1.3. MAIN RESULTS OF THE GOHBERG AND SIGAL THEORY 13

Proof. To prove that A is normal with respect to ∂V , it suﬃces to prove that

A(z) is invertible except at a finite number of points in V . To this end choose a

connected open set U with U ⊂ V so that A(z) is invertible in V \ U. Then, for

each ξ ∈ U, there exists a neighborhood Uξ of ξ in which the factorization (1.5)

holds. In Uξ, the kernel of A(z) has a constant dimension except at ξ. Since U is

compact, we can find a finite covering of U, i.e.,

U ⊂ Uξ1 ∪ ··· ∪ Uξk ,

for some points ξ1, . . . , ξk ∈ U. Therefore, dim Ker A(z) is constant in V \{ξ1, . . . , ξk},

and so A(z) is invertible in V \ {ξ1, . . . , ξk}.

Now, if A(z) is normal with respect to the contour ∂V and zi, i = 1, . . . , σ, are

all its characteristic values and poles lying in V , we put

M(A(z); ∂V ) =

σ

i=1

M(A(zi)). (1.9)

The full multiplicity M(A(z); ∂V ) of A(z) in V is the number of characteristic

values of A(z) in V , counted with their multiplicities, minus the number of poles

of A(z) in V , counted with their multiplicities.

Theorem 1.12 (Generalized argument principle). Suppose that the operator-

valued function A(z) is normal with respect to ∂V . Then we have

M(A(z); ∂V ) =

1

2

√

−1π

tr

∂V

A−1(z)

d

dz

A(z)dz. (1.10)

Proof. Let zj , j = 1, . . . , σ, denote all the characteristic values and all the

poles of A lying in V . The key of the proof lies in using the factorization (1.5) in

each of the neighborhoods of the points zj . We have

1

2

√

−1π

tr

∂V

A−1(z)

d

dz

A(z)dz =

σ

j=1

1

2

√

−1π

tr

∂Vj

A−1(z)

d

dz

A(z)dz, (1.11)

where, for each j, Vj is a neighborhood of zj . Moreover, in each Vj , the following

factorization of A holds:

A(z) =

E(j)(z)D(j)(z)F (j)(z), D(j)(z)

=

P0j) (

+

nj

i=1

(z − zj

)kij

Pi(j).

As for the matrix-valued case at the beginning of this chapter, it is readily verified

that

1

2

√

−1π

tr

∂Vj

A−1(z)

d

dz

A(z)dz =

1

2

√

−1π

tr

∂Vj

(D(j)(z))−1

d

dz

D(j)(z)dz

=

nj

i=1

kij = M(A(zj )).

Now, (1.10) follows by using (1.11).

The following is an immediate consequence of Lemma 1.11, identity (1.10), and

(1.4).