Question

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

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

Step-by-step solution

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

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

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