8.2: The Resolvent
The Transfer Function
One means by which to come to grips with \(R(s)\) is to treat it as the matrix analog of the scalar function
\[\frac{1}{s-b} \nonumber\]
This function is a scaled version of the even simpler function \(\frac{1}{1-z}\) This latter function satisfies the identity (just multiply across by \(1-z\) to check it)
\[\frac{1}{1-z} = 1+z+z^2+\cdots+z^{n-1}+\frac{z^n}{1-z} \nonumber\]
for each positive integer n. Furthermore, if \(|z| < 1\) then \(z^{n} \rightarrow 0\) as \(n \rightarrow \infty\) and so Equation becomes, in the limit,
\[\frac{1}{1-z} = \sum_{n = 0}^{\infty} z^n \nonumber\]
the familiar geometric series. Returning to Equation we write
\[\frac{1}{s-b} = \frac{\frac{1}{s}}{1-\frac{b}{s}} = \frac{1}{s}+\frac{b}{s^2}+ \cdots +\frac{b^{n-1}}{s^n}+\frac{b^n}{s^n} \frac{1}{s-b} \nonumber\]
and hence, so long as \(|s|>|b|\) we find,
\[\frac{1}{s-b} = \frac{1}{s} \sum_{n = 0}^{\infty} \left(\frac{b}{s}\right)^{n} \nonumber\]
This same line of reasoning may be applied in the matrix case. That is,
\[(sI-B)^{-1} = s^{-1} \left(I-\frac{B}{s}\right)^{-1} \left(\frac{1}{s}+\frac{B}{s^2}+ \cdots +\frac{B^{n-1}}{s^n}+\frac{B^n}{s^n}(sI-B)^{-1}\right) \nonumber\]
and hence, so long as \(|s| > ||B||\) where \(||B||\) is the magnitude of the largest element of \(B\)
\[\frac{1}{sI-B} = s^{-1} \sum_{n = 0}^{\infty} \left(\frac{B}{s}\right)^{n} \nonumber\]
Although Equation 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 however as an analytical rather than computational tool. More precisely, it facilitates the computation of integrals of \(R(s)\). However, \(C_{\rho}\) is the circle of radius \(\rho\) centered at the origin and \(\rho > ||B||\) then
\[\begin{align*} \int_{C_{\rho}} (sI-B)^{-1} ds &= \sum_{n = 0}^{\infty} B^n \int_{C_{\rho}} s^{-1-n} ds \\[4pt] &= 2 \pi i I \end{align*}\]
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
\[(s_{2}I-B)^{-1}-(s_{1}I-B)^{-1} = (s_{2}I-B)^{-1}(s_{1}I-B-s_{2}I+B)(s_{1}I-B)^{-1} \nonumber\]
that
\[R(s_{2})-R(s_{1}) = (s_{1}-s_{2})R(s_{2})R(s_{1}) \nonumber\]
The second identity is simply a rewriting of
\[(sI-B)(sI-B)^{-1} = (sI-B)^{-1}(sI-B) = I \nonumber\]
namely,
\[\begin{align*}BR(s) &=R(s)B \\[4pt] &= sR(s)-I \end{align*}\]