Classify the quadratic forms in Exercises 9–18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct \(P\) using the methods of Section 7.1.

12.\({\bf{ - }}x_{\bf{1}}^{\bf{2}}{\bf{ - 2}}{x_{\bf{1}}}{x_{\bf{2}}} - x_{\bf{2}}^{\bf{2}}\)

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

As the quadratic form is:

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

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

Now Find the characteristic polynomial.

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

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

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

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

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

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

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

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

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

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

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,

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

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