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Chapter 6: Q6.2-31E (page 331)

Question: Show that orthogonal projection of a vector y onto a line L through the origin in \({\mathbb{R}^{\bf{2}}}\) does not depend on the choice of the nonzero u in L used in the formula for \({\bf{\hat y}}\). To do this, suppose y and u are given and \({\bf{\hat y}}\) has been computed by formula (2) in this section. Replace u in that formula by \(c{\bf{u}}\), where c is an unspecified nonzero scalar. Show that the new formula gives the same \({\bf{\hat y}}\).

Short Answer

Expert verified

The new formula gives the same \(\widehat {\bf{y}}\) because y does not depend on the choice of nonzero u in L.

Step by step solution

01

Write orthogonal projection of y onto line L

The orthogonal projection of yonto the line L can be expressed as:

\(\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\)

02

Replace u with cu in the formula of \({\bf{\hat y}}\)

The formula \(\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot {\bf{u}}}}{{{\bf{u}} \cdot {\bf{u}}}}{\bf{u}}\) becomes:

\(\begin{array}{c}\widehat {\bf{y}} = \frac{{{\bf{y}} \cdot \left( {c{\bf{u}}} \right)}}{{\left( {c{\bf{u}}} \right) \cdot \left( {c{\bf{u}}} \right)}}\left( {c{\bf{u}}} \right)\\ = \frac{{c\left( {{\bf{y}} \cdot {\bf{u}}} \right)}}{{{c^2}\left( {{\bf{u}} \cdot {\bf{u}}} \right)}}\left( c \right){\bf{u}}\\ = \widehat {\bf{y}}\end{array}\)

Therefore, y does not depend on the choice of nonzero u in L.

Hence, it is proved that the new formula gives the same \(\widehat {\bf{y}}\).

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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 \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

A simple curve that often makes a good model for the variable costs of a company, a function of the sales level \(x\), has the form \(y = {\beta _1}x + {\beta _2}{x^2} + {\beta _3}{x^3}\). There is no constant term because fixed costs are not included.

a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\).

b. Find the least-squares curve of the form above to fit the data \(\left( {4,1.58} \right),\left( {6,2.08} \right),\left( {8,2.5} \right),\left( {10,2.8} \right),\left( {12,3.1} \right),\left( {14,3.4} \right),\left( {16,3.8} \right)\) and \(\left( {18,4.32} \right)\), with values in thousands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

a. Rewrite the data in Example 1 with new \(x\)-coordinates in mean deviation form. Let \(X\) be the associated design matrix. Why are the columns of \(X\) orthogonal?

b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, \(y = {\beta _0} + {\beta _1}x*\), where \(x* = x - 5.5\).

Suppose \(A = QR\), where \(Q\) is \(m \times n\) and R is \(n \times n\). Showthat if the columns of \(A\) are linearly independent, then \(R\) mustbe invertible.
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