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 $A$.

 
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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

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