Let $\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $ \langle \mathbf{v} , \mathbf{w} \rangle = \mathbf{v}^\trans \mathbf{w}$. A linear transformation $T : \R^2 \rightarrow \R^2$ is called an orthogonal transformation if for all $\mathbf{v} , \mathbf{w} \in \R^2$,
\[\langle T(\mathbf{v}) , T(\mathbf{w}) \rangle = \langle \mathbf{v} , \mathbf{w} \rangle.\]

For a fixed angle $\theta \in [0, 2 \pi )$ , define the matrix
\[ [T] = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \] and the linear transformation $T : \R^2 \rightarrow \R^2$ by
\[T( \mathbf{v} ) = [T] \mathbf{v}.\]

Prove that $T$ is an orthogonal transformation.

Solution.

Suppose we have vectors $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ and $\mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} $ . Then,
\[T(\mathbf{v}) = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) v_1 – \sin (\theta) v_2 \\ \sin(\theta) v_1 + \cos (\theta) v_2 \end{bmatrix},\] and
\[ T(\mathbf{w}) = \begin{bmatrix} \cos (\theta) & – \sin ( \theta ) \\ \sin ( \theta ) & \cos ( \theta ) \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} \cos(\theta) w_1 – \sin (\theta) w_2 \\ \sin(\theta) w_1 + \cos (\theta) w_2 \end{bmatrix}.\]

Then we find the inner product for these two vectors:
\begin{align*}
&\langle T(\mathbf{v} ) , T( \mathbf{w} ) \rangle \\
&= \begin{bmatrix} \cos(\theta) v_1 – \sin (\theta) v_2 & \sin(\theta) v_1 + \cos (\theta) v_2 \end{bmatrix} \begin{bmatrix} \cos(\theta) w_1 – \sin (\theta) w_2 \\ \sin(\theta) w_1 + \cos (\theta) w_2 \end{bmatrix} \\[6pt] &= \biggl( \cos(\theta) v_1 – \sin(\theta) v_2 \biggr) \biggl( \cos(\theta) w_1 – \sin ( \theta) w_2 \biggr) \\[6pt] &\qquad + \biggl( \sin (\theta) v_1 + \cos (\theta) v_2 \biggr) \biggl( \sin (\theta) w_1 + \cos(\theta) w_2 \biggr) \\[6pt] &= \cos^2(\theta) ( v_1 w_1 + v_2 w_2 ) + \sin(\theta) \cos(\theta) ( – v_1 w_2 – v_2 w_1 + v_1 w_2 + v_2 w_1 ) \\ &\qquad + \sin^2 (\theta) ( v_2 w_2 + v_1 w_1 ) \\[6pt] &= \left( \cos^2 ( \theta) + \sin^2 ( \theta ) \right) ( v_1 w_1 + v_2 w_2 ) \\
&= v_1 w_1 + v_2 w_2 \\
&= \langle \mathbf{v} , \mathbf{w} \rangle .
\end{align*}

This proves that $T$ is an orthogonal transformation. For the second-to-last equality, we used the Pythagorean identity $\sin^2 ( \theta ) + \cos^2 ( \theta ) = 1$.

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