Question

Assume \(A - s{I_n}\) is invertible and view (8) as a system of two matrix equations. Solve the top equation for \({\bf{x}}\) and substitute into the bottom equation. The result is an equation of the form \(W\left( s \right){\bf{u}} = {\bf{y}}\), where \(W\left( s \right)\) is a matrix that depends upon \(s\). \(W\left( s \right)\) is called the transfer function of the system because it transforms the input \({\bf{u}}\) into the output \({\bf{y}}\). Find \(W\left( s \right)\) and describe how it is related to the partitioned system matrix on the left side of (8). See Exercise 15.

Short answer

\({I_m} - C{\left( {A - s{I_n}} \right)^{ - 1}}B\), \(W\left( s \right)\) is the Schur complement of the matrix \(A - s{I_n}\).

Step-by-step solution

Rewrite equation (8)

The equation \(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\) can be expressed as

\(\left( {A - s{I_n}} \right){\bf{x}} + B{\bf{u}} = 0\) and \(C{\bf{x}} + {\bf{u}} = {\bf{y}}\).

Use the concept of inverse matrix for the reduced form of equation (8)

If the matrix \(\left( {A - s{I_n}} \right)\) is invertible, then by the equation \(\left( {A - s{I_n}} \right){\bf{x}} + B{\bf{u}}\),

\(\begin{array}{c}\left( {A - s{I_n}} \right){\bf{x}} + B{\bf{u}} = 0\\\left( {A - s{I_n}} \right){\bf{x}} = - B{\bf{u}}\\{\bf{x}} = - {\left( {A - s{I_n}} \right)^{ - 1}}B{\bf{u}}\end{array}\)

Substitute the value of \({\bf{x}}\) in the equation \(C{\bf{x}} + {\bf{u}} = {\bf{y}}\).

\(\begin{array}{c}C\left( { - {{\left( {A - s{I_n}} \right)}^{ - 1}}B{\bf{u}}} \right) + {\bf{u}} = {\bf{y}}\\{\bf{y}} = {I_m}{\bf{u}} - C{\left( {A - s{I_n}} \right)^{ - 1}}B{\bf{u}}\\ = \left( {{I_m} - C{{\left( {A - s{I_n}} \right)}^{ - 1}}B} \right){\bf{u}}\end{array}\)

Find the value of \(W\left( s \right)\)

As \({\bf{y}} = W\left( s \right){\bf{u}}\),

\(W\left( s \right) = {I_m} - C{\left( {A - s{I_n}} \right)^{ - 1}}B\).

So, the value of \(W\left( s \right)\) is \({I_m} - C{\left( {A - s{I_n}} \right)^{ - 1}}B\), and it represents the Schur complement of the matrix \(A - s{I_n}\).