Question

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Short answer

\({I_m} + C{\left( {A - BC - s{I_m}} \right)^{ - 1}}B\)

Step-by-step solution

Write the formula for Schur complement

The formula for Schur complement is \({A_{11}} - {A_{12}}A_{22}^{ - 1}{A_{21}}\).

Apply the Schur complement formula for the given matrix

For the matrix \(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\), apply the formula \({A_{11}} - {A_{12}}A_{22}^{ - 1}{A_{21}}\).

\(\begin{array}{c}S = {I_m} - \left( { - C} \right){\left( {A - BC - s{I_m}} \right)^{ - 1}}B\\ = {I_m} + C{\left( {A - BC - s{I_m}} \right)^{ - 1}}B\end{array}\)

So, the Schur complement is \({I_m} + C{\left( {A - BC - s{I_m}} \right)^{ - 1}}B\).