Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 6 - Linear Transformations - 6.1 Introduction to Linear Transformations - 6.1 Exercises - Page 300: 17

Answer

$T$ is a linear tranformation.

Work Step by Step

A transformation is said to be linear if $T(av+w)=aT(v)+T(w)$ for all $v$,$w$ in the given domain and for all scalars $a$. Here the domain is $M_{2,2}$; therefore, let $v = \left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right],w = \left[ {\begin{array}{*{20}{c}} e&f\\ g&h \end{array}} \right]$ be arbitrary elements of $M_{2,2}$ and $k$ be any scalar. Then, $\begin{array}{l} T(kv + w)\\ = T\left( {k\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} e&f\\ g&h \end{array}} \right]} \right)\\ = T\left( {\left[ {\begin{array}{*{20}{c}} {ka}&{kb}\\ {kc}&{kd} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} e&f\\ g&h \end{array}} \right]} \right)\\ = T\left( {\left[ {\begin{array}{*{20}{c}} {ka + e}&{kb + f}\\ {kc + g}&{kd + h} \end{array}} \right]} \right)\\ = (ka + e) + (kb + f) - (kc + g) + (kd + h)\\ = (ka + kb - kc + kd) + (e + f - g + h)\\ = k(a + - c + d) + (e + f - g + h)\\ = kT\left( {\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]} \right) + T\left( {\left[ {\begin{array}{*{20}{c}} e&f\\ g&h \end{array}} \right]} \right)\\ = kT(v) + T(w) \end{array}$ Since this holds for all $v,w$ in $M_{2,2}$ and for all scalars $k$, $T$ is a linear transformation.
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