Question

Consider the linear transformation U: R^3 rightarrow R^3

Alternate coordinate systems
ANSWERED
asked 2020-10-20

Consider the linear transformation \(\displaystyle{U}:{R}^{{3}}\rightarrow{R}^{{3}}\) defined by \(U \left(\begin{array}{c}x\\ y \\z \end{array}\right) = \left(\begin{array}{c} z - y \\ z + y \\ 3z - x - y \end{array}\right)\) and the bases \(\epsilon = \left\{ \left(\begin{array}{c}1\\ 0 \\0\end{array}\right), \left(\begin{array}{c}0\\ 1 \\ 0\end{array}\right), \left(\begin{array}{c}0\\ 0 \\ 1\end{array}\right) \right\}, \gamma = \left\{ \left(\begin{array}{c}1 - i\\ 1 + i \\ 1 \end{array}\right), \left(\begin{array}{c} -1\\ 1 \\ 0\end{array}\right), \left(\begin{array}{c}0\\ 0 \\ 1\end{array}\right) \right\}\), Compute the four coordinate matrices \(\displaystyle{{\left[{U}\right]}_{{\epsilon}}^{{\gamma}}},{{\left[{U}\right]}_{{\gamma}}^{{\gamma}}},\)

Answers (1)

2020-10-21

\(U\left(\begin{array}{c}1 - i\\ 1 + i \\ 1 \end{array}\right) = \left(\begin{array}{c} -i\\ 2 + i \\ 1 \end{array}\right) = 1 \cdot v_1 + 1 \cdot v_2 + 0 \cdot v_3\)
\(U\left(\begin{array}{c} -1\\ 1 \\ 0\end{array}\right) = \left(\begin{array}{c}-1\\ 1 \\ 0\end{array}\right) = 0 \cdot v_1 + 1 \cdot v_2 + 0 \cdot v_3\)
\(U\left(\begin{array}{c}0\\ 1 \\ 1\end{array}\right) = \left(\begin{array}{c}0\\ 2 \\ 2\end{array}\right) = 0 \cdot v_1 + 0 \cdot v_2 + 2 \cdot v_3\) So, \([U]_{\gamma}^{\gamma} = \begin{bmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\)
\(U\left(\begin{array}{c}1\\ 0 \\ 0\end{array}\right) = \left(\begin{array}{c}0\\ 0 \\ 1\end{array}\right). Since \left(\begin{array}{c}0\\ 0 \\ 1\end{array}\right) \not{\in} span \left\{ \gamma \right\}\) So, \([U]_{\epsilon}^{\gamma}\) does not exist \(U\left(\begin{array}{c}1 - i\\ 1 + i \\ 1\end{array}\right) = i_1 \cdot \epsilon_1 + (2 + i) \cdot \epsilon_2 + 1 \cdot \epsilon_2\)
\(U\left(\begin{array}{c} -1\\ 1 \\ 0\end{array}\right) = \left(\begin{array}{c}-1\\ 1 \\ 0\end{array}\right) = -1 \cdot \epsilon_1 + 1 \cdot \epsilon_2 + 0 \cdot \epsilon_3\)
\(U\left(\begin{array}{c}0\\ 1 \\ 1\end{array}\right) = \left(\begin{array}{c}0\\ 2 \\ 1\end{array}\right) = 0 \cdot \epsilon_1 + 2 \cdot \epsilon_2 + 1 \cdot \epsilon_3\) So, \([U]_{\epsilon}^{\gamma} = \begin{bmatrix}-i & -1 & 0 \\ 2 + i & 1 & 2 \\ 1 & 0 & 1 \end{bmatrix}\)

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