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Q17E

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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

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–Schmidt 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 \(R = {Q^T}A\).

a. False, because this statement is true when \(c \ne 0\).

b. True, because \({\rm{span}}\left\{ {{x_1}, \ldots ,{x_p}} \right\} = {\rm{span}}\left\{ {{v_1}, \ldots ,{v_p}} \right\}\).

c. True, by using the definition of \(QR\) factorization of a matrix.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: \(QR\) factorization of a Matrix

A matrix which has order \(m \times n\) can be written as the multiplication of a upper triangular matrix \(R\) and a matrix \(Q\) which is formed by applying Gram–Schmidt orthogonalization process to the \({\rm{col}}\left( A \right)\).

The matrix \(R\) can be found by the formula \({Q^T}A = R\).

Step 2: Checking whether the given statements are true of false

a.

This is false. Because this statement is true when \(c \ne 0\).

If \(c = 0\), then we will have \(\left\{ {{v_1},{v_2},0} \right\}\). We know that any set containing zero vector is not a linearly independent set.

b.

The given statement is true. From the Theorem 11, since \(\left\{ {{x_1}, \ldots ,{x_p}} \right\}\)gives the set of orthogonal vectors \(\left\{ {{v_1}, \ldots ,{v_p}} \right\}\), thus we have

\({\rm{span}}\left\{ {{x_1}, \ldots ,{x_p}} \right\} = {\rm{span}}\left\{ {{v_1}, \ldots ,{v_p}} \right\}\).

c.

The statement is true. From the definition of \(QR\) factorization of a matrix it is true.

We have \({Q^T}Q = I\), since the columns of the matrix \(Q\) are orthonormal.

\(\begin{aligned}{}A &= QR\\{Q^T}A &= {Q^T}QR\\{Q^T}A &= IR\\{Q^T}A &= R\end{aligned}\)

Thus, the statement is true.

Most popular questions for Math Textbooks

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 \(A = n\).

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.)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,{\rm{ }}{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}\)

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.)

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set. Verify the following equality by induction, beginning with \(p = 2\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_p}{{\bf{v}}_p}\), then

\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)

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