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 $\R^n$ be an inner product space with inner product $\langle \mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{\trans}\mathbf{y}$ for $\mathbf{x}, \mathbf{y}\in \R^n$.

A linear transformation $T:\R^n \to \R^n$ is called orthogonal transformation if for all $\mathbf{x}, \mathbf{y}\in \R^n$, it satisfies
\[\langle T(\mathbf{x}), T(\mathbf{y})\rangle=\langle\mathbf{x}, \mathbf{y} \rangle.\]

Prove that if $T:\R^n\to \R^n$ is an orthogonal transformation, then $T$ is an isomorphism.

 

We give two proofs.
The second one uses a fact about the injectivity of linear transformations.

Proof 1.

As $T$ is a linear transformation from $\R^n$ to itself, it suffices to show that $T$ is an injective linear transformation.

Suppose that $T(\mathbf{x})=T(\mathbf{y})$ for $\mathbf{x}, \mathbf{y}\in \R^n$.
We show that $\mathbf{x}=\mathbf{y}$.

We have
\begin{align*}
&\|\mathbf{x}-\mathbf{y}\|^2\\
&=(\mathbf{x}-\mathbf{y})^{\trans}(\mathbf{x}-\mathbf{y})\\
&=(\mathbf{x}^{\trans}-\mathbf{y}^{\trans})(\mathbf{x}-\mathbf{y})\\
&=\mathbf{x}^{\trans}\mathbf{x}-\mathbf{x}^{\trans}\mathbf{y}-\mathbf{y}^{\trans}\mathbf{x}+\mathbf{y}^{\trans}\mathbf{y}\\
&=\langle\mathbf{x}, \mathbf{x} \rangle-\langle\mathbf{x}, \mathbf{y} \rangle-\langle\mathbf{y}, \mathbf{x} \rangle+\langle\mathbf{y}, \mathbf{y} \rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{y}) \rangle-\langle T(\mathbf{y}), T(\mathbf{x}) \rangle+\langle T(\mathbf{y}), T(\mathbf{y}) \rangle\\
&\text{(since $T$ is an orthogonal transformation)}\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle+\langle T(\mathbf{x}), T(\mathbf{x}) \rangle\\
&\text{(since $T(\mathbf{x})=T(\mathbf{y})$)}\\
&=0.
\end{align*}
It follows that $\|\mathbf{x}-\mathbf{y}\|=0$ and hence $\mathbf{x}=\mathbf{y}$.
This proves that $T:\R^n\to \R^n$ is injective.

As $T$ is an injective linear transformation from the $n$-dimensional vector space $\R^n$ to itself, it is also surjective, and thus $T$ is an isomorphism.

Proof 2.

Recall that the linear transformation $T$ is injective if and only if the null space $\calN(T)=\{\mathbf{0}\}$, that is, $T(\mathbf{x})=\mathbf{0}$ implies that $\mathbf{x}=\mathbf{0}$.
(See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for the proof of this fact.)

We use this fact to show that $T$ is injective.
Suppose that $T(\mathbf{x})=\mathbf{0}$.
Then we have
\begin{align*}
\|\mathbf{x}\|^2&=\langle \mathbf{x}, \mathbf{x}\rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x})\rangle &&\text{as $T$ is orthogonal}\\
&=\langle \mathbf{0}, \mathbf{0}\rangle=0 &&\text{as $T(\mathbf{x})=\mathbf{0}$}.
\end{align*}

It follows that the length $\|\mathbf{x}\|=0$, and hence $\mathbf{x}=\mathbf{0}$.
This proves that the null space $\calN(T)=\{\mathbf{0}\}$ and $T$ is injective.

As $T$ is an injective linear transformation from $\R^n$ to itself, it is also surjective, and hence $T$ is an isomorphism.

The post An Orthogonal Transformation from $\R^n$ to $\R^n$ is an Isomorphism first appeared on Problems in Mathematics.