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The Partial Fraction Expansion of the Resolvent

Module by: Steven J. Cox. E-mail the author

Summary: (Blank Abstract)

Partial Fraction Expansion of the Transfer Function

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

R⁢s=∑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 =12⁢π⁢j⁢∫R⁢z⁢z− λ j k−1d z R j , k 1 2 j z C j R z z λ j k 1
(2)

Example 1: Concrete Example

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.

Proposition 1

R j , 1 2= R j , 1 R j , 1 2 R j , 1 as seen above in Equation 3.

Proof

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 =12⁢π⁢j⁢∫R⁢zd z =12⁢π⁢j⁢∫R⁢zd z R j , 1 1 2 j z Cj R z 1 2 j z Cj R z and so R j , 1 2=12⁢π⁢i2⁢∫R⁢zd z ⁢∫R⁢wd w R j , 1 2 1 2 2 z C j R z w Cj R w R j , 1 2=12⁢π⁢i2⁢∫∫R⁢z⁢R⁢wd w d z R j , 1 2 1 2 2 z C j w Cj R z R w R j , 1 2=12⁢π⁢i2⁢∫∫R⁢z−R⁢ww−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=12⁢π⁢i2⁢(∫R⁢z⁢∫1w−zd w d z −∫R⁢w⁢∫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=12⁢π⁢i⁢∫R⁢zd 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 =2⁢π⁢i 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,

Proposition 2

If j≠k j k then Pj ⁢ Pk =0 Pj Pk 0 .

Along the same lines we define Dj ≡ R j , 2 Dj R j , 2 and prove

Proposition 3

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 .

Proof

For kk and ll greater than or equal to one, R j , k + 1 ⁢ R j , l + 1 =12⁢π⁢i2⁢∫R⁢z⁢z− λj kd z ⁢∫R⁢w⁢w− 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 =12⁢π⁢i2⁢∫∫R⁢z⁢R⁢w⁢z− λj k⁢w− 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 =12⁢π⁢i2⁢∫∫R⁢z−R⁢ww−z⁢z− λj k⁢w− 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 =12⁢π⁢i2⁢∫R⁢z⁢z− λj k⁢∫w− λj lw−zd w d z −12⁢π⁢i2⁢∫R⁢w⁢w− 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 =12⁢π⁢i⁢∫R⁢z⁢z− 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 =2⁢π⁢i⁢z− 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 =12⁢π⁢i⁢∫R⁢z⁢z− 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

R⁢z=∑ 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.

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