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Problem 311

Show geometrically that for each angle $$\theta$$, the transformation $$\mathrm{T}_{\theta}: \mathrm{R}^{2} \rightarrow \mathrm{R}^{2}$$, defined by $(\mathrm{x}, \mathrm{y}) \mathrm{T}_{\theta}=(\mathrm{x} \cos \theta-\mathrm{y} \sin \theta, \mathrm{x} \sin \theta+\mathrm{y} \cos \theta)$, is an orthogonal transformation.

Expert verified
The transformation Tθ represented by the matrix $$\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$ is an orthogonal transformation. It is verified by showing that the magnitudes of the transformed vectors are equal to those of the original vectors (length preservation), and by confirming that the dot product of transformed vectors is equal to the dot product of original vectors (angle preservation).
See the step by step solution

Step 1: Define the Transformation Tθ

Write the given transformation of Tθ as a matrix multiplication and simplify: Given transformation, Tθ: ℝ² → ℝ² is defined as: $$(x, y)T_{\theta} = (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)$$ We can represent this transformation as a matrix multiplication: $$T_{\theta} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$

Step 2: Verify Length Preservation

Check if the transformation Tθ preserves vector lengths by finding the magnitude of the transformed vector and equating it with the original vector's magnitude. Let the input vector be $$\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$$. Then, the transformed vector is: $$T_{\theta} \vec{v} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix} = \vec{v'}$$ Now, compare the magnitudes of the original vector and the transformed vector: Magnitude of the original vector, $$| \vec{v} | = \sqrt{x^2 + y^2}$$ Magnitude of the transformed vector, $$| \vec{v'} | = \sqrt{(x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2}$$ By expanding and simplifying the expression for the magnitude of the transformed vector, we get: $$| \vec{v'} | = \sqrt{x^2 \cos^2 \theta - 2xy\cos \theta \sin \theta + y^2 \sin^2 \theta + x^2 \sin^2 \theta + 2xy \cos \theta \sin \theta + y^2 \cos^2 \theta}$$ $$| \vec{v'} | = \sqrt{x^2 (\cos^2 \theta + \sin^2 \theta) + y^2 (\sin^2 \theta + \cos^2 \theta)}$$ Since $$\cos^2 \theta + \sin^2 \theta = 1$$, we obtain: $$| \vec{v'} | = \sqrt{x^2 + y^2}$$ Hence, $$| \vec{v'} | = | \vec{v} |$$, and Tθ preserves vector lengths.

Step 3: Verify Angle Preservation

Check if the transformation Tθ preserves angles between vectors by finding the angle between the original vectors and the transformed vectors. For angle preservation, we should verify that the dot product of the transformed vectors is equal to the dot product of the original vectors: Let $$\vec{v_1} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$$ and $$\vec{v_2} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$$ be two original vectors in ℝ². Their dot product is: $$\vec{v_1} · \vec{v_2} = x_1x_2 + y_1y_2$$ Now, let $$\vec{v_1'} = T_{\theta} \vec{v_1}$$ and $$\vec{v_2'} = T_{\theta} \vec{v_2}$$ be their transformed vectors. Their dot product is: $$\vec{v_1'} · \vec{v_2'} = (x_1 \cos \theta - y_1 \sin \theta)(x_2 \cos \theta - y_2 \sin \theta) + (x_1 \sin \theta + y_1 \cos \theta)(x_2 \sin \theta + y_2 \cos \theta)$$ Expanding and simplifying this expression, we get: $$\vec{v_1'} · \vec{v_2'} = x_1x_2 (\cos^2 \theta + \sin^2 \theta) + y_1y_2 (\cos^2 \theta + \sin^2 \theta)$$ Since $$\cos^2 \theta + \sin^2 \theta = 1$$, we obtain: $$\vec{v_1'} · \vec{v_2'} = x_1x_2 + y_1y_2$$ Hence, $$\vec{v_1'} · \vec{v_2'} = \vec{v_1} · \vec{v_2}$$, which implies that Tθ preserves angles between vectors. We have verified that the transformation Tθ preserves both lengths and angles, meaning it is an orthogonal transformation, as required.

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