Question

Show that the trace of a rotation matrix equals $\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{+}\mathbf{1}$ where θ is the rotation angle, and the trace of a reflection matrix equals $\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{-}\mathbf{1}$. Hint: See equations (7.18) and (7.19), and Problem 10.

Short answer

The trace of rotation matrix is $Tr\text{A}=2\cos \theta +1$, and the trace of a reflection matrix is $TrB=2\cos \theta -1$.

$TrB=2\cos \theta -1$

Step-by-step solution

Given Information

$$\begin{gathered} A=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right) \\ B=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& -1\end{array}\right) \end{gathered}$$

Diagonal matrix

A Diagonal Matrix is a square matrix in which all elements are zero except the principal diagonal element.

Diagonal Matrix

The trace by the sum of the elements on the diagonals is $TrA=2\cos \theta +1$and $Tr\text{B}=2\cos \theta -1$. This can be verified by noting that the trace of a matrix equals the sum of its eigenvalues. The eigenvalues of A is obtained as shown below.

$TrA=2\cos \theta +1Tr\text{B}=2\cos \theta -1$

$$\begin{gathered} =\left|\begin{array}{ccc}\cos \theta -\lambda & -\sin \theta & 0\\ \sin \theta & \cos \theta -\lambda & 0\\ 0& 0& 1-\lambda \end{array}\right| \\ ={(\cos \theta -\lambda )}^2(1-\lambda )+{\sin }^2\theta (1-\lambda ) \\ =(1-\lambda )(1+{\lambda }^2-2.\cos \theta .x)=0 \end{gathered}$$

One eigenvalue is 1 , and the other two can be obtained from the quadratic equation $\cos \theta \pm i\sin \theta$. The sum of the eigenvalues is equal to $2\cos \theta +1$, as expected. For matrix $B$, the only thing that is different is the element $b_{33}$, and the characteristic equation can be changed as $(-1-\lambda )\left(1+{\lambda }^2-2·\cos \theta ·\lambda \right)=0$. The first eigenvalue changes to $-1$, while the other two remains the same. Therefore, the sum of the eigenvalues will be $2\cos \theta -1$, as expected.

$(-1-\lambda )\left(1+{\lambda }^2-2·\cos \theta ·\lambda \right)=0$

-1$2\cos \theta -1$