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.

17. \({\bf{11}}x_{\bf{1}}^{\bf{2}}{\bf{ + 11}}x_{\bf{2}}^{\bf{2}}{\bf{ + 11}}x_{\bf{3}}^{\bf{2}}{\bf{ + 11}}x_{\bf{4}}^{\bf{2}}{\bf{ + 16}}{x_{\bf{1}}}{x_{\bf{2}}}{\bf{ - 12}}{x_{\bf{1}}}{x_{\bf{4}}}{\bf{ + 12}}{x_{\bf{2}}}{x_{\bf{3}}}{\bf{ + 16}}{x_{\bf{3}}}{x_{\bf{4}}}\)

Consider \(11x_1^2 + 11x_2^2 + 11x_3^2 + 11x_4^2 + 16{x_1}{x_2} - 12{x_1}{x_4} + 12{x_2}{x_3} + 16{x_3}{x_4}\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{11}&8&0&{ - 6}\\8&{11}&6&0\\0&6&{11}&8\\{ - 6}&0&8&{11}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{aligned}} \right)\)

Therefore, the coefficient matrix of the quadratic form is \(A = \left( {\begin{aligned}{{}}{11}&8&0&{ - 6}\\8&{11}&6&0\\0&6&{11}&8\\{ - 6}&0&8&{11}\end{aligned}} \right)\).

Using the MATLAB command\({\rm{eigs}}\left( A \right)\), the eigenvalues are as:

\(\begin{aligned}{}{\lambda _1} = 21,\\{\lambda _2} = 21,\\{\lambda _3} = 1,\\{\lambda _4} = 1\end{aligned}\)

The eigenvectors are:

\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}4\\5\\3\\0\end{aligned}} \right)\), \({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{ - 5}\\{ - 4}\\0\\3\end{aligned}} \right)\), \({{\rm{v}}_3} = \left( {\begin{aligned}{{}}4\\{ - 5}\\3\\0\end{aligned}} \right)\), \({{\rm{v}}_4} = \left( {\begin{aligned}{{}}5\\{ - 4}\\0\\3\end{aligned}} \right)\)

Write the normalization of the vectors.

\({{\bf{u}}_1} = \left( {\begin{aligned}{{}}{\frac{{2\sqrt 2 }}{5}}\\{\frac{{\sqrt 2 }}{2}}\\{\frac{{3\sqrt 2 }}{{10}}}\\0\end{aligned}} \right)\),\({{\bf{u}}_2} = \left( {\begin{aligned}{{}}{ - \frac{{3\sqrt 2 }}{{10}}}\\0\\{\frac{{2\sqrt 2 }}{5}}\\{\frac{{\sqrt 2 }}{2}}\end{aligned}} \right)\),\({{\bf{u}}_3} = \left( {\begin{aligned}{{}}{\frac{{2\sqrt 2 }}{5}}\\{\frac{{ - \sqrt 2 }}{2}}\\{\frac{{3\sqrt 2 }}{{10}}}\\0\end{aligned}} \right)\),\({{\bf{u}}_4} = \left( {\begin{aligned}{{}}{\frac{{3\sqrt 2 }}{{10}}}\\0\\{ - \frac{{2\sqrt 2 }}{5}}\\{\frac{{\sqrt 2 }}{2}}\end{aligned}} \right)\)

Then the matrix\(P\)and\(D\)is shown below:

\(P = \left( {\begin{aligned}{{}}{\frac{{2\sqrt 2 }}{5}}&{ - \frac{{3\sqrt 2 }}{{10}}}&{\frac{{2\sqrt 2 }}{5}}&{\frac{{3\sqrt 2 }}{{10}}}\\{\frac{{\sqrt 2 }}{2}}&0&{ - \frac{{\sqrt 2 }}{2}}&0\\{\frac{{3\sqrt 2 }}{{10}}}&{\frac{{2\sqrt 2 }}{5}}&{\frac{{3\sqrt 2 }}{{10}}}&{ - \frac{{2\sqrt 2 }}{5}}\\0&{\frac{{\sqrt 2 }}{2}}&0&{\frac{{\sqrt 2 }}{2}}\end{aligned}} \right)\)and\(D = \left( {\begin{aligned}{{}}{21}&0&0&0\\0&{21}&0&0\\0&0&1&0\\0&0&0&1\end{aligned}} \right)\)

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

\(\begin{aligned}{}Q\left( {\rm{y}} \right) &= {{\rm{y}}^T}D{\rm{y}}\\ & = \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}&{{y_3}}&{{y_4}}\end{aligned}} \right)\left( {\begin{aligned}{{}}{21}&0&0&0\\0&{21}&0&0\\0&0&1&0\\0&0&0&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\\{{y_3}}\\{{y_4}}\end{aligned}} \right)\\ & = 21y_1^2 + 21y_2^2 + y_3^2 + y_4^2\end{aligned}\)

Thus, the new quadratic form is \(Q\left( {\rm{y}} \right) = 21y_1^2 + 21y_2^2 + y_3^2 + y_4^2\).