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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

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

14.\(A = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\), \(Q = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\)

The upper triangular matrix is \(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\).

And it is verified that upper triangular matrix is correct.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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

A matrix with order \(m \times n\) can be written as the product of an 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: Find the matrix \(R\)

The given matrices are:

\(A = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\)and \(Q = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\)

Find the upper triangular matrix by using \({Q^T}A = R\).

\(\begin{aligned}{}R & = {Q^T}A\\ & = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}&{\frac{2}{7}}&{\frac{4}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}&{\frac{{ - 4}}{7}}&{\frac{2}{7}}\end{aligned}} \right)\left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{\frac{{ - 2}}{7} \times \left( { - 2} \right) + \frac{5}{7} \times 5 + \frac{2}{7} \times 2 + \frac{4}{7} \times 4}&{\frac{{ - 2}}{7} \times 3 + \frac{5}{7} \times 7 + \frac{2}{7} \times \left( { - 2} \right) + \frac{4}{7} \times 6}\\{\frac{5}{7} \times \left( { - 2} \right) + \frac{2}{7} \times 5 + \left( {\frac{{ - 4}}{7}} \right) \times 2 + \frac{2}{7} \times 4}&{\frac{5}{7} \times \left( 3 \right) + \frac{2}{7} \times 7 + \left( {\frac{{ - 4}}{7}} \right) \times \left( { - 2} \right) + \frac{2}{7} \times 6}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\end{aligned}\)

Hence, the upper triangular matrix is,\(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\).

Step 3: Checking Whether \(R\) is correct or not

Now, check that whether the obtained matrix \(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\) is correct or not. Find the product of \(Q\) and \(R\), if the product is the same as matrix \(A\), then \(R = \left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\) is correct.

\(\begin{aligned}{}QR & = \left( {\begin{aligned}{{}{r}}{\frac{{ - 2}}{7}}&{\frac{5}{7}}\\{\frac{5}{7}}&{\frac{2}{7}}\\{\frac{2}{7}}&{\frac{{ - 4}}{7}}\\{\frac{4}{7}}&{\frac{2}{7}}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}7&7\\0&7\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{r}}{ - 2}&3\\5&7\\2&{ - 2}\\4&6\end{aligned}} \right)\\ & = A\end{aligned}\)

Hence verified, the obtained upper triangular matrix is correct.

Most popular questions for Math Textbooks

Suppose radioactive substance A and B have decay constants of \(.02\) and \(.07\), respectively. If a mixture of these two substances at a time \(t = 0\) contains \({M_A}\) grams of \(A\) and \({M_B}\) grams of \(B\), then a model for the total amount of mixture present at time \(t\) is

\(y = {M_A}{e^{ - .02t}} + {M_B}{e^{ - .07t}}\) (6)

Suppose the initial amounts \({M_A}\) and are unknown, but a scientist is able to measure the total amounts present at several times and records the following points \(\left( {{t_i},{y_i}} \right):\left( {10,21.34} \right),\left( {11,20.68} \right),\left( {12,20.05} \right),\left( {14,18.87} \right)\) and \(\left( {15,18.30} \right)\).

a. Describe a linear model that can be used to estimate \({M_A}\) and \({M_B}\).

b. Find the least-squares curved based on (6).

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

9. \(\left[ {\begin{aligned}{{}{}}3&{ - 5}&1\\1&1&1\\{ - 1}&5&{ - 2}\\3&{ - 7}&8\end{aligned}} \right]\)

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–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 \(R = {Q^T}A\).

Let \(U\) be an \(n \times n\) orthogonal matrix. Show that if \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_n}} \right\}\) is an orthonormal basis for \({\mathbb{R}^n}\), then so is \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_n}} \right\}\).

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.)
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