Exercise 2.2.3

Answers

Since

\begin{array}{l} T (1, 0) & = (1, 1, 2) = - \frac{1}{3} (1, 1, 0) + 0 (0, 1, 1) + \frac{2}{3} (2, 2, 3) \\ T (0, 1) & = (- 1, 0, 1) = - 1 (1, 1, 0) + 1 (0, 1, 1) + 0 (2, 2, 3) \\ T (1, 2) & = (- 1, 1, 4) = - \frac{7}{3} (1, 1, 0) + 2 (0, 1, 1) + \frac{2}{3} (2, 2, 3) \\ T (2, 3) & = (- 1, 2, 7) = - \frac{11}{3} (1, 1, 0) + 3 (0, 1, 1) + \frac{4}{3} (2, 2, 3) \end{array}

we have

{[T]}_{β}^{γ} = (\begin{matrix} - \frac{1}{3} & - 1 \\ 0 & 1 \\ \frac{2}{3} & 0 \end{matrix})

and

{[T]}_{α}^{γ} = (\begin{matrix} - \frac{7}{3} & - \frac{11}{3} \\ 2 & 3 \\ \frac{2}{3} & \frac{4}{3} \end{matrix}) .

jephianlin

2011-06-27 00:00