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Chapter 6: Q-6-7-SE (page 331)

A Householder matrix, or an elementary reflector, has the form \(Q = I - 2{\bf{u}}{{\bf{u}}^T}\) where u is a unit vector. (See Exercise 13 in the Supplementary Exercise for Chapter 2.) Show that Q is an orthogonal matrix. (Elementary reflectors are often used in computer programs to produce a QR factorization of a matrix A. If A has linearly independent columns, then left-multiplication by a sequence of elementary reflectors can produce an upper triangular matrix.)

Short Answer

Expert verified

It is proved that \(Q\) is an orthogonal matrix.

Step by step solution

01

Compute \({Q^T}\)

Consider \({\bf{u}}\)as a unit vector, and consider that \(Q = I - 2{\bf{u}}{{\bf{u}}^T}\).

02

Show that Q is an orthogonal matrix

Prove that Q is an orthogonal matrix as shown below:

\(\begin{array}{c}Q{Q^T} = {Q^2}\\ = {\left( {I - 2{\bf{u}}{{\bf{u}}^T}} \right)^2}\\ = I - 2{\bf{u}}{{\bf{u}}^T} - 2{\bf{u}}{{\bf{u}}^T} + 4\left( {{\bf{u}}{{\bf{u}}^T}} \right)\left( {{\bf{u}}{{\bf{u}}^T}} \right)\end{array}\)

Here, \({{\bf{u}}^T}{\bf{u}} = {\bf{u}} \cdot {\bf{u}} = 1\) because the unit vector is \({\bf{u}}\). Therefore,

\(\begin{array}{c}\left( {{\bf{u}}{{\bf{u}}^T}} \right)\left( {{\bf{u}}{{\bf{u}}^T}} \right) = {\bf{u}}\left( {{{\bf{u}}^T}{\bf{u}}} \right){{\bf{u}}^T}\\ = {\bf{u}}{{\bf{u}}^T}\end{array}\)

And,

\(\begin{array}{c}Q{Q^T} = I - 2{\bf{u}}{{\bf{u}}^T} - 2{\bf{u}}{{\bf{u}}^T} + 4\left( {{\bf{u}}{{\bf{u}}^T}} \right)\\ = I\end{array}\)

Hence, it is proved that \(Q\) is an orthogonal matrix.

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Most popular questions from this chapter

Let \(\overline x = \frac{1}{n}\left( {{x_1} + \cdots + {x_n}} \right)\), and \(\overline y = \frac{1}{n}\left( {{y_1} + \cdots + {y_n}} \right)\). Show that the least-squares line for the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) must pass through \(\left( {\overline x ,\overline y } \right)\). That is, show that \(\overline x \) and \(\overline y \) satisfies the linear equation \(\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x \). (Hint: Derive this equation from the vector equation \({\bf{y}} = X{\bf{\hat \beta }} + \in \). Denote the first column of \(X\) by 1. Use the fact that the residual vector \( \in \) is orthogonal to the column space of \(X\) and hence is orthogonal to 1.)

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)

In Exercises 11 and 12, find the closest point to \[{\bf{y}}\] in the subspace \[W\] spanned by \[{{\bf{v}}_1}\], and \[{{\bf{v}}_2}\].

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]

In Exercises 7鈥�10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]鈥檚, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]
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