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).
