# Rotation Matrix in Space and its Determinant and Eigenvalues

## Problem 218

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 $A$.

Add to solve later

Sponsored Links

Contents

## 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.

**Problem**.

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 $A$.

The solution is given in the post ↴

Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors

Add to solve later

Sponsored Links

## 1 Response

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