One means by which to come to grips with Rs
R s is
to treat it as the matrix analog of the scalar function
1s−b
1
sb
(1)
This function is a scaled version of the even simpler function
11−z
1
1z
.
This latter function satisfies the identity (just multiply across by
1−z1z
to check it)
11−z=1+z+z2+...+zn−1+zn1−z
1
1z
1
z
z2
...
z
n1
zn
1z
(2)
for each positive integer
nn. Furthermore, if
|z|<1
z
1
then
zn→0
z
n
0
as
n→∞
n
and so
Equation 2 becomes, in the limit,
11−z=∑n=0∞zn
1
1z
n
0
zn
the familiar geometric
series. Returning to
Equation 1 we write
1s−b=1s1−bs=1s+bs2+...+bn−1sn+bnsn1s−b
1
sb
1
s
1
bs
1s
b
s
2
...
b
n1
sn
b
n
s
n
1
sb
and hence, so long as
|s|>|b|
s
b
we find,
1s−b=1s∑n=0∞bsn
1
sb
1
s
n
0
b
s
n
This same line of reasoning may be applied in the matrix
case. That is,
sI−B-1=s-1I−Bs-11s+Bs2+...+Bn−1sn+BnsnsI−B-1
s
I
B
s
I
Bs
1
s
B
s2
...
B
n1
sn
Bn
sn
sI
B
(3)
and hence, so long as
|s|>∥B∥
s
B
where
∥B∥
B
is the magnitude of the
largest element of
BB, we find
sI−B-1=s-1∑n=0∞Bsn
sI
B
s
n
0
B
s
n
(4)
Although
Equation 4 is indeed a formula for the transfer
function you may, regarding computation, not find it any more
attractive than the Gauss-Jordan method. We view
Equation 4 however as an analytical
rather than computational tool. More precisely, it facilitates
the computation of integrals of
Rs
R
s
. For example, if
Cρ
Cρ is the circle of radius
ρρ centered at the origin
and
ρ>∥B∥
ρ
B
then
∫
Cρ
sI−B-1ds=∑n=0∞Bn∫
Cρ
s-1−nds=2πⅈI
s
Cρ
sI
B
n
0
B
n
s
Cρ
s
-1n
2
I
(5)
This result is essential to our study
of the eigenvalue problem. As are the two resolvent
identities. Regarding the first we deduce from the simple
observation
s2
I−B-1−
s1
I−B-1=
s2
I−B-1
s1
I−B−
s2
I+B
s1
I−B-1
s2
I
B
s1
I
B
s2
I
B
s1
I
B
s2
I
B
s1
I
B
that
R
s2
−R
s1
=
s1
−
s2
R
s2
R
s1
R
s2
R
s1
s1
s2
R
s2
R
s1
(6)
The second identity is simply a rewriting of
sI−BsI−B-1=sI−B-1sI−B=I
sI
B
sI
B
sI
B
sI
B
I
namely,
BRs=RsB=sRs−I
B
R
s
R
s
B
s
Rs
I
(7)
