Carry through the details of Example 2 to find the unit eigenvectors. Show that the resulting rotation matrix C is orthogonal. Hint: Find${\mathbf{\text{CC}}}^{\mathbf{\text{T}}}$.

$M=\left(\begin{array}{ccc}1& 3& 0\\ 3& -2& -1\\ 0& -1& 1\end{array}\right)$

A rotation matrix can be defined as a transformation matrix that operates on a vector and produces a rotated vector, such that the coordinate axes always remain fixed.

The value of $\lambda =1,-4,3$are known as the eigenvalues of matrix M.

Therefore, the eigenvectors corresponding to these eigenvalues$\lambda =1,-4,3$ can be mathematically presented as$\left(\frac{1}{\sqrt{10}},0,\frac{3}{\sqrt{10}}\right)\left(\frac{-3}{\sqrt{35}},\frac{5}{\sqrt{35}},\frac{1}{\sqrt{35}}\right)\left(\frac{-3}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{1}{\sqrt{14}}\right)$ respectively.

Hence, the rotation matrix can be written as shown below.

$C=\left(\begin{array}{ccc}\frac{1}{\sqrt{10}}& \frac{-3}{\sqrt{35}}& \frac{-3}{\sqrt{14}}\\ 0& \frac{5}{\sqrt{35}}& \frac{-2}{\sqrt{14}}\\ \frac{3}{\sqrt{10}}& \frac{1}{\sqrt{35}}& \frac{1}{\sqrt{14}}\end{array}\right)$

The transpose of matrix can be mathematically presented as.

$C^T=\left(\begin{array}{ccc}\frac{1}{\sqrt{10}}& 0& \frac{3}{\sqrt{10}}\\ \frac{-3}{\sqrt{35}}& \frac{5}{\sqrt{35}}& \frac{1}{\sqrt{35}}\\ \frac{-3}{\sqrt{14}}& \frac{-2}{\sqrt{14}}& \frac{1}{\sqrt{14}}\end{array}\right)$

The multiplication of matrix C and the transpose of it are evaluated as shown below.

$C{C}^T=\left(\begin{array}{ccc}\frac{70}{70}& 0& 0\\ 0& \frac{70}{70}& 0\\ 0& 0& \frac{70}{70}\end{array}\right)$

Hence, $C{C}^T=I$

Therefore, C is an orthogonal matrix.