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Chapter 6: Q27E (page 331)

Suppose a vector \({\bf{y}}\) is orthogonal to vectors \({\bf{u}}\) and \({\bf{v}}\). Show that \({\rm{y}}\) is orthogonal to the vector \({\bf{u}} + {\bf{v}}\).

Short Answer

Expert verified

It is verified that \({\bf{y}}\)is orthogonal to vector \({\bf{u}} + {\bf{v}}\).

Step by step solution

01

Definition of Orthogonal sets.

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} = 0\end{aligned}\).

02

 Computing the required values.

Since, \({\bf{y}}\) is orthogonal to vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\).

Therefore,

\(\begin{aligned}{l}{\bf{y}} \cdot {\bf{u}} = 0\\{\bf{y}} \cdot {\bf{v}} = 0\end{aligned}\)

Using the inner productrule, we have:

\(\begin{aligned}{c}{\bf{y}} \cdot \left( {{\bf{u}} + {\bf{v}}} \right) &= {\bf{y}} \cdot {\bf{u}} + {\bf{y}} \cdot {\bf{v}}\\ &= 0 + 0\\ &= 0\end{aligned}\).

Hence proved: \({\bf{y}}\)is orthogonal to vector \({\bf{u}} + {\bf{v}}\).

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Most popular questions from this chapter

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

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

16. Use a matrix inverse to solve the system of equations in (7) and thereby obtain formulas for \({\hat \beta _0}\) , and that appear in many statistics texts.

Given \(A = QR\) as in Theorem 12, describe how to find an orthogonal\(m \times m\)(square) matrix \({Q_1}\) and an invertible \(n \times n\) upper triangular matrix \(R\) such that

\(A = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\)

The MATLAB qr command supplies this “full” QR factorization

when rank \(A = n\).

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 \({\bf{\hat x}}\).

3. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{ - {\bf{2}}}\\{ - {\bf{1}}}&{\bf{2}}\\{\bf{0}}&{\bf{3}}\\{\bf{2}}&{\bf{5}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{4}}}\\{\bf{2}}\end{aligned}} \right)\)

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from \(t = 0\) to \(t = 12\). The positions (in feet) were: 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 294.5, 380.4, 471.1, 571.7, 686.8, 809.2.

a. Find the least-squares cubic curve \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\) for these data.

b. Use the result of part (a) to estimate the velocity of the plane when \(t = 4.5\) seconds.

Let \(X\) be the design matrix in Example 2 corresponding to a least-square fit of parabola to data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Suppose \({x_1}\), \({x_2}\) and \({x_3}\) are distinct. Explain why there is only one parabola that best, in a least-square sense. (See Exercise 5.)
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