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Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

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

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

Step by step solution

01

Find the value of \({W^T}W\)

Given, \(W = \left[ {\begin{array}{*{20}{c}}X&{{{\bf{x}}_0}}\end{array}} \right]\).

Then,

\(\begin{array}{c}{W^T}W = \left[ {\begin{array}{*{20}{c}}{{X^T}}\\{{\bf{x}}_0^T}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&{{{\bf{x}}_0}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{X^T}X}&{{X^T}{{\bf{x}}_0}}\\{{\bf{x}}_0^TX}&{{\bf{x}}_0^T{{\bf{x}}_0}}\end{array}} \right].\end{array}\)

02

Find the Schur complement \(S\)

By the formula \(S\)(Schur complement) from Exercise 15,

\(\begin{array}{c}S = {\bf{x}}_0^T{{\bf{x}}_0} - {\bf{x}}_0^TX{\left( {{X^T}X} \right)^{ - 1}}{X^T}{{\bf{x}}_0}\\ = {\bf{x}}_0^T\left( {{I_m} - X{{\left( {{X^T}X} \right)}^{ - 1}}{X^T}} \right){{\bf{x}}_0}\\ = {\bf{x}}_0^TM{{\bf{x}}_0}.\end{array}\)

So, the value of \(S\) is \({\bf{x}}_0^TM{{\bf{x}}_0}\).

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