuss symplectic
geometry.
We have
"dualed" a vector, a basis, and a complete vector space. The only
objectremaining is a linear transformation.
1.3.17. Definition.
Let
T : V --> 11be a linear transformation. Define
l'
:
11*
-->
V*
by13
[T*(g)J la) = g(T la»)
Via)
E V, 9 E
11*,
T* is called the
dual
or
pull
back,
ofT.
One can readily verify that
l'
E £'(11*,
VOl,
i.e., that T* is a linear operator
on
11*.
Some of the mapping properties
of
T* are tied to those of T. To see this
we first consider the kernel
of
1'.
Clearly, 9 is in the kernel
of
l'
if
and only
if
9
annihilates all vectors of the form T [a), i.e., all vectors in T(V).
I!
follows that 9 is
in T(V)o. In particular,
ifT
is surjective, T(V) = 11,and 9 annihilates all vectors in
11,
i.e., it is the zero linearfunctional. We conclude that
ker
T* = 0, and therefore,
T*is injective. Similarly, one can show that
if
T is injective, then T*is surjective.
We summarize the discussion above:
1.3.18. Proposition.
Let
T be a linear transformation
and
T* its pull back. Then
kerT* = T(V)o.
1fT
is surjective (injective), then T* is injective (surjective). In
particular, T" is an isomorphism
ifT
is.
I!
is useful to make a connection between the inner product and linear func-
tionals. To do this, consider a basis (Iat) , la2) ,
...
, IaN)) and
letaj
=
(al
aj). As
noted earlier, the set
of
scalars
(ail[:,t
defines a unique linear functional fa such
that
t, lai) = ai. Since (a I
aj)
is also equal to cq,
ilis
natural to identify fa with
duals
and
inner
products
13Donotconfusethis"*,,withcomplex
conjugation.
40 1.
VECTORS
AND
TRANSFORMATIONS
tihe
symbol (ai, and write
T:
fa
>-->
(al where T is tiheidentification map.
It
is also convenient to introduce
tihe
notation'f
(Ia))t
es
(ai,
(1.12)
dagger
ofa
linear
combination
of
vectors
where
tihe
symbolt means "dual, or daggerof." Now we ask: How does
tihis
dagger
operation act on a linear combination
of
vectors?
Let
]c) =
ala)
+
fJ
Ih) and take
tihe
inner product
of
[c) witihan arbitrary vector Ix} using linearity iu
tihe
second
factor:
(xl
e) = a
(xl
a) +
fJ
(xl
h). Now complex conjugate botihsides and use
tihe
(sesqui)symmetry
of
tihe
inner product:
(LHS)*
=
(xle)*
=
(cl
x},
(RHS)*
=
a*
(xl
a)*
+
fJ*
(xl
h)* =
a*
(al x) +
fJ*
(hi x)
=
(a*
(al +
fJ*
(hi) [x) .
Since
tihis
is true for all [x), we musthave
(Ie))t
es (c] =
a*
(al+fJ*
(hi.Therefore,
in a duality "operation" tihecomplex scalars must be conjugated. So, we have
(a la) +
fJ
Ih})t =
a*
(al +
fJ*
(hi.
(1.13)
Thus, unlike tiheassociation la) .... la which is linear,
tihe
association la .... (al is
uotlinear,
but sesquilinear, i.e., tiheideutification map T mentioned above is also
sesquilinear:
T(ala +fJlb) =
a*
(al +
fJ*
(hi =
a*T(l
a)
+fJ*T(lb);
It
is convenient to represent lu) E
en
as a column vector
Then
tihe
definition
of
the complex innerproduct suggests tihat
tihe
dual
of
la) must
be represented as a row vector witihcomplex conjugate entries:
Compare
(1.14)
with
the
comments
after
(1.9).
The
complex
conjugation
in
(1.14)
is
the
resuit
of
the
sesquilinearily
of
the
association
la)<+ (s],
(al =
(aj
ai
...
a~),
and tiheinner product can be written as
tihe
(matrix) product
(al h) =
(aj
ai
(1.14)
14The
significance of this
notation
will becomeclearinSection2.3.