91Ó°ÊÓ

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\(A = \left( {\begin{aligned}{}{ - 5}&{}&{ - 5}\\5&{}&{ - 5}\end{aligned}} \right)\)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:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{}{ - 5}&{}&{ - 5}\\5&{}&{ - 5}\end{aligned}} \right) - \lambda \left( {\begin{aligned}{}1&0\\0&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 5 - \lambda }&{}&{ - 5}\\5&{}&{ - 5 - \lambda }\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{}{ - 5 - \lambda }&{}&{ - 5}\\5&{}&{ - 5 - \lambda }\end{aligned}} \right)\\ &= \left( { - 5 - \lambda } \right)\left( { - 5 - \lambda } \right) + 25\\ &= {\lambda ^2} - 10\lambda + 50\end{aligned}\)

So, the characteristic equation of the matrix \(A\) is \({\lambda ^2} - 10\lambda + 50 = 0\).

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - 10\lambda + 50 = 0\), 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 \({\lambda ^2} - 10\lambda + 50 = 0\) is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 10\lambda + 50 &= 0\\\lambda &= \frac{{ - \left( {10} \right) \pm \sqrt {{{\left( {10} \right)}^2} - 4\left( {50} \right)} }}{2}\\ &= \frac{{ - 10 \pm \sqrt { - 100} }}{2}\\ &= - 5 \pm 5i\end{aligned}\)

The eigenvalues of \(A\) are \(\lambda = - 5 \pm 5i\) .

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\(\lambda = - 5 \pm 5i\), \(\left( {a,b} \right) = \left( {5,5} \right)\). Find the scale factor \(r\) as follows:

\(\begin{aligned}{}r &= \left| \lambda \right|\\ &= \left| { - 5 \pm 5i} \right|\\ &= \sqrt {{{\left( { - 5} \right)}^2} + {{\left( 5 \right)}^2}} \\ &= 5\sqrt 2 \end{aligned}\)

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

\(\begin{aligned}{}\phi &= {\tan ^{ - 1}}\left( {\frac{5}{{ - 5}}} \right)\\ &= \frac{{3\pi }}{4}\,{\rm{radians}}\end{aligned}\)

Thus, the angle of rotation is \(\varphi = \frac{{3\pi }}{4}\,{\rm{radians}}\) and the scale factor \(r = 5\sqrt 2 \).