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Linear Algebra and its Applications
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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

Suppose T and U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?

It is proved that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \), for all x in \({\mathbb{R}^n}\), is true.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Show that the mapping is identity mapping

Consider A and B are the standard matrices of T and U. Then, by matrix multiplication, AB is the standard matrix of the mapping \(x \mapsto T\left( {U\left( x \right)} \right)\). The mapping is identity mapping. Thus, according to the hypothesis, \(AB = I\).

Step 2: Determine whether \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \), for all x, is true

The matrices are invertible because, according to the invertible matrix theorem, A and B are square matrices and \(B = {A^{ - 1}}\). Therefore, \(BA = I\). It follows that the mapping \(x \mapsto T\left( {U\left( x \right)} \right)\) is identity mapping. This indicates that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\).

Thus, it is proved that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\).

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