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

11.\(\left( {\begin{aligned}{}{\,\,\,.1}&{}&{.1}\\{ - .1}&{}&{.1}\end{aligned}} \right)\)

If \(A\) is a \(n \times n\) matrix, then \(det\left( {A - \lambda I} \right) = 0\), is called the characteristic equation of a matrix \(A\).

It is given that\(A = \left( {\begin{aligned}{}{\,\,\,.1}&{}&{.1}\\{ - .1}&{}&{.1}\end{aligned}} \right)\)and\(I = \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\)is the identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{}{\,\,\,.1}&{}&{.1}\\{ - .1}&{}&{.1}\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{.1 - \lambda }&{}&{.1}\\{ - .1}&{}&{.1 - \lambda }\end{aligned}} \right)\end{aligned}\)

Now calculate the determinant of the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{.1 - \lambda }&{}&{.1}\\{ - .1}&{}&{.1 - \lambda }\end{aligned}} \right)\\ &= \left( {.1 - \lambda } \right)\left( {.1 - \lambda } \right) + .01\\ &= {\lambda ^2} - .2\lambda + .02\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - .2\lambda + .02 = 0\).

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - .2\lambda + .02 = 0\), as follows:

For the quadratic equation, \(a{x^2} + bx + c = 0\) , the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\) .

Thus, the solution of the characteristic equation \({\lambda ^2} - .2\lambda + .02 = 0\) is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - .2\lambda + .02 &= 0\\\lambda &= \frac{{ - \left( { - .2} \right) \pm \sqrt {{{\left( { - .2} \right)}^2} - 4\left( {0.2} \right)} }}{2}\\ &= \frac{{.2 \pm \sqrt { - .04} }}{2}\\ &= .1 \pm .1i\end{aligned}\)

The eigenvalues of \(A\) are \(\lambda = .1 \pm .1i\) .

For the Eigenvalue,\({\lambda _i} = a \pm bi\), the scale factor is\(r = \left| \lambda \right|\)and the angle of rotation is\(\varphi = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)\).

For the Eigenvalue\(\lambda = .1 \pm .1i\), \(\left( {a,b} \right) = \left( {.1,.1} \right)\). Find the scale factor \(r\) as follows:

\(\begin{aligned}{}r &= \left| \lambda \right|\\ &= \left| {.1 \pm .1i} \right|\\ &= \sqrt {{{\left( {.1} \right)}^2} + {{\left( {.1} \right)}^2}} \\ &= .1\left( {\sqrt 2 } \right)\\ &= \frac{{\sqrt 2 }}{{10}}\end{aligned}\)

Find the angle of rotation \(\varphi \)as follows:

\(\begin{aligned}{}\phi &= {\tan ^{ - 1}}\left( {\frac{{.1}}{{.1}}} \right)\\ &= \frac{\pi }{4}\,{\rm{radians}}\end{aligned}\)

Thus, the angle of rotation is \(\varphi = \frac{\pi }{4}\,{\rm{radians}}\) and the scale factor \(r = \frac{{\sqrt 2 }}{{10}}\).