Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations
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4.1
DIRECT
SUMS
111
verified that P] = Pj and P
jPk
= 0 for j
i'
k. Thus, the P
/s
are (not necessarily
hermitian) projection operators. Furthermore, for an arbitrary vector
lu), we have
¥Iu)
E V.
Since this is true for all vectors, we have the identity
(4.2)
orthogonal
complement
ofa
subspace
4.1.5.
Definition.
Let
V be an innerproduct space.
Let
JV(be any subspace
ofV.
Denote by
JV(.L
the set
of
all vectors in Vorthogonal to all the vectors in
:M.
JV(.L
(pronounced "em
perp")
is called the orthogonalcomplementofJV(.
4.1.6.
Proposition.
JV(.L
is a subspace
ofV.
Proof In fact, if la),
Ib)
E
JV(.L,
then for any vector [c) E JV(,we have
=0 =0
-
---
(cl (a la) +
f3lb»
= a (c] a) +f3 (c] b) =
O.
So,
ala)
+f3lb} E
JV(.L
for arbitrary a,
f3
E C and la} , Ib) E
JV(.L.
o
If
V
of
Equation (4.1) is an inner product space, and the subspaces are mutually
orthogonal, then for arbitrary
lu) , Iv} E V,
whichshows that P
j is hermitian.InChapter2, we assnmedthe projectionoperators
to be hermitian. Now we see that only
in an ioner product space (and only
if
the subspaces
of
a direct
sum
are orthogonal) do we recover the hermiticity
of
projection operators.
4.1.7.
Example.
Consideran orthonormalbasis BM = !Iei)):"=l for M, and extendit
to a basis
B =
!Iei))~l
for V. Now coostruct a (hermitian) projection operator P =
L~l
lei) {ei
I·
Thisis theoperatorthat
projects
an
arbitrary
vector
inVanta the
subspace
M.
It
is straightforwardto showthat1- P isthe projectionoperatorthatprojectsontoM.L
(seeProblem4.2).
An
arbitrary
vector
la) E Vcanbe
written
as
la) =(P +1 - P)la) =Pia) +(1 - P) la) .
'-.-'
in
.M:
in:M.l
Furthermore,
theonly
vector
that
canbeinboth
:M
andMl..is thezero
vector,
because
itis
the onlyvectororthogonalto itself.
III
112
4.
SPECTRAL
DECOMPOSITION
Fromthis example and the remarks immediatelypreceding it we may conclude
the following:
4.1.8. Proposition.
If
V is an inner product space, then V = M
Ell
M.Lfor any
subspace
M. Furthermore, the projection operators corresponding to M and M.L
are hermitian.
4.2 Invariant Subspaces
invariant
subspace:
reduction
ofan
operator
matrix
representation
ofan
operator
in
a
subspace
block
diagonal
matrix
defined
This section explores the possibility of obtaioiog subspaces by means
of
the action
of
a linearoperator on vectors of an N-dimensional vector space V. Let
10)
be any
vector 10V, and A a linearoperator on V. The vectors
10)
,A
10)
,A
2
1a) ,
...
,AN
10)
are linearly dependent (there are N + I of them!). Let M es Span{A
k
10)
Jf=o'
It
follows that, m es dim M
="
dim V, and M has the property that for any vector
[x) E M the vectorA[x) also belongs to M (show this!).
In
otherwords, no vector
10M "leaves" the subspace wheu acted on by A.
4.2.1. Definition.
A subspace M is an invariant subspace
of
the operator A ifA
transforms vectors
ofM
into vectors
ofM.
This iswrittensuccinctlyas A(M) C M.
We say that M reduces A ifboth M and M.L are invariant subspaces
of
A.
Startiog with a basis
of
M, we can extend it to a basis B = {Iai)
J!:,
of V
whose first
m vectors span M. The matrix representation
of
A 10such a basis is
given by the relation
Alai)
=
I:f=''''jilaj),i
=
1,2,
...
,N.lfi
="
m,then
",
ji
= 0 for j > m, because
Alai)
belongs to M when i
="
m and therefore can
be written as a linear combioation of only
{Ia,)
, 102),
...
, lam)). Thus, the matrix
representation of A10
B will have the form
A =
(An
A12),
02'
A22
where
An
is
anm
x m matrix, A'2
anm
x
(N
- m) matrix, 021the
(N
- m) x m
zero matrix, and
A22
an
(N
- m) x
(N
- m) matrix. We say that
An
represents
the operator A10the m-dimensional subspace M.
It may also happen that the snbspace spanned by the remaioiog basis vectors
10
B, namely la
m
+,) , la
m
+2) ,
...
, ION),is also an invarianl subspace of A.Then
A12
will be zero, and A will take a block
diagonal
form:'
A =
(An
0)
o
A22
A matrix representation of an operator that can be brought ioto this form by a
reducible
and
suitablechoice
of
basis is said 10 be reducible; otherwise, it is called irreducible.
irreducible
matrices
1Fromnowon,we shalldenoteallzero
matrices
bythesamesymbol
regardless
of their
dimensionality.
4,2
INVARIANT
SUBSPACES
113
A reducible matrix A is denoted in two different waysr'
(4.3)
condition
for
invariance
For
example, when M reduces A and one chooses a basis the first m vectors
of
which are in M and the remaining ones in M1-, then A is reducible. We have
seen on a number of occasions the significance
of
the hermitian conjugate of an
operator (e.g., in relation to hermitian and unitary operators). The importance of
this operator will be borne out further whenwe study the spectral theoremlater in
this chapter. Let us now investigate some properties of the adjoint
of
an operator
in the context of invariant subspaces.
4.2.2.
Lemma.
A subspace M
of
an inner product space V is invariant under the
linear operator
A
if
and only
ifM
1- is invariant underAt,
Proof
The proofis left as a problem.
D
Animmediate consequence
of
the above lemmaand the two identities (At)t =
A and
(M
1-)1- = M is contained in the following theorem.
4.2.3.
Theorem.
A subspace
of
V reduces A
if
and
only
if
it is invariant under
both A and
At.
4.2.4.
Lemma.
Let
M be a subspace
of
V
and
P the hermitianprojection operator
onto
M.
Then M is invariant under the linear operatorA
if
and
only
if
AP = PAP,
Proof
Suppose M is invariant. Then for any [x) in V, we have
Pix)
EM·=}
AP Ix)
EM=}
PAP Ix) = AP Ix) .
Since the last equality holds for arbitrary [x), we bave AP = PAP,
Conversely, suppose AP = PAP. For any IY) E
M,
we have
Ply)
= IY) =} AP Iy) = A Iy) = P(AP
IY»
E M.
'-.-'
=PAP
Therefore, M is invariant under A.
D
4.2.5.
Theorem.
Let
M be a subspace
of
V, P the hermitian projection operator
of
Vonto M, and
Aa
linear operator on V.Then M reduces A
ifand
only
if
A
and
P
commute.
21tis
common
to use a singlesubscriptfor submatrices
of
a
block
diagonalmatrix,
just
as it is
common
to use a single subscript
for entries
of
a diagonal matrix.
114 4.
SPECTRAL
DECOMPOSITION
Proof
Suppose M reduces A.Then by Theorem 4.2.3, M is invariant under both
Aand At. Lemma4.2.4 then implies
AP = PAP and
Atp
= PAtp. (4.4)
Taking the adjoint of the second equation yields
(Atp)t
=
(PAtp)t,
or PA =PAP.
This equation together with the first equation
of
(4.4) yields PA = AP.
Conversely, suppose that PA
= AP. Then p
2A
= PAP, whence PA = PAP.
Taking adjoints gives
Atp
= PAtp, because P is hermitian. By Lemma 4.2.4, M
is invariant under At. Similarly, from PA
= AP, we get PAP = Ap
2
,
whence
PAP
= AP. Once again by Lemma 4.2.4, M is invariant under A. By Theorem
4.2.3, M rednces A. D
The main goal of the remaining part
of
this chapter is to prove that certain
operators, e.g. hermitian operators, are diagonalizable, that is, that we can always
find an (orthonormal) basis in which they are represented by a diagonal matrix.
4.3 Eigenvalues
and
Eigenvectors
eigenvalue
and
eigenvector
Let us begin by considering eigenvaluesand eigenvectors, which are generaliza-
tions
of
familiar concepts in two and three dimensions. Consider the operation of
rotation aboutthe z-axis by an angle 0 denotedby Rz(O).Such a rotationtakes any
vector
(x, y) in the xy-plane to a new vector (x cos 0 - y sin 0, x sin 0 +y cos 0).
Thus, unless (x, y) =
(0,0)
or 0 is an integer multiple of
2n,
the vector will
change. Is there a nonzero vector that is so special (eigen,in German) that it does
not change when acted on by Rz(O)? As long as we confine ourselves to two di-
mensions, the answer is no. But
if
we lift ourselves up from the two-dimensional
xy-plane, we encounter many such vectors, alI
of
which lie along the z-axis.
The foregoing example can be generalized to any rotation (normally specified
by EuIer angles).
In fact, the methods developed in this section can be used to
show that a general rotation, given by Euler angles, always has an unchanged
vector lying along the axis around which the rotation takes place. This concept is
further generalized in the following definition.
4.3.1. Definition. A scalar A is an eigenvalue and a nonzero vector la} is an
eigenvector
of
the linear transformation A E
J:.,
(V) if
Ala}
=
Ala).
(4.5)
4.3.2. Proposition.
Add
the zerovector to the set
of
all eigenvectors
of
Abelonging
to the same eigenvalue
A,and denote the span
of
the resulting set by M
A
. Then
M
A
is a subspace
of
V,
and
every (nonzero) vector in M
A
is an eigenvector
of
A
with eigenvalue
A.
Proof
The prooffollows immediately from the above definition and the definition
of a subspace. D
4.3
EIGENVALUES
AND
EIGENVECTORS
115
eigenspace
4.3.3. Definition. The subspace
JVe,.
is referred to as the eigenspace
of
A corre-
sponding to the eigenvalue
),..
Its dimension is called the geometric multiplicity
of
),..
An
eigenvalue is called simple ifits geometric multiplicity is 1. The set
of
spectrum
eigenvalues
of
Ais called the spectrum
of
A.
By their very construction, eigenspaces corresponding to differenteigenvalues
have no vectors in common except the zero vector. This can be demonstrated by
noting that
if
Iv)
E.
M,
n MILfor ),
f=
1-'-,
then
0=
(A -
),.1)
Iv) = A Iv) - ),.Iv) = I-'-Iv) - ),.Iv) =
(I-'-
-),.)
Iv) '* Iv) =
O.
'--.-'
,.0
Let us rewrite Equation (4.5) as (A -
),,1)
la) =
O.
This equation says that la)
is an eigenvector of A
if
and only
if
la) belongs to the kernel of A -
),,1.
If
the
latter is invertible, then its kernel will consist
of
only the zero vector, which is
not acceptable as a solution of Equation (4.5). Thus,
if
we are to obtain nontrivial
solutions, A
-),,1
musthave no inverse. This is true
if
and only
if
det(A -
),,1)
=
O.
(4.6)
characteristic
polynomial
and
characteristic
roots
ofan
operator
The
determinantin Equation (4.6) is a polynomialin)", calledthe
characteris-
tic polynomial of A.The roots
of
this polynomial are called
characteristic
roots
and are simply the eigenvalues
of
A. Now, any polynomial
of
degree greater than
or equal to I has at least one (complex) root, which yields the following theorem.
4.3.4.
Theorem.
Every operator on afinite-dimensional vectorspace over
iC
has
at least one eigenvalue
and
therefore at least one eigenvector.
Let
X},),.2,
•.•
,),.p
be the distinct roots of the characteristic polynomial of A,
and let x
j occurmj times. Then 3
p
det(A -
),,1)
= (),.I - ),.)m
1
•••
(),.p - ),.)m
p
= D(),.j _ ),.)mj.
j~1
For),. = 0, this gives
p
detA
- 'l.rn t , m2
'l.mp_D'l.mj
-
11.1
11.2
•••
"-p
-
Aj.
j=l
(4.7)
(4.8)
Equation (4.8) states that the determinant of an operator is the product of all
its eigenvalues.
In particular,
if
one
of
the eigenvalues is zero, then the operator is
not invertible.
3mj is called the algebraic multiplicity of
J..
i-
116 4.
SPECTRAL
DECDMPDSITIDN
4.3.5.
Example.
Letusfiudtheeigenvaluesof a projectionoperatorP.1f la} isan eigen-
vector,then Pia) = Ala}. ApplyingP onboth sides again, weobtain
p2la)
=
AP
la}
=
1.(1.
la})=
Alia).
But p2 =P; thus,Pia} =Alia).
It
follows
thaUlla)
=A[e},or (1.
1
-
A)
la} =
O.
Since
la}
i'
0, wemustbaveA(A-I) = 0, or
1.=
0, 1.Thus,theonlyeigenvalues of aprojection
operator
are0 and1.The
presence
of zeroas an
eigenvalue
of Pis an
indication
that
Pis
notinvertible. II
4.3.6. Example. Tobeableto seethe
difference
between
algebraic
and
geometric
multi-
plicities,considerthematrixA=
(A
1
),
whosecharacteristicpolynomialis (I - 1.)1. Thus,
the
matrix
hasonlyone
eigenvalue,
A= I,
with
algebraic
multiplicity ml = 2.
However,
themostgeneralvectorla) satisfying
(A-1)
la) = 0 is easilyshownto beofthefonn
(~).
Thisshows
that
M).=l is
one-dimensional,
i.e., the
geometric
multiplicity of Ais 1.
l1li
diagonalizable
operators
As mentioned at the beginning
of
this chapter, it is useful to represent an
operator by as simple a matrix as possible.
The
simplest matrix is a diagoual
matrix. This motivates the following definition:
4.3.7. Definition.
A linearoperatorAon a vector spaceV is said to bediagonal-
izable
ifthereis a basisfor Vall
of
whosevectorsare eigenvectors
of
A.
4.3.8.
Theorem.
Let A be a diagonalizable operator on a vector space V with
distincteigenvalues{Aj
}j~t.
Thereare (notnecessarilyhermitian)projectionop-
eratorsPj on Vsuchthat
r
(I)
1 =
LPj,
j=l
(2)
PiPj
= 0 for i
t=
i.
r
(3) A =
I>jPj.
j=l
Proof
Let
JY(j denote the eigenspace corresponding to the eigenvalue Ai- Since
the eigenvectors spanV and the only
common
vector
of
two different eigenspaces
is the zero vector (see comments after Definition 4.3.3), we have
This immediately gives
(I)
and
(2)
if
we use Equations (4.1) and (4.2) where
Pj
is the projection operatoronto JY(
s-
Toprove (3), let Iv}
be
an arbitraryvectorin V.
Then
[u)
can
be
writtenuniquely
as a sum
of
vectors each coming from
one
eigenspace. Therefore,
A [u) =
~A
IVj) =
~Aj
IVj) =
(~AjPj)
Iv}.
Since this equality holds for all vectors Iv}, (3) follows.
D
4.4
SPECTRAL
DECOMPOSITION
117
4.4 Spectral Decomposition
This section derives
one
of
the
most
powerful theorems in the theory
of
linear
operators, the spectral decomposition theorem. We shall derive the theorem for
operators that generalize hermitian and unitary operators.
normal
operator
4.4.1. Definition. A
normal
operator is an operator on an inner product space
defined
that commutes with its adjoint.
An important consequence
of
this definition is that
if
and only
if
Ais
normal. (4.9)
4.4.2.
Proposition.
Let
A be a normal operator on V. Then
Ix}
is an eigenvector
of
Awith eigenvalue).
ifand
only
iflx}
is an eigenvector
of
At with eigenvalue):*.
Proof By Equation(4.9),
thefactthat(A-)"1)t
= At -),,*1,and the factthatA-)"1
isnonnal
(reader, verify), we have
II
(A-)"1)xll = 0
ifandonly
if
II
(At
-)..
*1)xll =
O.
Since it is only the zero vector thathas the zero
nonn,
we get
(A - ),,1)
Ix}
= 0
if and only
if
(At -
),,*1)
[x) =
o.
This proves the proposition.
D
We obtainausefulconsequence
of
this propositionby applyingit to ahermitian
operator
H and a unitary operator" U.
In
the first case, we get
Therefore, Ais real.Inthe secondcase, we write
Ix) =
1jx)
=
UU
t
[x) =
U()..*
Ix)) =
)..*U
Ix) =
U*
Ix)
=>
U*
=
I.
Therefore,
)..
is unimodular (has absolute value equal to 1). We summarize the
foregoing discussion:
4.4,3.
Corollary.
The eigenvalues
of
a hermitian operatorare real. The eigenval-
ues
of
a unitary operator have unit absolute value.
4.4.4.
Example.
Letus find the eigenvalues and eigenvectors of the hermitianmatrix
H
=
(~r})'
Wehave
(
- A
det(H
-Al)
= det i
Thus,
the
eigenvalues,
Al = 1and
A2
=
-I,
are
real,
as
expected.
"Obviously, both
are
normal
operators.
118 4.
SPECTRAL
DECOMPOSITION
Tofindtheeigenvectors, we write
-i)
("I)
=
(--:"1
-
i"2)
-1
cz
la'i
- cz
oraz =
ial.
whichgives lal) =
(iC:X
1
r
)
=
al
(}),
wherecq isan arbitrarycomplexnumber.
Atso,
-i)(Ih) =(fh - ifh)
1 fh
zfh
+fh
Always
normalize
the
eigenvectors!
or fh =
-ilh,
whichgives
la2}
=
(!'~1)
=PI
<-~;l,
where PI is an arbitrarycomplex
number.
Itis
desirable,
inmost
situations,
to
orthonormalize
theeigenvectors. Inthe
present
case,
theyare
already
orthogonal.
Thisis a
property
shared
byall
eigenvectors
of a
hermitian
(in
fact,
normal)
operator
stated
in thenext
theorem.
We
therefore
needto merely
normalize
the eigenvectors:
or
1"11
= 1/.J2and
"I
=
ei~
1.J2
for some'PE
lll..
A similarresullis obtainedfor PI. The
choiceip = 0 yields
and III
The
following theoremprovesfor all normal operators the orthogonality prop-
erty
of
their eigenvectors illnstrated in the example above for a simple hermitian
operator.
4.4.5. Theorem. An eigenspace
of
a normaloperatorreducesthat operator. More-
over, eigenspaces
of
a normal
operator
are mutually orthogonal.
Proof
The
first part
of
the theorem is a trivial consequence
of
Proposition 4.4.2
and Theorem 4.2.3.
To prove the second part, let Iu) E
JY[A
and Iv) E
JY[I'
with
A
i'
1-'-
Then, using Theorem 4.2.3 once more, we obtain
A(vlu)
=
(vIAu)
=
(vIAu)
=
(Atvlu)
=
(Il*vlu)
=Il(vlu)
It
follows that (A-
Il)
(vi
u) = 0 and since
Ai'
u,
(vi
u) = O.
D
spectral
decomposition
theorem
4.4.6. Theorem. (Spectral Decomposition Theorem)
Let
Abe a normaloperator
on afinite-dimensional complex inner product space
V.
Let
AI, A2,
...
, A, be its
distinct eigenvalues. Then there exist nonzero (hermitian) projection operators
PI, Pz,
...
, P
r
such that
Vi
i'
j ..
4.4
SPECTRAL
DECDMPDSITIDN
119
3.
L;'=I
AiPi = A.
Proof
Let
Pi be the operator that projects onto the eigenspace Mi corresponding
to eigenvalue
Ai. By conuuents after Proposition 4.1.6, these projection operators
are hermitian. Because
of
Theorem 4.4.5, the only vector conuuon to any two
distinct eigenspaces is the zero vector. So, it makes sense to talk about the direct
sum oftheseeigenspaces.
Let M =
MI
E9M2E9·
. 'E9M
r
and P =
Lr=1
Pi, where
P is the orthogonal projection operator onto
M.
Since Aconuuutes with every Pi
(Theorem4.2.5), it conuuuteswith P. Hence, by Theorem4.2.5, M reduces A,i.e.,
M1-is also invariant underA. Now regardthe resttiction of A to M1- as an operator
in its own right on the finite-dimensional vector space M
1-. Theorem 4.3.4 now
forces
A to have at least one eigenvector in M1-.
But
this is impossible because all
eigenvectors of A have been accounted for in its eigenspaces. The only resolution
is for M
1- to be zero. This gives
and
The second equation follows from the first and Equations (4.1) and (4.2). The
remaining partof the theorem follows from arguments sintilar to those used in the
proof
of
Theorem4.3.8. D
We can now establish the counection between the diagonalizability of a normal
operatorand the spectral theorem.
In
each subspaceMi' we choosean orthonormal
basis. The union of all these bases is clearly a basis for the whole space
V.Let
us label these basis vectors
lei), where the subscript indicates the subspace and
the superscript indicates the particnlarvectorin that subspace. Clearly,
(ei lei:) =
8",8
jj
, and
Pj
=
L;~llei){eil.
Noting that Pk
len
=8kj' lei:),we can obtain
the mattix elements of
A in such a basis:
Only the diagonal elements are nonzero. We note that for eachsubscript
j we have
mj orthonormal vectors
lei},
where mj is the dimension
of
Mi- Thus, Aj occurs
mj times as a diagonal element. Therefore, in such an orthonormal basis, A will
be represented by
diag(AI,
...
, AI,
A2,
...
,
A2,
...
, A
r
,
•••
, A
r
).
'"-.,,-"
.,
.
ml
times
m2 times m- times
Let us sunuuarize the foregoing discussion:
4.4.7. Corollary.
If
A E £'(V) isnormal,thenVhasanorthonormalbasisconsist-
ing
of
eigenvectors
of
A.
Therefore,
a normaloperatoron a complexinnerproduct
spaceis diagonalizable.
120 4.
SPECTRAL
DECOMPOSITION
Computation
of
the
largest
and
the
smallest
eigenvalues
ofa
normal
operator
Using this corollary, the readermay show the following:
4.4.8. Corollary. A hermitianoperatorispositiveifandonlyifallitseigenvalues
are positive.
4.4.9.Example.
COMPUTATION
OF
LARGEST
AND
SMALLEST
EIGENVALUES
There
is an
elegant
technique
that
yieldsthe
largest
andthe
smallest
(in
absolute
value)
eigenvalues
of a
normal
operator
A in a
straightforward
wayif the
eigenspaces
of
these
eigenvalues
are
one
dimensional.
For
convenience,
assume
that
the
eigenvalues
are
labeled
in
order
of
decreasing
absolute
values:
IAII
>
IA21
>
...
>
IArl
;6
O.
Let (laknt'=1 be a basis of Vconsistingof eigenvectorsof A,and Ix) =
Lt'=1
~k
lak) an
arbitrary
vectorinV.
Then
AmIx) = f
~kAm
lak) = f
~kAT
lak) = A'i'
[~llal)
+
i:
~k
e
flak)]
.
k=1 k=1 k=2 I
Inthelimitm --+
00,
the
summation
inthe
brackets
vanishes.
Therefore,
and
A
hermitian
matrix
can
be
diagonalized
bya
unitary
matrix.
forany
Iy}
E V.
Taking
the
ratio
of
this
equation
and
the
corresponding
oneform + 1,we
obtain
. (yl A
m
+
1
Ix)
lim AI.
m-e-co
(yl Am Ix)
Notehow
crucially
this
relation
depends
onthefact
that
Al is
nondegenerate,
i.e.,
that
JV(1
is
one-dimensional.
By
taking
larger
and
larger
values
form, we can
obtain
a
better
and
better
approximation
tothe
largest
eigenvalue.
Assuming
that
zero isnotthesmallesteigenvalueAr-and
therefore
notaneigenvalue-
of A, we can findthe smallesteigenvalueby reptacingA with
A-I
and Al with
liAr.
The
details
areleftasan
exercise
forthe
reader.
III
Any given hermitian matrix H can be thought
of
as the representation
of
a
hermitianoperatorin the standard orthonormalbasis. We
can
find a unitarymatrix
Uthat Cantransform the standard basis to the orthonormal basis consisting
of
lej),
the eigenvectors
of
the hermitian operator.
The
representation
of
the hermitian
operatorin the new basis is
UHUt, as discussedin Section 3.3. However, the above
argumentshowedthat the new matrix is diagonal. We therefore have the following
result.
4.4.10. Corollary. A hermitianmatrixcanalwaysbe broughtto diagonalform by
means
of
a unitarytransformation
matrix.
4.4.11.Example. Letus
consider
the
diagonalization
of the
hermitian
matrix
o
o
-I-i
1-
i
-I+i
-I+i
o
o
-I-i)
1+
i
o .
o