Chapter 2: Q22Q (page 93)
Show that if the columns of Bare linearly dependent, then so are the columns of AB.
The columns of ABare linearly dependent.
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In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.
Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).
Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?
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 “input” to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the “output” and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the “state” 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.)
Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).
Use the inverse found in Exercise 1 to solve the system
\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{6}}{{\bf{x}}_{\bf{2}}} = {\bf{2}}\\{\bf{5}}{{\bf{x}}_{\bf{1}}} + {\bf{4}}{{\bf{x}}_{\bf{2}}} = - {\bf{1}}\end{aligned}\)
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