Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 6 - Linear Transformations - 6.5 The Matrix of a Linear Transformation - Problems - Page 427: 8

Answer

See below

Work Step by Step

a) Obtain $T(\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix})=\begin{bmatrix} 2 & 0\\ 0 & 0 \end{bmatrix}-\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\\ T(\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix})=\begin{bmatrix} 0 & 2\\ 0 & 0 \end{bmatrix}-\begin{bmatrix} 0 & 0\\ 1 & 0 \end{bmatrix}=\begin{bmatrix} 0& 2\\ -1 & 0 \end{bmatrix} \\ T(\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix})=\begin{bmatrix} 0 & 0\\ 2 & 0 \end{bmatrix}-\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 & -1\\ 2 & 0 \end{bmatrix} \\ T(\begin{bmatrix}0 & 0\\ 0 & 1 \end{bmatrix})=\begin{bmatrix} 0 & 0\\ 0 & 2 \end{bmatrix}-\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0& 0\\ 0 & 1 \end{bmatrix}$ then we have: $T(E_{11})=\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}=1.E_{11}+0.E_{12}+0.E_{21}+0.E_{22}\\ T(E_{12})=\begin{bmatrix} 0 & 2\\ -1 & 0 \end{bmatrix}=0.E_{11}+2.E_{12}-1.E_{21}+0.E_{22}\\ T(E_{21})=\begin{bmatrix} 1 & -1\\ 2 & 0 \end{bmatrix}=0.E_{11}-1.E_{12}+2.E_{21}+0.E_{22}\\ T(E_{22})=\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}=0.E_{11}+0.E_{12}+0.E_{21}+1.E_{22}$ as $[T(E_{11})]_C=\begin{bmatrix} 1\\ 0 \\ 0 \\ 0 \end{bmatrix}\\ [T(E_{12})]_C=\begin{bmatrix} 0 \\ 2 \\ -1 \\ 0 \end{bmatrix} \\ [T(E_{21})]_C=\begin{bmatrix} 0 \\ -1 \\ 2 \\ 0 \end{bmatrix} \\ [T(E_{22})]_C=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$ Hence, $[T]_B^C=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$ b) Obtain $T(\begin{bmatrix} -1 & -2\\ -2 & -3 \end{bmatrix})=\begin{bmatrix} -2 & -4\\ -4 & -6 \end{bmatrix}-\begin{bmatrix} -1 & -2\\ -2 & -3 \end{bmatrix}=\begin{bmatrix} -1 & -2\\ -2 & -3 \end{bmatrix}\\ T(\begin{bmatrix} 1 & 1\\ 2 & 2 \end{bmatrix})=\begin{bmatrix} 2 & 2\\ 4 & 4 \end{bmatrix}-\begin{bmatrix} 1 & 2\\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 1& 0\\ 3 & 2 \end{bmatrix} \\ T(\begin{bmatrix} 0 & -3\\ 2& -2 \end{bmatrix})=\begin{bmatrix} 0 & -6\\ 4 & -4 \end{bmatrix}-\begin{bmatrix} 0 & 2\\ -3 & -2 \end{bmatrix}=\begin{bmatrix} 0 & -8\\ 7 & -2 \end{bmatrix} \\ T(\begin{bmatrix}0 & 4\\ 1 & 0 \end{bmatrix})=\begin{bmatrix} 0 & 8\\ 2 & 0 \end{bmatrix}-\begin{bmatrix} 0 & 1\\ 4 & 0 \end{bmatrix}=\begin{bmatrix} 0& 7\\ -2 & 0 \end{bmatrix}$ then we have: $T(\begin{bmatrix} -1 & -2\\ -2 & -3 \end{bmatrix})=\begin{bmatrix} -1 & -2\\ -2 & -3 \end{bmatrix}=-1.E_{11}-2.E_{12}-2.E_{21}-3.E_{22}\\ T(\begin{bmatrix} 1 & 1\\ 2 & 2 \end{bmatrix})=\begin{bmatrix} 1 & 0\\ 3 & 2 \end{bmatrix}=1.E_{11}+0.E_{12}+3.E_{21}+2.E_{22}\\ T(\begin{bmatrix} 0 & -3\\ 2 & -2 \end{bmatrix})=\begin{bmatrix} 0 & -8\\ 7 & -2 \end{bmatrix}=0.E_{11}-8.E_{12}+7.E_{21}-2.E_{22}\\ T(\begin{bmatrix} 0 & 4\\ 1 & 0 \end{bmatrix})=\begin{bmatrix} 0 &7\\ -2 & 0 \end{bmatrix}=0.E_{11}+7.E_{12}-2.E_{21}+0.E_{22}$ as $[T(\begin{bmatrix} -1 & -2 \\ -2 & -3 \end{bmatrix})]_C=\begin{bmatrix} -1\\ -2 \\ -2 \\ -3 \end{bmatrix}\\ [T(\begin{bmatrix} 1 & 1\\ 2 & 2 \end{bmatrix})]_C=\begin{bmatrix} 1 \\ 0 \\ 3 \\ 2 \end{bmatrix} \\ [T(\begin{bmatrix} 0 & -3\\ 2 & -2 \end{bmatrix})]_C=\begin{bmatrix} 0 \\ -8 \\ 7 \\ -2 \end{bmatrix} \\ [T(\begin{bmatrix} 0 & 4\\ 1 & 0 \end{bmatrix})]_C=\begin{bmatrix} 0 \\ 7 \\ -2 \\ 0 \end{bmatrix}$ Hence, $[T]_B^C=\begin{bmatrix} -1 & 1 & 0 & 0 \\ -2 & 0 & -8 & 7 \\-2 & 3 & 7 & -2 \\ -3 & 2 & -2 & 0\end{bmatrix}$
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