Question

Given u in \({\mathbb{R}^n}\) with \({{\mathop{\rm u}\nolimits} ^T}u = 1\), let \(P = u{{\mathop{\rm u}\nolimits} ^T}\) (an outer product) and \(Q = I - 2P\). Justify statements (a), (b), and (c).

1. \({P^2} = P\)
2. \({P^T} = P\)
3. \({Q^2} = I\)

The transformation \({\mathop{\rm x}\nolimits} \mapsto P{\mathop{\rm x}\nolimits} \) is called a projection and \({\mathop{\rm x}\nolimits} \mapsto Q{\mathop{\rm x}\nolimits} \) is called a Householder reflection. Such reflections are used in computer programs to create multiple zeros in a vector (usually a column of a matrix).

Short answer

1. It is proved that \({P^2} = P\).
2. It is proved that \({P^T} = P\).
3. It is proved that \({Q^2} = I\).

Step-by-step solution

Show that \({P^2} = P\)

a)

It is given that u is in \({\mathbb{R}^n}\) with \({{\mathop{\rm u}\nolimits} ^T}u = 1\). Let \(P = u{{\mathop{\rm u}\nolimits} ^T}\) and \(Q = I - 2P\).

\(\begin{array}{c}{P^2} = \left( {{{{\mathop{\rm uu}\nolimits} }^T}} \right)\left( {{{{\mathop{\rm uu}\nolimits} }^T}} \right)\\ = {\mathop{\rm u}\nolimits} \left( {{u^T}{\mathop{\rm u}\nolimits} } \right){{\mathop{\rm u}\nolimits} ^T}\\ = {\mathop{\rm u}\nolimits} \left( 1 \right){{\mathop{\rm u}\nolimits} ^T}\\ = P\end{array}\)

Since u satisfies \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm u}\nolimits} = 1\), it is proved that \({P^2} = P\).

Show that \({P^T} = P\)

\(\begin{array}{c}{P^T} = {\left( {{{{\mathop{\rm uu}\nolimits} }^T}} \right)^T}\\ = {\left( {{{\mathop{\rm u}\nolimits} ^T}} \right)^T}{{\mathop{\rm u}\nolimits} ^T}\\ = {\mathop{\rm u}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\\ = P\end{array}\)

Thus, it is proved that \({P^T} = P\).

Show that \({Q^2} = I\)

\(\begin{array}{c}{Q^2} = \left( {I - 2P} \right)\left( {I - 2P} \right)\\ = I - I\left( {2P} \right) - 2PI + 2P\left( {2P} \right)\\ = I - 4P + 4{P^2}\\ = I - 4P + 4P\,\,\,\left( {{\mathop{\rm Since}\nolimits} \,\,{P^2} = P} \right)\\ = I\end{array}\)

Thus, it is proved that \({Q^2} = I\).