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,

$$\begin{align*}BR(s) &=R(s)B \\[4pt] &= sR(s)-I \end{align*}$$