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Chapter 2: Q14SE (page 93)

Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\0\\1\end{aligned}} \right)\) and \(x = \left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\). Determine \(P\) and \(Q\) as in Exercise 13, and compute \(Px\) and \(Qx\). The figure shows that \(Qx\) is the reflection of the x through \({x_1}{x_2}\)-plane.

Short Answer

Expert verified

\(P = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right),Q = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\), \(P{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\0\\3\end{aligned}} \right)\), \(Q{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}1\\5\\{ - 3}\end{aligned}} \right)\).

Step by step solution

01

Determine \(P\) and \(Q\) as in Exercise 13

In Exercise 13, it is given that u is in \({\mathbb{R}^n}\) with \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm u}\nolimits} = 1\). Let \(P = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) (an outer product) and \(Q = I - 2P\).

Calculate \(P = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) as shown below.

\(\begin{aligned}{c}P = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\\ = \left( {\begin{aligned}{*{20}{c}}0\\0\\1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right)\end{aligned}\)

Calculate \(Q = I - 2P\) as shown below.

\(\begin{aligned}{c}Q = I - 2P\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right) - 2\left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&2\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\end{aligned}\)

Thus, \(P = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right),Q = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\).

02

Determine \(Px\) and \(Qx\)

It is given that \(x = \left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\).

Calculate \(Px\) a shown below.

\(\begin{aligned}{c}P{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0&0&0\\0&0&0\\0&0&1\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{0 + 0 + 0}\\{0 + 0 + 0}\\{0 + 0 + 3}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0\\0\\3\end{aligned}} \right)\end{aligned}\)

Calculate \(Qx\) as shown below.

\(\begin{aligned}{c}Q{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1\\5\\3\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{1 + 0 + 0}\\{0 + 5 + 0}\\{0 + 0 - 3}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1\\5\\{ - 3}\end{aligned}} \right)\end{aligned}\)

Thus, \(P{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\0\\3\end{aligned}} \right)\),

\(Q{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}1\\5\\{ - 3}\end{aligned}} \right)\).

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Most popular questions from this chapter

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.

Show that if ABis invertible, so is B.

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.

Suppose \(AD = {I_m}\) (the \(m \times m\) identity matrix). Show that for any b in \({\mathbb{R}^m}\), the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has a solution. (Hint: Think about the equation \(AD{\mathop{\rm b}\nolimits} = {\mathop{\rm b}\nolimits} \).) Explain why Acannot have more rows than columns.

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]
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