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Q2.3-39Q

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Linear Algebra and its Applications
Found in: Page 93
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 \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be an invertible linear transformation, and let S and U be functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\) such that \(S\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) and \(\)\(U\left( {T\left( {\mathop{\rm x}\nolimits} \right)} \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\). Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\). This will show that T has a unique inverse, as asserted in theorem 9. [Hint: Given any v in \({\mathbb{R}^n}\), we can write \({\mathop{\rm v}\nolimits} = T\left( {\mathop{\rm x}\nolimits} \right)\) for some x. Why? Compute \(S\left( {\mathop{\rm v}\nolimits} \right)\) and \(U\left( {\mathop{\rm v}\nolimits} \right)\)].

It is proved that \(U\left( v \right) = S\left( v \right)\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Show that T is onto mapping

For any v in \({\mathbb{R}^n}\), you can write \({\mathop{\rm v}\nolimits} = T\left( x \right)\) for some x (since \(T\) is onto mapping).

Step 2: Show that \(U\left( v \right) = S\left( v \right)\) for all v in \({\mathbb{R}^n}\)

According to the assumed properties of S and U, \(S\left( v \right) = S\left( {T\left( x \right)} \right) = x\) and \(U\left( v \right) = U\left( {T\left( x \right)} \right) = x\). Therefore, \[S\left( v \right)\] and \[U\left( v \right)\] are equal for any v.

This means that S and U are the same functions from \({\mathbb{R}^n}\) into \({\mathbb{R}^n}\).

Thus, it is proved that \(U\left( v \right) = S\left( v \right)\).

Most popular questions for Math Textbooks

Let A be an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If A is larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

Suppose the last column of AB is entirely zero but B itself has no column of zeros. What can you say about the columns of A?

Suppose A is a \(3 \times n\) matrix whose columns span \({\mathbb{R}^3}\). Explain how to construct an \(n \times 3\) matrix D such that \(AD = {I_3}\).

Let U be the \({\bf{3}} \times {\bf{2}}\) cost matrix described in Example 6 of Section 1.8. The first column of U lists the costs per dollar of output for manufacturing product B, and the second column lists the costs per dollar of output for product C. (The costs are categorized as materials, labor, and overhead.) Let \({q_1}\) be a vector in \({\mathbb{R}^{\bf{2}}}\) that lists the output (measured in dollars) of products B and C manufactured during the first quarter of the year, and let \({q_{\bf{2}}}\), \({q_{\bf{3}}}\) and \({q_{\bf{4}}}\) be the analogous vectors that list the amounts of products B and C manufactured in the second, third, and fourth quarters, respectively. Give an economic description of the data in the matrix UQ, where \(Q = \left( {\begin{aligned}{*{20}{c}}{{{\bf{q}}_1}}&{{{\bf{q}}_2}}&{{{\bf{q}}_3}}&{{{\bf{q}}_4}}\end{aligned}} \right)\).

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).
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