In Exercises 7-12, use Example 6 to list the eigenvalues of \({\bf{A}}\). In each case, the transformation \({\bf{x}} \mapsto A{\bf{x}}\) is the composition of a rotation and a scaling. Give the angle \(\varphi \) of the rotation, where \( - \pi < \varphi < \pi \), and give the scale factor \(r\).

\(\left( {\begin{aligned}0&{}&{.3}\\{ - .3}&{}&0\end{aligned}} \right)\)

Let \(A\) be a matrix that has a complex eigenvalue \(\lambda \) of the form \(\lambda = a + bi\).

Then the formula for the scaling factor \(r\) is given by,

\(\begin{aligned}{}r &= \left| \lambda \right|\\ \Rightarrow r &= \sqrt {{a^2} + {b^2}} \end{aligned}\).

And the angle of rotation \(\varphi \) will be \(\varphi = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)\).

Given that\(A = \left( {\begin{aligned}{}0&{}&{.3}\\{ - .3}&{}&0\end{aligned}} \right)\).

Comparing this matrix to the matrix \(C\) from example 6 we have,

\(a = 0,b = - 0.3, - b = 0.3\).

Hence using the example \(6\) the eigenvalues of the matrix \(A\) are \(\lambda = 0 \pm 0.3i\).

Find the scaling factor for this matrix by using the formula \(r = \sqrt {{a^2} + {b^2}} \).

\(\begin{aligned}{}r &= \left| \lambda \right|\\r &= \sqrt {0 + {{\left( {0.3} \right)}^2}} \\r &= \sqrt {{{\left( {0.3} \right)}^2}} \\r &= 0.3\end{aligned}\)

The scaling factor is \(r = 0.3\).

Find the angle of rotation by using the formula \(\varphi = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)\).

\(\begin{aligned}{}\varphi &= {\tan ^{ - 1}}\left( {\frac{{ - 0.3}}{0}} \right)\\\varphi &= {\tan ^{ - 1}}\left( { - \infty } \right)\\\varphi &= - \frac{\pi }{2}\end{aligned}\)

Hence the angle of rotation is \(\varphi = - \frac{\pi }{2}\).