Question

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 \(r\).

7. \(\left( {\begin{aligned}{}{\sqrt 3 }&{}&{ - 1}\\1&{}&{\,\sqrt 3 }\end{aligned}} \right)\)

Short answer

The angle of rotation is \(\varphi = \frac{\pi }{6}\,{\rm{radians}}\) and the scale factor \(r = 2\).

Step-by-step solution

Find the characteristic equation

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

It is given that\(A = \left( {\begin{aligned}{}{\sqrt 3 }&{}&{ - 1}\\1&{}&{\,\,\sqrt 3 }\end{aligned}} \right)\)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:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{}{\sqrt 3 }&{}&{ - 1}\\1&{}&{\,\,\sqrt 3 }\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&{}&0\\0&{}&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{\sqrt 3 - \lambda }&{}&{ - 1}\\1&{}&{\sqrt 3 - \lambda }\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{\sqrt 3 - \lambda }&{}&{ - 1}\\1&{}&{\sqrt 3 - \lambda }\end{aligned}} \right)\\ &= \left( {\sqrt 3 - \lambda } \right)\left( {\sqrt 3 - \lambda } \right) + 1\\ &= {\lambda ^2} - 2\sqrt 3 \lambda + 4\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - 2\sqrt 3 \lambda + 4 = 0\).

Find the Eigenvalues

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - 2\sqrt 3 \lambda + 4 = 0\), 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 \({\lambda ^2} - 2\sqrt 3 \lambda + 4 = 0\) is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 2\sqrt 3 \lambda + 4 &= 0\\\lambda &= \frac{{ - \left( { - 2\sqrt 3 } \right) \pm \sqrt {{{\left( { - 2\sqrt 3 } \right)}^2} - 4\left( 4 \right)} }}{2}\\ &= \frac{{2\sqrt 3 \pm \sqrt { - 4} }}{2}\\ &= \sqrt 3 \pm i\end{aligned}\)

The eigenvalues of \(A\) are \(\lambda = \sqrt 3 + i\) and \(\lambda = \sqrt 3 - i\).

Find the angle of rotation and scale factor

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 \(\lambda = \sqrt 3 \pm i\), \(\left( {a,b} \right) = \left( {\sqrt 3 ,\,1} \right)\). Find the scale factor \(r\) as follows:

\(\begin{aligned}{}r &= \left| \lambda \right|\\ &= \left| {\sqrt 3 \pm i} \right|\\ &= \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {1^2}} \\ &= 2\end{aligned}\)

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

\(\begin{aligned}{}\phi &= {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)\\ &= \frac{\pi }{6}\,{\rm{radians}}\end{aligned}\)

Thus, the angle of rotation is \(\varphi = \frac{\pi }{6}\,{\rm{radians}}\) and the scale factor \(r = 2\).