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.

Consider the $2\times 2$ matrix
\[A=\begin{bmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta \end{bmatrix},\] where $\theta$ is a real number $0\leq \theta < 2\pi$.

(a) Find the characteristic polynomial of the matrix $A$.

(b) Find the eigenvalues of the matrix $A$.

(c) Determine the eigenvectors corresponding to each of the eigenvalues of $A$.

Proof.

(a) Find the characteristic polynomial of the matrix $A$.

The characteristic polynomial $p(t)$ of $A$ is computed as follows.
Let $I$ be the $2\times 2$ identity matrix.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
\cos \theta -t & -\sin \theta\\
\sin \theta& \cos \theta -t
\end{vmatrix}\\
&=(\cos \theta -t)^2+\sin^2 \theta \tag{*}\\
&=t^2-(2\cos \theta) t+\cos^2 \theta+\sin^2 \theta\\
&=t^2-(2\cos \theta) t+1
\end{align*}
by the trigonometry identity $\cos^2 \theta+\sin^2 \theta=1$.

Hence the characteristic polynomial of $A$ is
\[p(t)=t^2-(2\cos \theta) t+1.\]

(b) Find the eigenvalues of the matrix $A$.

The eigenvalues of $A$ are roots of the characteristic polynomial $p(t)$.
So let us solve
\[p(t)=t^2-(2\cos \theta) t+1=0.\] By the quadratic formula, we have
\begin{align*}
t&=\frac{2\cos \theta \pm \sqrt{(2\cos \theta)^2-4}}{2}\\[6pt] &=\cos \theta \pm \sqrt{\cos^2 \theta -1}\\
&=\cos \theta \pm \sqrt{-\sin^2 \theta}\\
&=\cos \theta \pm i \sin \theta =e^{\pm i\theta}.
\end{align*}

Hence eigenvalues of $A$ are
\[\cos \theta \pm i \sin \theta =e^{\pm i\theta}.\]

Alternatively, we could have used the equation (*).
Then we have
\begin{align*}
&p(t)=(\cos \theta -t)^2+\sin^2 \theta=0\\
&\Leftrightarrow (\cos \theta-t)^2=-\sin^2 \theta\\
&\Leftrightarrow \cos \theta -t=\pm i \sin \theta
\end{align*}
and thus, we obtain the eigenvalues
\[\cos \theta \pm i \sin \theta =e^{\pm i\theta}.\]

(c) Determine the eigenvectors

Let us first find the eigenvectors corresponding to the eigenvalue $\lambda=\cos \theta +i \sin \theta$.
We have
\begin{align*}
A-\lambda I=\begin{bmatrix}
-i \sin \theta & -\sin \theta\\
\sin \theta& -i \sin \theta
\end{bmatrix}.
\end{align*}

If $\theta=0, \pi$, then $\sin \theta=0$ and we have
\[A-\lambda I=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}\] and thus each nonzero vector of $\R^2$ is an eigenvector.

If $\lambda \neq 0, \pi$, then $\sin \theta \neq 0$.
Thus we have by elementary row operations
\begin{align*}
A-\lambda I=\begin{bmatrix}
-i \sin \theta & -\sin \theta\\
\sin \theta& -i \sin \theta
\end{bmatrix}
\xrightarrow[\frac{1}{\sin \theta} R_2]{\frac{i}{\sin \theta} R_1}
\begin{bmatrix}
1 & -i\\
1& -i
\end{bmatrix}
\xrightarrow{R_2-R_1} \begin{bmatrix}
1 & -i\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors for $\lambda$ are
\[\begin{bmatrix}
i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.

The other eigenvalue is the conjugate $\bar{\lambda}$ of $\lambda$.
Since $A$ is a real matrix, the eigenvectors of $\bar{\lambda}$ are the conjugate of those of $\lambda$.
Hence the eigenvectors corresponding to $\bar{\lambda}$ is
\[\begin{bmatrix}
-i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.

In summary, when $\theta=0, \pi$, the eigenvalues are $1, -1$, respectively, and every nonzero vector of $\R^2$ is an eigenvector.

If $\theta \neq 0, \pi$, then the eigenvectors corresponding to the eigenvalue $\cos \theta +i\sin \theta$ are
\[\begin{bmatrix}
i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.
The eigenvectors corresponding to the eigenvalue $\cos \theta -i\sin \theta$ are
\[\begin{bmatrix}
-i \\
1
\end{bmatrix}t,\] for any nonzero scalar $t$.

Related Question.

For a similar proble, try the following.

The solution is given in the post ↴
Rotation Matrix in Space and its Determinant and Eigenvalues

The post Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors first appeared on Problems in Mathematics.

For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by
\[A=\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}.\]

(a) Find the determinant of the matrix $A$.

(b) Show that $A$ is an orthogonal matrix.

(c) Find the eigenvalues of $A$.

Solution.

(a) The determinant of the matrix $A$

By the cofactor expansion corresponding to the third row, we compute
\begin{align*}
\det(A)&=\begin{vmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{vmatrix}\\
&=0\cdot \begin{vmatrix}
-\sin \theta & 0\\
\cos \theta& 0
\end{vmatrix}-0\cdot \begin{vmatrix}
\cos \theta & 0\\
\sin \theta& 0
\end{vmatrix}+1\cdot \begin{vmatrix}
\cos \theta & -\sin \theta\\
\sin \theta& \cos \theta
\end{vmatrix}\\
&=\cos^2 \theta +\sin^2 \theta\\
&=1.
\end{align*}
The last step follows from the famous trigonometry identity
\[\cos^2 \theta +\sin^2 \theta=1.\] Thus we have
\[\det(A)=1.\]

(b) The matrix $A$ is an orthogonal matrix

We give two solutions for part (b).

The first solution of (b)

The first solution computes $A^{\trans}A$ and show that it is the identity matrix $I$.
We have
\begin{align*}
A^{\trans}A&=\begin{bmatrix}
\cos\theta & \sin\theta & 0 \\
-\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta &\cos\theta &0 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
\cos^2 \theta +\sin^2\theta & 0 & 0 \\
0 &\cos^2 \theta+\sin^2 \theta &0 \\
0 & 0 & 1
\end{bmatrix}\\
&=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}=I.
\end{align*}
Similarly, you can check that $AA^{\trans}=I$. Thus $A$ is an orthogonal matrix.

The second solution of (b)

The second proof uses the following fact: a matrix is orthogonal if and only its column vectors form an orthonormal set.
Let
\[A_1=\begin{bmatrix}
\cos \theta \\
\sin \theta \\
0
\end{bmatrix}, A_2=\begin{bmatrix}
-\sin\theta \\
\cos \theta \\
0
\end{bmatrix}, A_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\] be the column vectors of the matrix $A$. The length of these vectors are all $1$. For example, we have
\begin{align*}
||A_1||=\sqrt{(\cos\theta)^2+(\sin \theta)^2+0^2}=\sqrt{1}=1.
\end{align*}
Similarly, we have $||A_2||=||A_3||=1$.
The dot (inner) product of $A_1$ and $A_2$ is
\begin{align*}
A_1\cdot A_2=\cos \theta \cdot (-\sin \theta)+\sin \theta \cdot \cos \theta +0\cdot 0=0.
\end{align*}
Similarly, we have $A_1\cdot A_3=A_2\cdot A_3=0$.
Therefore, the column vectors $A_1, A_2, A_3$ are orthonormal vectors. Hence by the above fact, the matrix $A$ is orthogonal.

(c) The eigenvalues of $A$

We compute the characteristic polynomial $p(t)=\det(A-tI)$ as follows.
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
\cos\theta-t & -\sin\theta & 0 \\
\sin\theta &\cos\theta -t&0 \\
0 & 0 & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
\cos \theta -t & -\sin \theta\\
\sin \theta& \cos \theta-t
\end{vmatrix} \text{ by the third row cofactor expansion}\\
&=(1-t)(\cos^2 \theta -2t \cos \theta +t^2 +\sin^2 \theta)\\
&=(1-t)(t^2-(2\cos \theta)t+1).
\end{align*}

The eigenvalues are roots of the characteristic polynomial $p(t)$, hence we solve
\[p(t)=(1-t)(t^2-(2\cos \theta)t+1)=0.\] One solution is $t=1$. The other solutions come from the quadratic polynomial in $p(t)$.
By the quadratic formula, those solutions are
\begin{align*}
t&=\cos\theta \pm\sqrt{\cos^2 \theta -1}\\
&=\cos\theta \pm \sqrt{-\sin^2 \theta}\\
&=\cos \theta \pm i \sin \theta
\end{align*}
since $\sin \theta\geq 0$ since $0 \leq \theta \leq \pi$.
Therefore the eigenvalues of the matrix $A$ are
\[1, \cos \theta \pm i \sin \theta.\]

Related Question.

The following problem treats the rotation matrix in the plane.

The solution is given in the post ↴
Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors

The post Rotation Matrix in Space and its Determinant and Eigenvalues first appeared on Problems in Mathematics.