4

Spectral Theory

A SHORT SUMMARY of Chapter 4 is to be added here.

4.1 Spectrum of a Bounded Operator

Let X be a normed space, and let A ∈ BL(X). We shall associate some sets
of scalars with A that naturally arise while solving the operator equation
A(x) − k x = y, where y ∈ X and k ∈ K are given, and x ∈ X is to be found.
Let us fix k ∈ K. We want to investigate when the above equation has a
unique solution for every ‘free term’ y ∈ X, and whether this unique solution
depends continuously on y. It is easy to see that there exits a solution of this
equation for every y ∈ X if and only if R(A − kI) = Y , that is, the operator
A − kI is onto. Also, when there is a solution for a given y ∈ X, it is unique if
and only if the only solution of ‘homogeneous’ equation A(x) − k x = 0 is the
zero solution x = 0, that is, the operator A − kI is one-one. Further, when
there is a unique solution for every y ∈ X, it will depend continuously on y
if and only if the linear map (A − kI)−1 from X to X is continuous. Keeping
these facts in mind, we now make the following definition.
An bounded operator A on a normed space X is called invertible (in
BL(X)) if A is one-one and onto, and if the inverse map A−1 is in BL(X).
We give a necessary and sufficient condition for the invertibility of A. Let us
recall from Section 2.1 that A is bounded below if there is β > 0 such that
β#x# ≤ #A(x)# for all x ∈ X.

Proposition 4.1. Let X be a normed space, and let A ∈ BL(X). Then A is

invertible if and only if A is bounded below and onto.

Proof. Suppose A is invertible. Then A is bounded below since #x# =

#A−1 ((A(x))# ≤ #A−1 ##A(x)# for all x ∈ X by the basic inequality given in
Theorem 2.10 (i). Conversely, suppose A is bounded below and onto. Clearly,
A is one-one. Also, the inverse map A−1 : X → X is linear. Further, there
152 4 Spectral Theory

is β > 0 such that β#A−1 (y)# ≤ #A(A−1 (y))|| = #y# for all y ∈ X. Thus
A−1 ∈ BL(X) and #A−1 # ≤ 1/β. (Compare Proposition 2.5.) &
'

It follows that A ∈ BL(X) fails to be invertible if (i) A is not one-one, (ii)

A is one-one, but not bounded below, (iii) A is bounded below, but not onto.
We are, therefore, led to associate the following subsets of K with A while
attempting to solve the operator equation A(x) − k x = y.
Let A ∈ BL(X). The set σ(A) := {λ ∈ K : A − λI is not invertible} is
called the spectrum of A. A scalar in σ(A) is called a spectral value of A.
The set σe (A) := {λ ∈ K : A − λI is not one-one} is called the eigenspec-
trum of A, and the set σa (A) := {λ ∈ K : A − λI is not bounded below} is
called the approximate eigenspectrum of A. Clearly, σe (A) ⊂ σa (A) and
σa (A) ⊂ σ(A). Then it is easy to see that λ ∈ σe (A) if and only if there is
x ∈ X such that A(x) = λx and #x# = 1, and λ ∈ σa (A) if and only if there is
a sequence (xn ) in X such that A(xn ) − λxn → 0 and #xn # = 1 for all n ∈ N.
This explains the adjective ‘approximate’.

Eigenspectrum
A scalar λ ∈ σe (A) is called an eigenvalue of A. A nonzero x ∈ X satisfying
A(x) = λx is called an eigenvector of A corresponding to the eigenvalue λ
of A, and subspace Eλ := {x ∈ X : A(x) = λx} is called the corresponding
eigenspace of A. If λ1 , . . . , λm are distinct eigenvalues of A, and xj ∈ Eλj is
nonzero for j = 1, . . . , m, then it is easy to see that {x1 , . . . , xm } is a linearly
independent subset of X.
Although the eigenspectrum of A seems to to be more tractable than the
rest of the spectrum of A, it is by no means easy to find eigenvalues of A. This
is partly because both sides of the eigenequation A(x) = λx, x )= 0, involve
unknown elements, unlike the operator equation A(x) = y, where y ∈ X is
known and x ∈ X is to be found. If, however, a scalar λ is known to be
an eigenvalue of A, then finding the corresponding eigenvectors is reduced
to solving the ‘homogeneous’ operator equation A(x) − λx = 0. Similarly, if
a nonzero element x of X is known to be an eigenvector of A, then we can
find the corresponding eigenvalue λ by considering a linear functional f on X
such that f (x) )= 0, and let λ := f (A(x))/f (x). In particular, if X is an inner
product space, then we need only find y ∈ X such that *x, y+ = ) 0, and let
λ := *A(x), y+/*x, y+.
Let λ be an eigenvalue of A. If x is a corresponding eigenvector of A, then
|λ| #x# = #λx# = #A(x)# ≤ #A# #x#. Hence |λ| ≤ #A#.

Proposition 4.2. let X be a normed space, and A ∈ BL(X).

(i) If X is finite dimensional, then σ(A) = σe (A).
(ii) If X is infinite dimensional and A is of finite rank, then σ(A) = σe (A),
and 0 ∈ σe (A).
4.1 Spectrum of a Bounded Operator 153

Proof. Since σe (A) ⊂ σ(A), we need only show that K \ σe (A) ⊂ K \ σ(A).
(i) Suppose dim X := n < ∞. Let k ∈ K \ σe (A). By the rank-nullity
theorem stated in Section 0.2, dim Z(A − kI) + dim R(A − kI) = n. Since
A − kI is one-one, we see that dim Z(A − kI) = {0}. Hence R(A − kI) is an n
dimensional subspace of X, that is, R(A − kI) = X. Thus A − kI is one-one
and onto. By Proposition 2.6 (i), (A − kI)−1 is continuous. Hence k )∈ σ(A).
(ii) Suppose X is infinite dimensional, and A is of finite rank. Let
{x1 , x2 , . . .} be a linearly independent subset of X. Assume for a moment
that A is one-one. Then {A(x1 ), A(x2 ), . . .} would be an infinite linearly in-
dependent subset of the finite dimensional subspace R(A) of X. This shows
that A cannot be one-one, that is, 0 ∈ σe (A).
Let k ∈ K \ σe (A) and k )= 0. Let B : R(A) → R(A) denote the restriction
of A − kI to R(A). Since A − kI is one-one, so is B. Also, since R(A) is finite
dimensional, B is onto by (i) above. Consider y ∈ X. Then A(y) ∈ R(A),
and there is u ∈ R(A) with B(u) = A(y), that is, A(u) − ku = A(y). Since
A(u − y) = ku, we let x := (u − y)/k, and obtain A(x) = u = kx + y, that is,
(A − kI)(x) = y. Thus A − kI is onto.
Next, assume for a moment that A − kI is not bounded below. Then there
is a sequence (xn ) in X such that A(xn )−kxn → 0 and #xn # = 1 for all n ∈ N.
Now (A(xn )) is a bounded sequence in the finite dimensional normed space
R(A). By Theorem 1.9, there is a subsequence (A(xnj )) such that A(xnj ) → y
in R(A). Then kxnj → y as well. As a result,

A(y) = A( lim kxnj ) = k lim A(xnj ) = ky.

j→∞ j→∞

Since A − kI is one-one, y = 0. Thus kxnj → 0. But k )= 0 and #xnj # = 1 for

each j ∈ N, and we arrive at a contradiction. Hence A − kI is bounded below.
By Proposition 4.1, A − kI is invertible, that is, k )∈ σ(A). &
'

Remark 4.3. (i) Let dim X := n < ∞, and x1 , . . . , xn be a basis for X. As

we have seen in Example 0.3 (i), there are functionals f1 , . . . , fn on X such
that fi (xj ) = δi,j for all i, j = 1, . . . , n and x = f1 (x)x1 + · · · + fn (x)xn for
every x ∈ X. Define Φ : X → Kn by Φ(x) := (f1 (x), . . . , fn (x)) for x ∈ X,
and note that Φ is linear, one-one and onto. Consider an operator A : X → X.
Let ki,j := fi (A(xj )) for i, j = 1, . . . , n, and let M := [ki,j ]. Then the n×n
matrix M defines the operator A. Since A(xj ) = k1,j x1 + · · · + kn,j xn for each
j = 1, . . . , n, we see that M [Φ(x)]t = [Φ(A(x))]t for all x ∈ X. It follows that
A(x) = λx if and only if M [Φ(x)]t = λ[Φ(x)]t for x ∈ X and λ ∈ K. Hence
the problem of finding eigenvalues and the corresponding eigenvectors of A is
reduced to a matrix eigenvalue problem.
(ii) Let an operator A : X → X be of finite rank. Then there are x1 , . . . , xn
in X and functionals g1 , . . . , gn on X such that A(x) = g1 (x)x1 +· · ·+gn (x)xn
for all x ∈ X. (For example, if X is finite dimensional, and x1 , . . . , xn
is basis!nfor X along with the functionals f1 , . . . , fn as in (i) above, then
gi = j=1 ki,j fj for i = 1, . . . , n.) We show that the problem of finding
154 4 Spectral Theory

nonzero eigenvalues and the corresponding eigenvectors of A can be reduced

to a matrix eigenvalue problem. Let M denote the n×n matrix [gi (xj )]. Then
for x ∈ X, u := (u(1), . . . , u(n)) ∈ Kn and nonzero λ ∈ K, we prove
n
1"
A(x) = λx and u = (g1 (x), . . . , gn (x)) ⇐⇒ M ut = λut and x = u(j)xj .
λ j=1
!n
Suppose A(x) = λx and u = (g1 (x), . . . , gn (x)). Then j=1 gj (x)xj = λx,
!n !n
and hence j=1 gj (x)gi (xj ) = λgi (x), that is, j=1 u(j)gi (xj ) = λu(i) for
# !n $
each i = 1, . . . , n. Thus M ut = λut . Also, x = j=1 u(j)x j /λ. Conversely,
# ! n $ ! n
suppose M ut = λut and x = j=1 u(j)xj /λ. Then j=1 gi (xj )u(j) =
λu(i) for each i = 1, . . . , n, and so
"n n % n &
1" "
A(x) = gi (x)xi = u(j)gi (xj ) xi
i=1
λ i=1 j=1
n n
1" "
= (λu(i))xi = u(i)xi = λx.
λ i=1 i=1
!n # !n $
Also, λu(i) = j=1 gi (xj )u(j) = gi j=1 u(j)xj = gi (λx) = λgi (x), and so
u(i) = gi (x) for all i = 1, . . . , n since λ )= 0, that is, u = (g1 (x), . . . , gn (x)).
If x ∈ X and u ∈ Kn are related as above, then we see that x = 0 if and
only if u = 0. This completes the reduction of the eigenvalue problem. !
Examples 4.4. (i) Let n ∈ N, X := Kn with a given norm, and let M be an
n×n matrix with entries in K. For x := (x(1), . . . , x(n)) ∈ Kn , define A(x)
to be the matrix multiplication of M and the column vector [x(1), . . . , x(n)]t .
Then A : Kn → Kn is linear, and it is continuous by Proposition 2.6. For
k ∈ K, the operator A − kI is invertible if and only if the matrix M − kI is
nonsingular, that is, det(M − kI) =
) 0. Thus λ ∈ σ(A) if and only if λ is a root
of the characteristic polynomial p(t) = det(M − t I). Since the characteristic
polynomial of M is of degree n, there are at most n distinct eigenvalues of
A. If K = C, then the fundamental theorem of algebra shows that there is at
least one eigenvalue of A. On the other hand, if K := R, then A may not ' have (
0 1
an eigenvalue. A simple example is provided by the matrix M :=
−1 0
whose characteristic polynomial is t2 + 1.
If M is a triangular matrix and λ1 , . . . , λn are the diagonal entries, then

det(M − tI) = (λ1 − t) · · · (λn − t),

and hence σ(A) = σe (A) = {λ1 , . . . , λn }. If M is not a triangular matrix,

then the problem of finding the eigenvalues of A poses great difficulties. Al-
gorithms have been developed to reduce M to an ‘approximately triangular’
matrix by similarity transformations. The most notable among these is the
QR algorithm. (See, for example, [23, 4.6].)
4.1 Spectrum of a Bounded Operator 155

If λ is an eigenvalue of A, then |λ| ≤ #A#, where # · # is the operator norm

on BL(Kn ) induced by the given norm on Kn . Various choices of the norms
on Kn yield upper bounds for the eigenspectrum of A. Let M = [ki,j ], i, j =
1, . . . , n. Let α1 denote the maximum of the column sums of the matrix |M |,
α∞ denote the maximum of the row sums of the matrix |M |, and let β2 :=
# !n !n $
2 1/2
i=1 j=1 |ki,j | . Then |λ| ≤ min{α1 , α∞ , β2 } for every eigenvalue λ of
A. Exercise 2 of this chapter gives a well-known inclusion result for eigenvalues
of A due to Gershgorin.
(ii) Let M denote the n×n matrix with all entries equal to 1, and let A
denote the operator on Kn defined by M . We observe that x1 := (1, . . . , 1) is
an eigenvector of A with A(x1 ) = nx1 , and so the corresponding eigenvalue
of A is n. Also, A has rank 1, and hence its nullity is n − 1. The eigenspace
of A corresponding to the eigenvalue of 0 is the n − 1 dimensional subspace
{(x(1), . . . , x(n) ∈ Kn : x(1) + · · · + x(n) = 0} of Kn . It follows that σ(A) =
{0, 1}. We observe that the Helmert basis described in Exercise 17 of Chapter
1 constitutes an orthonormal basis for Kn consisting of eigenvectors of A.
(iii) Let X := L2 ([0, 1]), and for x ∈ X, define
) 1
A(x)(s) := min{s, t}x(t)dm(t), s ∈ [0, 1].
0

Then A ∈ BL(X). Let x1 (t) := sin πt/2 for t ∈ [0, 1]. Suppose we know that
x1 ∈ X is an eigenvector of A. Then for s ∈ [0, 1],
) s ) 1
πt πt
A(x1 )(s) = t sin dt + s sin dt
0 2 s 2
% ) sπ/2 ) π/2 &
2 2
= u sin u du + s sin u du
π π 0 sπ/2
% &
2 2* sπ sπ sπ + sπ
= sin − cos + s cos
π π 2 2 2 2
4 sπ
= 2 sin .
π 2
Thus A(x1 ) = (4/π 2 )x1 , and so 4/π 2 is the eigenvalue of A corresponding
to the eigenvector x1 . Similarly, if we let xj (t) := sin(2j − 1)πt/2, t ∈ [0, 1],
then we can see that A(xj ) = (4/(2j − 1)2 π 2 )xj , and so 4/(2j − 1)2 π 2 is the
eigenvalue of A corresponding to the eigenvector xj for j = 2, 3, . . . !

Approximate Eigenspectrum
Suppose (λn ) is a sequence of eigenvalues of A and λn → λ in K. Then λ
may not be an eigenvalue of A. (See Examples 4.6 (i).) However, λ is certainly
in the approximate eigenspactrum of A. This follows by considering xn ∈ X
such that A(xn ) = λn xn and #xn # = 1 for n ∈ N, and noting that
156 4 Spectral Theory

A(xn ) − λxn = A(xn ) − λn xn + (λn − λ)xn = (λn − λ)xn → 0.

On the other hand, not every scalar in σa (A) is a limit of a sequence of scalars
in σe (A). (See Examples 4.6 (ii) and 4.12 of this chapter.)

Proposition 4.5. Let X be a normed space, and let A ∈ BL(X). Then the
approximate eigenspectrum σa (A) of A is a bounded and closed subset of K.

Proof. Let λ ∈ σa (A). Then there is a sequence (xn ) in X such that #xn # = 1
for all n ∈ N and A(xn ) − λxn → 0. Now for all n ∈ N,

|λ| = #λxn # ≤ #λxn − A(xn )# + #A(xn )# ≤ #A(xn ) − λxn # + #A#.

Let n → ∞ to obtain |λ| ≤ #A#. Hence λ ∈ σa (A) is a bounded subset of K.

(See Exercise 6 of this chapter for a possibly sharper bound.)
To show that λ ∈ σa (A) is a closed subset of K, consider a sequence (λn )
in σa (A) such that λn → λ in K. Assume for a moment that λ )∈ σa (A). Then
there is β > 0 such that #A(x) − λx# ≥ β#x# for all x ∈ X. Since λn → λ,
there is n0 ∈ N such that #λ − λn0 | < β. Then for all x ∈ X,

#A(x) − λn0 x# ≥ #A(x) − λx# − |λ − λn0 | #x# ≥ (β − |λ − λn0 |)#x#,

and so, λn0 )∈ σa (A). This contradiction shows that σa (A) is closed. &
'

Examples 4.6. (i) Let X := 'p with p ∈ {1, 2, ∞}. Consider A ∈ BL('p )
defined by
* x(2) x(3) +
A(x) := x(1), , ,··· for x := (x(1), x(2), . . .) ∈ X.
2 3
Then A ∈ BL(X) and #A# = 1. Since A(ej ) = ej /j, we see that 1/j is an
eigenvalue of A, and ej is a corresponding eigenvector for each j ∈ N. Since
A(x) = 0 if and only if x = 0, we see that 0 is not an eigenvalue of A. However,
since #A(ej )#p → 0, and #ej #p = 1 for all j ∈ N, we see that A is not bounded
below, that is, 0 is an approximate eigenvalue of A.
Now #fix k ∈ K \ {0, 1, 1/2, 1/3, . . .}.$ For y := (y(1), y(2), . . .) ∈ X, define
B(y) := y(1)/(1 − k), y(2)/( 12 − k), . . . . Find δ > 0 such that |(1/j) − k| ≥ δ
for all j ∈ N. Then
|y(j)| 1
|B(y)(j)| = ≤ |y(j)| for all j ∈ N.
|(1/j) − k| δ

It follows that B(y) ∈ X for all y ∈ X. Also, #B# ≤ (1/δ), and (A − kI)B =
I = B(A − kI). Hence A − kI is invertible. Thus

σe (A) = {1, 1/2, 1/3, . . .} and σa (A) = {0, 1, 1/2, 1/3, . . .} = σ(A).

(ii) Let X := (c00 , # · #p ) with p ∈ {1, 2, ∞}. Consider the right shift
operator A on X defined by
4.1 Spectrum of a Bounded Operator 157

A(x) := (0, x(1), x(2), . . .) for x := (x(1), x(2), . . .) ∈ X.

Since #A(x)#p = #x#p for all x ∈ X, we see that A ∈ BL(X) and #A# = 1.
Since A is one-one, 0 )∈ σe (A). Let now k ∈ K \ {0}. If A(x) − kx = 0, then
−kx(1) = 0, x(1) − kx(2) = 0, x(2) − kx(3) = 0, . . ., and so x(j) = 0 for all
j ∈ N, that is, x = 0. Hence A − kI is one-one. Thus σe (A) = ∅.
If k ∈ K and |k| )= 1, then #A(x) − kx#p ≥ |1 − |k|| #x# for all x ∈ X,
and so A − kI is not bounded below, that is, k )∈ σa (A). Now let λ ∈ K with
|λ| = 1. First let p ∈ {1, 2}, and for n ∈ N, define
1
xn := (1, λ, (λ)2 , . . . , (λ)n−1 , 0, 0, . . .).
n1/p
Then xn ∈ c00 , #xn #p = 1 for all n ∈ N, and

1 21/p
#A(xn ) − λxn #p = 1/p
#(−λ, 0, . . . , 0, (λ)n−1 , 0, 0, . . .)#p = → 0.
n n1/p
Hence λ ∈ σa (A). Similarly, if p = ∞, then for n ∈ N, define
1
xn := (1, 2λ, . . . , n(λ)n−1 , (n − 1)(λ)n , . . . , 2(λ)2n−3 , (λ)2n−2 , 0, 0, . . .).
n
Then xn ∈ c00 , #xn #∞ = 1 and #A(xn ) − λxn #∞ is equal to
1 1
#(−λ, −1, −, λ, . . . , −(λ)n−2 , −(λ)n−1 , . . . , −(λ)2n−2 , 0, 0, . . .)#∞ = → 0.
n n
Hence λ ∈ σa (A). Thus in both cases, σa (A) = {λ ∈ K : |λ| = 1}.
Lastly, we show that A − λI is not onto for any λ ∈ K. In fact, if λ ∈ K,
then there is no x ∈ c00 such that (A − λI)(x) = e1 . This is obvious if λ = 0.
If λ )= 0 and (A − λI)(x) = (1, 0, 0, . . .), then −λx(1) = 1, x(1) − λx(2) = 0,
x(2) − λx(3) = 0, . . ., and so x(j) = −1/λj for all j ∈ N. But then x )∈ c00 .
Hence σ(A) = K.
(iii) Let X := C([a, b]) with the sup norm. Fix x0 ∈ X, and consider the
multiplication operator defined by

A(x) := x0 x for all x ∈ X.

Then A ∈ BL(X) and #A# = #x0 #∞ . Let E := {x(t) ∈ K : t ∈ [a, b]}. We

show that σa (A) = E = σ(A).
Suppose λ ∈ E. Then there is t0 ∈ [a, b] such that λ = x0 (t0 ). For n ∈ N,
define xn ∈ X by
,
1 − n|t − t0 | if t ∈ [a, b] and |t − t0 | ≤ 1/n,
xn (t) :=
0 otherwise.

Then #xn #∞ = 1 for all n ∈ N. We show that #A(xn )−λxn #∞ → 0. Let ( > 0.
Since x0 is continuous at t0 , there is δ > 0 such that |x0 (t) − x0 (t0 )| < ( for all
158 4 Spectral Theory

t ∈ [a, b] with |t − t0 | < δ. Choose n0 such that n0 > 1/δ. Then xn (t) = 0 for
all n ≥ n0 and t ∈ [a, b] with |t − t0 | ≥ δ. Hence for all n ≥ n0 and t ∈ [a, b],

|A(xn )(t) − λxn (t)| = |x0 (t) − x0 (t0 )| |xn (t)| < (,

and so #A(xn ) − λxn #∞ ≤ ( for all n ≥ n0 . This shows that E ⊂ σa (A).

Next, suppose k )∈ E, that is, k )= x0 (t) for any t ∈ [a, b]. Then the
function 1/(x0 − k) belongs to X. For y ∈ X, define B(y) := y/(x0 − k). Then
B ∈ BL(X) and (A − kI)B = I = B(A − kI). This shows that σ(A) ⊂ E.
Since σa (A) ⊂ σ(A), we obtain σa (A) = E = σ(A). In Exercise 5 of this
chapter, we describe σe (A), and in Exercises 9 and 11 of this chapter, we treat
the multiplication operators on '2 and L2 ([a, b]). !

Spectral Values of Bounded Operators on Banach Space

Let us now consider the spectrum of a bounded operator defined on a Banach
space X. In this case, Proposition 4.1 can be improved as follows.

Proposition 4.7. Let X be a Banach space, and let A ∈ BL(X). Then the
following conditions are equivalent.
(i) A is invertible.
(ii) A is bounded below and the range R(A) of A is dense in X.
(iii) A is one-one and onto.

Proof. Proposition 4.1 shows that (i) implies (ii), and it is obvious that (i)
implies (iii). Also, the bounded inverse theorem (Theorem 2.31) shows that
(iii) implies (i). Thus it remains to show that (ii) implies (i).
Let A be bounded below and R(A) be dense in X. In view of Proposition
4.1, it is enough to show that R(A) = X. Let β > 0 be such that β#x# ≤
#A(x)# for all x ∈ X. Consider y ∈ X. There is a sequence (xn ) in X such
that A(xn ) → y in X. In particular, (A(xn )) is a Cauchy sequence in X. Now

β#xn − xm # ≤ #A(xn − xm )# = #A(xn ) − A(xm )#

for all n, m ∈ N. Hence (xn ) is a Cauchy sequence in X. Since X is a Banach

space, there is x ∈ X such that xn → x in X. Then A(xn ) → A(x) by the
continuity of A. Thus y = A(x) ∈ R(A). Thus R(A) = X. &
'

Let X be a Banach space, and let A ∈ BL(X). The condition (ii) of

the above result shows that λ ∈ σ(A) if and only if either A − λI is
not bounded below or the range of A − λI is not dense in X. The set
{λ ∈ K : the range of A − λI is not dense in X} is called the compression
spectrum of A. By Proposition 3.23, this set is the same as σe (A$ ). Thus
σ(A) = σa (A) ∪ σe (A$ ), so that each spectral value of A is either in the ap-
proximate eigenspecrum of A or in the eigenspectrum of A$ .
4.1 Spectrum of a Bounded Operator 159

We shall now give a sufficient condition for the invertibility of the operator
I − A. It would enable us to obtain an expansion of the inverse of I − A as a
power series in A, which generalizes the geometric series expansion
∞
"
1
= kn for k ∈ K with |k| < 1.
1 − k n=0

This will lead us to prove some important properties of the spectrum of a

bounded operator on a Banach space.
Lemma 4.8. Let X be a Banach space, and let A ∈ BL(X). If #A# < 1, then
I − A is invertible,
∞
" 1
(I − A)−1 = An and #(I − A)−1 # ≤ .
n=0
1 − #A#
!∞
Proof. Consider the series n=0 An with terms in the normed space BL(X).
This series is absolutely summable since #A# < 1, and
∞
" ∞
" 1
#An # ≤ #A#n = .
n=0 n=0
1 − #A#

Since X is a Banach space, BL(X) is also a Banach !∞space by Theorem

n
2.10 (iii). Now Theorem 1.22 shows that the series ! n=0 A is summable
∞
in BL(X),! that is, there is B ∈ BL(X) such that B = n=0 An . For m ∈ N,
m
let Bm := n=0 An . Then Bm → B in BL(X). It is easy to see that

Bm (I − A) = I − Am+1 = (I − A)Bm for each m ∈ N.

Since #Am+1 # ≤ #A#m+1 → 0 as m → ∞, we obtain B(I −A) = I = (I −A)B.

Hence I − A is one-one and onto. By Proposition 4.7, I − A is invertible and
∞
"
(I − A)−1 = B = An .
n=0
!∞
As a result, #(I − A)−1 # ≤ n=0 #An # ≤ 1/(1 − #A#). &
'

Proposition 4.9. Let X be a Banach space, and let A ∈ BL(X). If λ ∈ σ(A),

then |λ| ≤ #A#. In fact, |λ| ≤ inf{#An #1/n : n ∈ N} for every λ ∈ σ(A).
Consequently, σ(A) is a bounded subset of K.

Proof. Suppose k ∈ K and |k| > #A#. Let B := A/k ∈ BL(X). Then #B# < 1,
and so I−B is invertible by Lemma 4.8. Hence A−kI = −k(I−B) is invertible,
that is, k )∈ σ(A). Thus if λ ∈ σ(A), then |λ| ≤ #A#.
More generally, let n ∈ N, and suppose k ∈ K with |k| > #An #1/n . Then
|k | > #An #, that is, #(A/k)n # < 1. By Lemma 4.8, I − (A/k)n is invertible.
n

We show that, in fact, I − (A/k) = I − B is invertible.

160 4 Spectral Theory

Let C := (I − (A/k)n )−1 = (I − B n )−1 . Then (I − B n )C = I = C(I − B n ),

that is, (I − B)(I + B + · · · + B n−1 )C = I = C(I − B)(I + B + · · · + B n−1 ).
Let D := (I + B + · · · + B n−1 )C. Since BC = CB, we see that

C(I − B)(I + B + · · · + B n−1 ) = (I + B + · · · + B n−1 )C(I − B) = D(I − B).

Hence (I − B)D = I = D(I − B). This shows that I − B is invertible, and

its inverse is D ∈ BL(X). As before, A − kI = −k(I − B) is invertible,
that is, k )∈ σ(A). Thus if λ ∈ σ(A), then |λ| ≤ #An #1/n . It follows that
|λ| ≤ inf{#An #1/n : n ∈ N} for every kλ ∈ σ(A). &
'
Theorem 4.10. Let X be a Banach space, and let A, B ∈ BL(X). Suppose
A is invertible and #(A − B)A−1 # < 1. Then B is invertible, and
∞
" # $n
B −1 = A−1 (A − B)A−1 .
n=0

Further, if ( := #(A − B)A−1 #, then

1 (
#B −1 # ≤ #A−1 # and #B −1 − A−1 # ≤ #A−1 #.
1−( 1−(
Proof. Lemma 4.8 shows that I − (A − B)A−1 = BA−1 is invertible, and
∞
" # $n
(BA−1 )−1 = (A − B)A−1 .
n=0

Since A is invertible, it follows that B = (BA−1 )A is invertible, and

∞
" # $n
B −1 = A−1 (BA−1 )−1 = A−1 (A − B)A−1 .
n=0

Again, since ( = #(A − B)A−1 # < 1, we obtain

∞
" 1
#B −1 # ≤ #A−1 # #(A − B)A−1 #n = #A−1 #.
n=0
1−(
!∞ # $n
Also, B −1 − A−1 = A −1
n=1 (A − B)A−1 , and so
∞
" (
#B −1 − A−1 # ≤ #A−1 # #(A − B)A−1 #n = #A−1 #.
n=1
1−(

&
'
Corollary 4.11. Let X be a Banach space. The set of all invertible operators
is open in BL(X). Consequently, σ(A) is a closed subset of K for every A in
BL(X). Further, the inversion map A 3−→ A−1 is continuous on the set of all
invertible operators.
4.1 Spectrum of a Bounded Operator 161

Proof. If A, B ∈ BL(X), A is invertible and #A − B# < 1/#A−1 #, then

#(A − B)A−1 # ≤ #A − B# #A−1 # < 1,

and so B is invertible by Theorem 4.10. Hence the open ball in BL(X) about
A and of radius 1/#A−1 # is contained in the set of all invertible operators.
This set is, therefore, open in BL(X).
Let A ∈ BL(X), and let (λn ) be a sequence in σ(A) such that λn → λ
in K. Then A − λn I → A − λI in BL(X). Since the set of all noninvertible
operators is closed in BL(X), we see that A − λI is not invertible, that is,
λ ∈ σ(A). Thus σ(A) is a closed subset of K.
Further, let A ∈ BL(X) be invertible, and consider a sequence (An ) of in-
vertible operators such that An → A in BL(X). Then (n := #(A−An )A−1 # ≤
#A − An # #A−1 # → 0. Hence there is n0 such that (n < 1 for all n ≥ n0 . Then
(n
#A−1
n −A
−1
#≤ #A−1 # for all n ≥ n0
1 − (n

by Theorem 4.10. Hence A−1 n → A

−1
in BL(X). Thus the inversion map is
continuous on the set of invertible operators. &
'

Example 4.12. Let X := 'p with p ∈ {1, 2, ∞}. Consider the right shift op-
erator A on X defined by A(x) := (0, x(1), x(2), . . .) for x := (x(1), x(2), . . .)
in X. Clearly, #A# = 1. By Proposition 4.9, |λ| ≤ #A# = 1 for every λ ∈ σ(A).
Hence σ(A) ⊂ {λ ∈ K : |λ| ≤ 1}.
Let λ ∈ K and |λ| < 1. We show that A − λI is not onto; in fact, there is
no x ∈ 'p such that (A − λI)(x) = e1 . This is obvious if λ = 0. If λ )= 0 and
(A − λI)(x) = (1, 0, 0, . . .), then −λx(1) = 1, x(1) − λx(2) = 0, x(2) − λx(3) =
0, . . ., and so x(j) = −1/λj for all j ∈ N. But then x )∈ 'p since x(j) → ∞.
Hence {λ ∈ K : |λ| < 1} ⊂ σ(A). By Corollary 4.11, σ(A) is a closed subset
of K, and so {λ ∈ K : |λ| ≤ 1} ⊂ σ(A). Thus σ(A) = {λ ∈ K : |λ| ≤ 1}.
As in Example 4.6, we see that σe (A) = ∅ and σa (A) = {λ ∈ K : |λ| = 1}.
The left shift operator B on X defined by B(x) := (x(2), x(3), . . .) for
x := (x(1), x(2), . . .) ∈ X. See Exercise 15 of this chapter for σ(B). !

Proposition 4.9 and Corollary 4.11 show that the spectrum of a bounded
operator on a Banach space is a bounded and closed subset of K. Example
4.12 illustrates how these properties can be used to determine σ(A). However,
it is not clear whether σ(A) is a nonempty subset of K. In Example 4.4 (i),
we have seen that if K := R, then σ(A) can very well be empty. On the other
hand, the Gelfand-Mazur theorem says that if X is a nonzero Banach
space over C, then σ(A) )= ∅ for every A ∈ BL(X); in fact, the spectral
radius formula of Gelfand, says that max{|λ| : λ ∈ σ(A)} = inf{#An #1/n :
n ∈ N} = limn→∞ #An #1/n .1 (Compare Proposition 4.9.)
1
If X is a nonzero normed space over C, and A ∈ BL(X), then by considering the
completion of X, σ(A) "= ∅ and inf{$An $1/n : n ∈ N} ≤ sup{|λ| : λ ∈ σ(A)}.