The Gauss-Jordan method informs us that RR will be a matrix of rational
functions with a common denominator. In keeping with the
notation of the previous chapters, we assume the denominator
to have the hh distinct roots,
λj
j=1…h
λj
j
1
…
h
with associated multiplicities
mj
j=1…h
mj
j
1
…
h
.
Now, assembling the partial fraction expansions of each
element of RR we
arrive at
Rs=∑j=1h∑k=1
mj
R
j
,
k
s−
λj
k
Rs
j
1
h
k
1
mj
R
j
,
k
s
λj
k
(1)
where, recalling the equation from
Cauchy's Theorem, the matrix
R
j
,
k
R
j
,
k
equals the following:
R
j
,
k
=12πj∫Rzz−
λ
j
k−1d
z
R
j
,
k
1
2
j
z
C
j
R
z
z
λ
j
k
1
(2)
As we look at this example with respect to the eigenvalue
problem eqn1 and eqn2, we find
R
1
,
1
=(
100
010
000
)
R
1
,
2
=(
010
000
000
)
and
R
2
,
1
=(
000
000
001
)
R
1
,
1
100
010
000
R
1
,
2
010
000
000
and
R
2
,
1
000
000
001
One notes immediately that these matrices enjoy some amazing
properties. For example
R
1
,
1
2=
R
1
,
1
,
R
2
,
1
2=
R
2
,
1
,
R
1
,
1
R
2
,
1
=0
, and
R
2
,
1
2=0
R
1
,
1
2
R
1
,
1
,
R
2
,
1
2
R
2
,
1
,
R
1
,
1
R
2
,
1
0
, and
R
2
,
1
2
0
(3)
Below we will now show that this is no accident. As a
consequence of
Equation 2 and
the first resolvent identity, we shall find that these results
are true in general.
R
j
,
1
2=
R
j
,
1
R
j
,
1
2
R
j
,
1
as seen above in Equation 3.
Recall that the
Cj
Cj
appearing in Equation 2 is any
circle about
λj
λj
that neither touches nor encircles any other root. Suppose
that
Cj
Cj
and
Cj
′
Cj
are two such circles and
Cj
′
Cj
encloses
Cj
Cj
. Now
R
j
,
1
=12πj∫Rzd
z
=12πj∫Rzd
z
R
j
,
1
1
2
j
z
Cj
R
z
1
2
j
z
Cj
R
z
and so
R
j
,
1
2=12πi2∫Rzd
z
∫Rwd
w
R
j
,
1
2
1
2
2
z
C
j
R
z
w
Cj
R
w
R
j
,
1
2=12πi2∫∫RzRwd
w
d
z
R
j
,
1
2
1
2
2
z
C
j
w
Cj
R
z
R
w
R
j
,
1
2=12πi2∫∫Rz−Rww−zd
w
d
z
R
j
,
1
2
1
2
2
z
C
j
w
Cj
R
z
R
w
w
z
R
j
,
1
2=12πi2(∫Rz∫1w−zd
w
d
z
−∫Rw∫1w−zd
z
d
w
)
R
j
,
1
2
1
2
2
z
C
j
R
z
w
Cj
1
w
z
w
Cj
R
w
z
C
j
1
w
z
R
j
,
1
2=12πi∫Rzd
z
=
R
j
,
1
R
j
,
1
2
1
2
z
C
j
R
z
R
j
,
1
We used the first resolvent identity, This Transfer
Function eqn, in moving from the second to the third
line. In moving from the fourth to the fifth we used only
∫1w−zd
w
=2πi
w
Cj
1
w
z
2
(4)
and
∫1w−zd
z
=0
z
C
j
1
w
z
0
The latter integrates to zero because
Cj
Cj
does not encircle
ww.
From the definition of orthogonal projections, which states that matrices
that equal their squares are projections, we adopt the
abbreviation
Pj
≡
R
j
,
1
Pj
R
j
,
1
With respect to the product
Pj
Pk
Pj
Pk
, for
j≠k
j
k
, the calculation runs along the same lines. The
difference comes in Equation 4 where, as
Cj
Cj
lies completely outside of
Ck
Ck
, both integrals are
zero. Hence,
If
1≤k≤
m
j
−1
1
k
m
j
1
then
D
j
k=
R
j
,
k
+
1
D
j
k
R
j
,
k
+
1
.
D
j
m
j
=0
D
j
m
j
0
.
For kk and
ll greater than or equal to
one,
R
j
,
k
+
1
R
j
,
l
+
1
=12πi2∫Rzz−
λj
kd
z
∫Rww−
λj
ld
w
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
C
j
R
z
z
λj
k
w
Cj
R
w
w
λj
l
R
j
,
k
+
1
R
j
,
l
+
1
=12πi2∫∫RzRwz−
λj
kw−
λj
ld
w
d
z
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
C
j
w
Cj
R
z
R
w
z
λj
k
w
λj
l
R
j
,
k
+
1
R
j
,
l
+
1
=12πi2∫∫Rz−Rww−zz−
λj
kw−
λj
ld
w
d
z
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
Cj
w
C
j
R
z
R
w
w
z
z
λj
k
w
λj
l
R
j
,
k
+
1
R
j
,
l
+
1
=12πi2∫Rzz−
λj
k∫w−
λj
lw−zd
w
d
z
−12πi2∫Rww−
λj
l∫z−
λj
kw−zd
z
d
w
R
j
,
k
+
1
R
j
,
l
+
1
1
2
2
z
C
j
R
z
z
λj
k
w
Cj
w
λj
l
w
z
1
2
2
w
Cj
R
w
w
λj
l
z
C
j
z
λj
k
w
z
R
j
,
k
+
1
R
j
,
l
+
1
=12πi∫Rzz−
λj
k+ld
z
=
R
j
,
k
+
l
+
1
R
j
,
k
+
1
R
j
,
l
+
1
1
2
z
C
j
R
z
z
λj
k
l
R
j
,
k
+
l
+
1
because
∫w−
λj
lw−zd
w
=2πiz−
λj
l
w
Cj
w
λj
l
w
z
2
z
λj
l
(5)
and
∫z−
λj
kw−zd
z
=0
z
C
j
z
λj
k
w
z
0
With
k=l=1
k
l
1
we have shown
R
j
,
2
2=
R
j
,
3
R
j
,
2
2
R
j
,
3
,
i.e.,
Dj
2=
R
j
,
3
Dj
2
R
j
,
3
. Similarly, with
k=1
k
1
and
l=2
l
2
we find
R
j
,
2
R
j
,
3
=
R
j
,
4
R
j
,
2
R
j
,
3
R
j
,
4
,
i.e.,
Dj
3=
R
j
,
4
Dj
3
R
j
,
4
. Continuing in this fashion we find
R
j
,
k
R
j
,
k
+
1
=
R
j
,
k
+
2
=j
R
j
,
k
R
j
,
k
+
1
R
j
,
k
+
2
j
, or
Dj
k+1=
R
j
,
k
+
2
Dj
k
1
R
j
,
k
+
2
. Finally, at
k=
m
j
−1
k
m
j
1
this becomes
Dj
m
j
=
R
j
,
m
j
+
1
=12πi∫Rzz−
λj
m
j
d
z
=0
Dj
m
j
R
j
,
m
j
+
1
1
2
z
C
j
R
z
z
λj
m
j
0
by Cauchy's Theorem.
With this we now have the sought after expansion
Rz=∑
j
=1h1z−
λj
Pj
+∑
k
=1
mj
−11z−
λj
k+1
Dj
k
R
z
j
1
h
1
z
λj
Pj
k
1
mj
1
1
z
λj
k
1
Dj
k
(6)
along with the verification of a number of the properties laid out
in
Complex
Integration eqns 1-3.
