8.3: The Partial Fraction Expansion of the Resolvent
- Page ID
- 21849
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Partial Fraction Expansion of the Transfer Function
The Gauss-Jordan method informs us that \(R\) 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 \(h\) distinct roots, \(\{\lambda_{j} | j = \{1, \cdots, h\}\}\) with associated multiplicities \(\{m_{j} | j = \{1, \cdots, h\}\}\)
Now, assembling the partial fraction expansions of each element of \(R\) we arrive at
\[R(s) = \sum_{j = 1}^{h} \sum_{k=1}^{m_{j}} \frac{R_{j,k}}{(s-\lambda_{j})^k} \nonumber\]
where, recalling the equation from Cauchy's Theorem, the matrix \(R_{j,k}\) equals the following:
\[R_{j,k} = \frac{1}{2\pi j} \int R(z)(z-\lambda_{j})^{k-1} dz \nonumber\]
As we look at this example in the introduction, we find
\[\begin{array}{ccc} {R_{1,1} = \begin{pmatrix} {1}&{0}&{0}\\ {0}&{1}&{0}\\ {0}&{0}&{0} \end{pmatrix}}&{R_{1,1} = \begin{pmatrix} {0}&{0}&{0}\\ {1}&{0}&{0}\\ {0}&{0}&{0} \end{pmatrix}}&{R_{2,1} = \begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{1} \end{pmatrix}} \end{array} \nonumber\]
One notes immediately that these matrices enjoy some amazing properties. For example
\[\begin{array}{ccccc} {R_{1,1}^{2} = R_{1,1}}&{R_{1,2}^{2} = R_{1,2}}&{R_{1,1} R_{2,1} = 0}&{and}&{R_{2,1}^{2} = R_{2,1}} \end{array} \nonumber\]
Below we will now show that this is no accident. As a consequence of Equation and the first resolvent identity, we shall find that these results are true in general.
\(R_{j,1}^{2} = R_{j,1}\) as seen above.
Recall that the \(C_{j}\) appearing in Equation is any circle about \(\lambda_{j}\) that neither touches nor encircles any other root. Suppose that \(C_{j}\) and \(C_{j}'\) are two such circles and \(C_{j}'\) encloses \(C_{j}\). Now,
\[R_{j,1} = \frac{1}{2\pi j} \int R(z) dz = \frac{1}{2\pi j} \int R(z) dz \nonumber\]
and so
\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} \int R(z) dz = \frac{1}{2\pi j} \int R(w) dw \nonumber\]
\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} \int \int R(z) R(w) dw dz \nonumber\]
\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} \int \int \frac{R(z)-R(w)}{w-z} dw dz \nonumber\]
\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} (\int R(z)- \int \frac{1}{w-z} dw dz - \int R(w)- \int \frac{1}{w-z} dz dw) \nonumber\]
\[R_{j,1}^2 = \frac{1}{2\pi i} \int R(z) dz = R_{j,1} \nonumber\]
We used the first resolvent identity, This Transfer Function equation, in moving from the second to the third line. In moving from the fourth to the fifth we used only
\[\int \frac{1}{w-z} dw = 2 \pi i \nonumber\]
and
\[\int \frac{1}{w-z} dz = 0 \nonumber\]
The latter integrates to zero because \(C_{j}\) does not encircle ww
From the definition of orthogonal projections, which states that matrices that equal their squares are projections, we adopt the abbreviation
\[P_{j} \equiv R_{j,1} \nonumber\]
With respect to the product \(P_{j}P_{k}\), for \(j \ne k\), the calculation runs along the same lines. The difference comes in Equation where, as \(C_{j}\) lies completely outside of \(C_{k}\), both integrals are zero. Hence,
If \(j \ne k\) then \(P_{j}P_{k} = 0\)
Along the same lines we define
\[D_{j} \equiv R_{j,2} \nonumber\]
and prove
If \(1 \le k \le m_{j}-1\) then \(D_{j}^{k} = R_{j,k+1}\) \cdot D_{j}^{m_{j}} = 0\)
For \(k\) and \(l\) greater than or equal to one,
\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int R(z)(z-\lambda_{j})^{k} dz \int R(w)(w-\lambda_{j})^{l} dw \nonumber\]
\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int \int R(z) R(w)(z-\lambda_{j})^{k} (w-\lambda_{j})^{l} dw dz \nonumber\]
\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int \int \frac{R(z)-R(w)}{w-z} (z-\lambda_{j})^{k} (w-\lambda_{j})^{l} dw dz \nonumber\]
\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int R(z) (z-\lambda_{j})^{k} \int \frac{(w-\lambda_{j})^{l}}{w-z} dw dz-\frac{1}{(2\pi i)^2} \int R(w) (w-\lambda_{j})^{k} \int \frac{(z-\lambda_{j})^{k}}{w-z} dz dw \nonumber\]
\[R_{j,k+1} R_{j,l+1} = \frac{1}{2\pi i} \int R(z) (z-\lambda_{j})^{k+l} dz = R_{j,k+l+1} \nonumber\]
because
\[\int \frac{(w-\lambda_{j})^{l}}{w-z} dw = 2 \pi i (z-\lambda_{j})^{l} \nonumber\]
and
\[\int \frac{(z-\lambda_{j})^{k}}{w-z} dw = 0 \nonumber\]
With \(k = l = 1\) we have shown \(R_{j,2}^{2} = R_{j,3}\) i.e., \(D_{j}^{2} = R_{j,3}\). Similarly, with \(k = 1\) and \(l = 2\) we find \(R_{j,2} R_{j,3} = R_{j,4}\) i.e., \(D_{j}^{3} = R_{j,4}\). Continuing in this fashion we find \(R_{j,k} R_{j,k+1} = R_{j,k+2} = j\), or \(D_{j}^{k+1} = R_{j,k+2}\). Finally, at \(k = m_{j-1}\) this becomes
\[D_{j}^{m_{j}} = R_{j, m_{j}+1} = \frac{1}{2\pi i} \int R(z)(z-\lambda_{j})^{m_{j}} dz = 0 \nonumber\]
by Cauchy's Theorem.
With this we now have the sought after expansion
\[R(z) = \sum_{j = 1}^{h} \frac{1}{z-\lambda_{j}} P_{j}+\sum_{k = 1}^{m_{j-1}} \frac{1}{(z-\lambda_{j})^{k+1}} D_{j}^{k} \nonumber\]
along with the verification of a number of the properties laid out in Complex Integration Equations.