Bounded Linear Operators on a Hilbert Space

Self-adjoint and unitary operators 197
S = T ∗ , we have ind S = −1 and ind T = 1. The multiplication operator in
Example 8.20 is not Fredholm because it does not have closed range.
It is possible to prove that if A is Fredholm and K is compact, then A + K
is Fredholm, and ind (A + K) = ind A. Thus the index of a Fredholm operator
is unchanged by compact perturbations. In particular, compact perturbations of
the identity are Fredholm operators with index zero, so they satisfy the Fredholm
alternative in Definition 8.19. We will prove a special case of this result, for compact,
self-adjoint perturbations of the identity, in Theorem 9.26.
8.4 Self-adjoint and unitary operators
Two of the most important classes of operators on a Hilbert space are the classes
of self-adjoint and unitary operators. We begin by defining self-adjoint operators.
Definition 8.23 A bounded linear operator A : H → H on a Hilbert space H is
self-adjoint if A ∗ = A.
Equivalently, a bounded linear operator A on H is self-adjoint if and only if
〈x, Ay〉 = 〈Ax, y〉 for all x, y ∈ H.
Example 8.24 From Example 8.14, a linear map on R n with matrix A is selfadjoint
if and only if A is symmetric, meaning that A = A T , where A T is the
transpose of A. A linear map on C n with matrix A is self-adjoint if and only if A
is Hermitian, meaning that A = A ∗ .
Example 8.25 From Example 8.16, an integral operator K : L2 ([0, 1]) → L2 ([0, 1]),
1
Kf(x) =
is self-adjoint if and only if k(x, y) = k(y, x).
0
k(x, y)f(y) dy,
Given a linear operator A : H → H, we may define a sesquilinear form
a : H × H → C
by a(x, y) = 〈x, Ay〉. If A is self-adjoint, then this form is Hermitian symmetric, or
symmetric, meaning that
a(x, y) = a(y, x).
It follows that the associated quadratic form q(x) = a(x, x), or
q(x) = 〈x, Ax〉, (8.13)