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Chapter 6: Q6.2-34E (page 331)

Question: Given \({\bf{u}} \ne {\bf{0}}\) in \({\mathbb{R}^n}\), let \(L = {\bf{Span}}\left\{ {\bf{u}} \right\}\). For y in \({\mathbb{R}^n}\), the reflection of y in L is the point \({\bf{ref}}{{\bf{l}}_L}y\) defined by

\({\bf{ref}}{{\bf{l}}_L}{\bf{y}} = {\bf{2}} \cdot {\bf{pro}}{{\bf{j}}_L}{\bf{y}} - {\bf{y}}\)

See the figure, which shows that \({\bf{ref}}{{\bf{l}}_L}{\bf{y}}\) is the sum of \({\bf{\hat y}} = {\bf{pro}}{{\bf{j}}_L}{\bf{y}}\) and \(\widehat {\bf{y}} - {\bf{y}}\). Show that the mapping \({\bf{y}} \mapsto {\bf{ref}}{{\bf{l}}_L}{\bf{y}}\) is a linear transformation.

Short Answer

Expert verified

It is proved that the mapping \({\bf{y}} \mapsto {\rm{ref}}{{\rm{l}}_L}{\bf{y}}\) is a linear transformation.

Step by step solution

01

Write the projection for x

The projection for x is given by;

\({\rm{pro}}{{\rm{j}}_L}{\bf{x}} = \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\)

The transformation becomes:

\(\begin{array}{c}T\left( {\bf{x}} \right) = {\rm{pro}}{{\rm{j}}_L}{\bf{x}}\\ = \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\end{array}\)

02

Use the inner product for the transformation

Apply an inner product for the transformation \(T\left( {\bf{x}} \right)\) by using the transformation \(T\left( {\bf{x}} \right) = \frac{{{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\).

\(\begin{array}{c}T\left( {a{\bf{x}} + b{\bf{y}}} \right) = \frac{{\left( {a{\bf{x}} + b{\bf{y}}} \right) \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \frac{{b{\bf{x}} \cdot {\bf{u}} + b{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = \frac{{a{\bf{x}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}} + \frac{{b{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\\ = aT\left( {\bf{x}} \right) + bT\left( {\bf{y}} \right)\end{array}\)

Therefore, \(T\left( {\bf{x}} \right)\) is a linear transformation.

03

Check whether \({T_{\bf{1}}}\) is a linear transformation

Let, \({T_1}\left( {\bf{y}} \right) = {\rm{ref}}{{\rm{l}}_L}{\bf{y}} = 2 \cdot {\rm{pro}}{{\rm{j}}_L}{\bf{y}} - {\bf{y}}\).

Check inner product for the transformation \({T_1}\) as shown below:

\(\begin{array}{c}{T_1}\left( {c{\bf{y}} + d{\bf{z}}} \right) = {\rm{re}}{{\rm{f}}_L}\left( {c{\bf{y}} + d{\bf{z}}} \right)\\ = 2 \cdot {\rm{pro}}{{\rm{j}}_L}\left( {c{\bf{y}} + d{\bf{z}}} \right) - \left( {c{\bf{y}} + d{\bf{z}}} \right)\\ = 2 \cdot \left( {\left( c \right){\rm{pro}}{{\rm{j}}_L}{\bf{y}} + \left( d \right){\rm{pro}}{{\rm{j}}_L}{\bf{z}}} \right) - c{\bf{y}} - d{\bf{z}}\\ = 2c \cdot \left( {{\rm{pro}}{{\rm{j}}_L}{\bf{y}}} \right) + 2d \cdot \left( {{\rm{pro}}{{\rm{j}}_L}{\bf{z}}} \right) - c{\bf{y}} - d{\bf{z}}\\ = c\left( {2 \cdot {\rm{pro}}{{\rm{j}}_L}{\bf{y}} - {\bf{y}}} \right) + d\left( {2 \cdot {\rm{pro}}{{\rm{j}}_L}{\bf{z}} - {\bf{z}}} \right)\\ = c\left( {{\rm{ref}}{{\rm{l}}_L}{\bf{y}}} \right) + d\left( {{\rm{ref}}{{\rm{l}}_L}{\bf{z}} - {\bf{z}}} \right)\\ = c{T_1}\left( {\bf{y}} \right) + d{T_1}\left( {\bf{z}} \right)\end{array}\)

The above equation shows that \({T_1}\) is a linear transformation.

Thus, it is proved that \({\bf{y}} \mapsto {\rm{ref}}{{\rm{l}}_L}{\bf{y}}\) is a linear transformation.

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Most popular questions from this chapter

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1. Show that the Cauchy-Schwarz inequality holds for \({\bf{x}} = \left( {{\bf{3}}, - {\bf{2}}} \right)\) and \({\bf{y}} = \left( { - {\bf{2}},{\bf{1}}} \right)\). (Suggestion: Study \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\) may be written in the form

\(\begin{aligned}{c}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

Derive the normal equations (7) from the matrix form given in this section.

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram鈥揝chmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

  1. \(\left( {\begin{aligned}{{}{}}3\\0\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}8\\5\\{ - 6}\end{aligned}} \right)\)
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