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 $\sin^2 ( \theta ) + \cos^2 ( \theta ) = 1$.

The post The Rotation Matrix is an Orthogonal Transformation first appeared on Problems in Mathematics.

Let $\mathbf{v}$ be a vector in an inner product space $V$ over $\R$.
Suppose that $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is an orthonormal basis of $V$.
Let $\theta_i$ be the angle between $\mathbf{v}$ and $\mathbf{u}_i$ for $i=1,\dots, n$.

Prove that
\[\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1.\]

Definition (Angle between Vectors).

Let $\langle\mathbf{a}, \mathbf{b}\rangle$ denote the inner product of vectors $\mathbf{a}$ and $\mathbf{b}$ in $V$.

Recall that the angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is defined as the unique number $\theta$ between $0$ and $\pi$ satisfying
\[\cos \theta=\frac{\langle\mathbf{a}, \mathbf{b}\rangle}{\|\mathbf{a}\| \|\mathbf{b}\|}.\]

Proof.

Express the vector $\mathbf{v}$ as a linear combination of the basis vectors as
\[\mathbf{v}=a_1\mathbf{u}_1+\dots+a_n\mathbf{u}_n\] for some real numbers $a_1, \dots, a_n$.

The length of the vector $\mathbf{v}$ is given by
\[\|\mathbf{v}\|=\sqrt{a_1^2+\cdots+a_n^2}. \tag{*}\]

For each $i$, we have using the properties of the inner product
\begin{align*}
\langle \mathbf{v}, \mathbf{u}_i\rangle&=\langle a_1\mathbf{u}_1+\dots+a_n\mathbf{u}_n, \mathbf{u}_i\rangle\\
&=a_1\langle\mathbf{u}_1, \mathbf{u}_i\rangle+\cdots +a_n \langle\mathbf{u}_n, \mathbf{u}_i \rangle\\
&=a_i \tag{**}
\end{align*}
since $\langle\mathbf{u}_i, \mathbf{u}_i\rangle=1$ and $\langle\mathbf{u}_j, \mathbf{u}_i\rangle=0$ if $j\neq i$ as $\{\mathbf{u}_1, \dots, \mathbf{u}_n\}$ is orthonormal.

By definition of the angle, we have
\begin{align*}
\cos \theta_i&=\frac{\langle\mathbf{v}, \mathbf{u}_i\rangle}{\|\mathbf{v}\| \|\mathbf{u}_i\|}=\frac{\langle\mathbf{v}, \mathbf{u}_i\rangle}{\|\mathbf{v}\| } && \text{since $\|\mathbf{u}_i\|=1$.}
\end{align*}
It follows that
\begin{align*}
\cos ^2\theta_1+\cdots+\cos^2 \theta_n &=\frac{\langle\mathbf{v}, \mathbf{u}_1\rangle^2}{\|\mathbf{v}\|^2 }+\cdots+\frac{\langle\mathbf{v}, \mathbf{u}_n\rangle^2}{\|\mathbf{v}\|^2 }\\[6pt] &=\frac{1}{\|\mathbf{v}\|^2}(a_1^2+\cdots a_n^2) &&\text{by (**)}\\[6pt] &=\frac{1}{\|\mathbf{v}\|^2}\cdot \|\mathbf{v}\|^2 &&\text{by (*)}\\[6pt] &=1.
\end{align*}

Thus we obtain
\[\cos ^2\theta_1+\cdots+\cos^2 \theta_n=1\] as required.

The post The Sum of Cosine Squared in an Inner Product Space first appeared on Problems in Mathematics.