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Q14E

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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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.)

It is verified that, \(\bar y\) and \(\) satisfies the linear equation \(\bar y = {\beta _0} + \bar x{\beta _1}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

The General Linear Model

The equation of the general linear model is defined as:

\({\bf{y}} = X\beta + \in \)

Here, \({\bf{y}} = \left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right)\) is an observational vector, \(X = \left( {\begin{aligned}1&{{x_1}}& \cdots &{x_1^n}\\1&{{x_2}}& \cdots &{x_2^n}\\ \vdots & \vdots & \ddots & \vdots \\1&{{x_n}}& \cdots &{x_n^n}\end{aligned}} \right)\) is the design matrix, \(\beta = \left( {\begin{aligned}{{\beta _1}}\\{{\beta _2}}\\ \vdots \\{{\beta _n}}\end{aligned}} \right)\) is parameter vector, and \( \in = \left( {\begin{aligned}{{ \in _1}}\\{{ \in _2}}\\ \vdots \\{{ \in _n}}\end{aligned}} \right)\) is a residual vector.

Determine whether \(\overline x \) and \(\overline y \) satisfies the given linear equation 

Write the design matrix as \(X = \left( {\begin{aligned}1&{\bf{x}}\end{aligned}} \right)\).

As the equation of the general linear model is given by \({\bf{y}} = X\beta + \in \). Isolate residual vector from the equation.

\( \in = {\bf{y}} - X{\bf{\hat \beta }}\)

As the residual vector is orthogonal to columns of \(X\), so \(1 \cdot \in = 0\).

\(\begin{aligned}1 \cdot \left( {{\bf{y}} - X{\bf{\hat \beta }}} \right) = 0\\{1^T}{\bf{y}} - \left( {{1^T}X} \right){\bf{\hat \beta }} = 0\\\left( {\begin{aligned}1\\1\\ \vdots \\1\end{aligned}} \right)\left( {\begin{aligned}{{y_1}}\\{{y_2}}\\ \vdots \\{{y_n}}\end{aligned}} \right) - {\left( {\begin{aligned}1\\1\\ \vdots \\1\end{aligned}} \right)^T}\left( {\begin{aligned}{{x_1}}\\{{x_2}}\\ \vdots \\{{x_n}}\end{aligned}} \right) \cdot {\bf{\hat \beta }} = 0\\\left( {{y_1} + {y_2} + \cdots + {y_n}} \right) - \left( {\begin{aligned}{1 + 1 + \ldots n{\rm{ times}}}&{{x_1} + {x_2} + \cdots + {x_n}}\end{aligned}} \right) \cdot {\bf{\hat \beta }} = 0\\\sum y - \left( {\begin{aligned}n&{\sum x }\end{aligned}} \right)\left( {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right) = 0\\\sum y - \left( {n{\beta _0} + \sum x {\beta _1}} \right) = 0\end{aligned}\)

Write \(\sum y \) as \(n\bar y\) and \(\sum x \) as \(n\bar x\).

\(\begin{aligned}n\bar y - \left( {n{\beta _0} + n\bar x{\beta _1}} \right) = 0\\n\bar y - n{\beta _0} - n\bar x{\beta _1} = 0\\\bar y - {\beta _0} - \bar x{\beta _1} = 0\\\bar y = {\beta _0} + \bar x{\beta _1}\end{aligned}\)

This implies that \(\bar y\) and \(\bar x\) satisfies the linear equation \(\bar y = {\beta _0} + \bar x{\beta _1}\).

Most popular questions for Math Textbooks

In Exercises 9-12 find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

12. \(A = \left[ {\begin{array}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{ - {\bf{1}}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{1}}&{ - {\bf{1}}}\end{array}} \right]\), \({\bf{b}} = \left( {\begin{array}{{}{}}{\bf{2}}\\{\bf{5}}\\{\bf{6}}\\{\bf{6}}\end{array}} \right)\)

[M] Let \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) be the fourth-order and fifth order Fourier approximations in \(C\left[ {{\bf{0}},{\bf{2}}\pi } \right]\) to the square wave function in Exercise 10. Produce separate graphs of \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) on the interval \(\left[ {{\bf{0}},{\bf{2}}\pi } \right]\), and produce graph of \({f_{\bf{5}}}\) on \(\left[ { - {\bf{2}}\pi ,{\bf{2}}\pi } \right]\).

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1, and let \({\bf{x}} = \left( {{\bf{1}},{\bf{1}}} \right)\) and \({\bf{y}} = \left( {{\bf{5}}, - {\bf{1}}} \right)\).

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

b. Describe all vectors \(\left( {{z_{\bf{1}}},{z_{\bf{2}}}} \right)\), that are orthogonal to y.

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

In Exercises 5-14, the space is \(C\left[ {0,2\pi } \right]\) with inner product (6).

7. Show that \({\left\| {\cos kt} \right\|^2} = \pi \) and \({\left\| {\sin kt} \right\|^2} = \pi \) for \(k > 0\).
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