Exercise 5.B.8

Answers

Proof. Define $T \in L (ℝ^{2})$ by

T (x, y) = \frac{1}{\sqrt{2}} (x - y, x + y)

Then, for $(a, b) \in ℝ^{2}$ , we have

\begin{aligned} T^{4} (a, b) & = \frac{1}{\sqrt{2}} T^{3} (a - b, a + b) \\ = \frac{1}{2} T^{2} (- 2 b, 2 a) \\ = \frac{1}{2 \sqrt{2}} T (- 2 a - 2 b, 2 a - 2 b) \\ = \frac{1}{4} (- 4 a, - 4 b) \\ = - (a, b) \end{aligned}

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celiopassos

2017-10-06 00:00