91Ó°ÊÓ

Consider \(2x_1^2 - 4{x_1}{x_2} - x_2^2\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}2&{ - 2}\\{ - 2}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right)\)

Therefore, thecoefficient matrix of the quadratic form is\(A = \left( {\begin{aligned}{{}}2&{ - 2}\\{ - 2}&{ - 1}\end{aligned}} \right)\).

Now Find the characteristic polynomial.

\(\begin{aligned}{}\left| {\begin{aligned}{{}}{2 - \lambda }&{ - 2}\\{ - 2}&{ - 1 - \lambda }\end{aligned}} \right| & = 0\\\left( {2 - \lambda } \right)\left( { - 1 - \lambda } \right) - 4 = 0\\\left( {2 - \lambda } \right)\left( { - 1 - \lambda } \right) & = 4\\\lambda & = - 2,3\end{aligned}\)

\(\begin{aligned}{}\left( {A - \lambda I} \right){{\rm{u}}_1} & = 0\\\left( {\begin{aligned}{{}}4&{ - 2}\\{ - 2}&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = 0\\4{x_1} - 2{x_2} & = 0\\ - 2{x_1} + {x_2} & = 0\end{aligned}\)

Thus, the eigen vector is \({{\rm{u}}_1} = \left( {\begin{aligned}{{}}1\\2\end{aligned}} \right)\).

\(\begin{aligned}{}\left( {A - \lambda I} \right){{\rm{u}}_2} & = 0\\\left( {\begin{aligned}{{}}{ - 1}&{ - 2}\\{ - 2}&{ - 4}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) & = 0\\ - {x_1} - 2{x_2} & = 0\\ - 2{x_1} - 4{x_2} & = 0\end{aligned}\)

Thus, the eigenvector is \({{\rm{u}}_2} = \left( {\begin{aligned}{{}}{ - 2}\\1\end{aligned}} \right)\).

\(\begin{aligned}{}{P_1} & = \frac{1}{{\left\| {{{\rm{u}}_1}} \right\|}}{{\rm{u}}_1}\\ & = \frac{1}{{\sqrt {1 + 4} }}\left( {\begin{aligned}{{}}1\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\end{aligned}} \right)\end{aligned}\)

\(\begin{aligned}{}{P_1} &= \frac{1}{{\left\| {{{\rm{u}}_2}} \right\|}}{{\rm{u}}_2}\\ & = \frac{1}{{\sqrt {1 + 4} }}\left( {\begin{aligned}{{}}{ - 2}\\1\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{\frac{{ - 2}}{{\sqrt 5 }}}\\{\frac{1}{{\sqrt 5 }}}\end{aligned}} \right)\end{aligned}\)

Therefore, the matrix\(P\)and\(D\)are shown below:

\(\begin{aligned}{}P & = \left( {\begin{aligned}{{}}{{P_1}}&{{P_2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 5 }}}&{\frac{{ - 2}}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}}\end{aligned}} \right)\end{aligned}\)

\(D = \left( {\begin{aligned}{{}}{ - 2}&0\\0&3\end{aligned}} \right)\)

Now write the new quadratic form by using the transformation\({\rm{x}} = P{\rm{y}}\).

\(\begin{aligned}{}Q\left( {\rm{x}} \right) & = {{\rm{x}}^T}A{\rm{x}}\\ & = {\left( {P{\rm{y}}} \right)^T}A\left( {P{\rm{y}}} \right)\\ & = {{\rm{y}}^T}{P^T}AP{\rm{y}}\\ & = {{\rm{y}}^T}D{\rm{y}}\\ & = Q\left( {\rm{y}} \right)\end{aligned}\)

Therefore,

\(\begin{aligned}{}{{\rm{y}}^T}D{\rm{y}} & = \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{ - 2}&0\\0&3\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}}{ - 2{y_1}}&{3{y_2}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\end{aligned}} \right)\\ & = - 2y_1^2 + 3y_2^2\end{aligned}\)

Thus, the new quadratic form is \(Q\left( {\rm{y}} \right) = - 2y_1^2 + 3y_2^2\).