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.

16. \({\bf{4}}x_{\bf{1}}^{\bf{2}}{\bf{ + 4}}x_{\bf{2}}^{\bf{2}}{\bf{ + 4}}x_{\bf{3}}^{\bf{2}}{\bf{ + 4}}x_{\bf{4}}^{\bf{2}}{\bf{ + 8}}{x_{\bf{1}}}{x_{\bf{2}}}{\bf{ + 8}}{x_{\bf{3}}}{x_{\bf{4}}}{\bf{ - 6}}{x_{\bf{1}}}{x_{\bf{4}}}{\bf{ + 6}}{x_{\bf{2}}}{x_{\bf{3}}}\)

Consider \(4x_1^2 + 4x_2^2 + 4x_3^2 + 4x_4^2 + 8{x_1}{x_2} + 8{x_3}{x_4} - 6{x_1}{x_4} + 6{x_2}{x_3}\).

As the quadratic form is:

\({{\rm{x}}^T}A{\rm{x}} = \left( {\begin{aligned}{{}}{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}\end{aligned}} \right)\left( {\begin{aligned}{{}}4&4&0&{ - 3}\\4&4&3&0\\0&3&4&4\\{ - 3}&0&4&4\end{aligned}} \right)\left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{aligned}} \right)\)

Therefore, the coefficient matrix of the quadratic form is \(A = \left( {\begin{aligned}{{}}4&4&0&{ - 3}\\4&4&3&0\\0&3&4&4\\{ - 3}&0&4&4\end{aligned}} \right)\).

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

\(\begin{aligned}{}{\lambda _1} = 9,\\{\lambda _2} = 9,\\{\lambda _3} = - 1,\\{\lambda _4} = - 1\end{aligned}\)

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\(D = \left( {\begin{aligned}{{}}9&0&0&0\\0&9&0&0\\0&0&{ - 1}&0\\0&0&0&{ - 1}\end{aligned}} \right)\)

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

\(\begin{aligned}{}Q\left( {\rm{y}} \right) &= {{\rm{y}}^T}D{\rm{y}}\\ &= \left( {\begin{aligned}{{}}{{y_1}}&{{y_2}}&{{y_3}}&{{y_4}}\end{aligned}} \right)\left( {\begin{aligned}{{}}9&0&0&0\\0&9&0&0\\0&0&{ - 1}&0\\0&0&0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}}{{y_1}}\\{{y_2}}\\{{y_3}}\\{{y_4}}\end{aligned}} \right)\\ &= 9y_1^2 + 9y_2^2 - y_3^2 - y_4^2\end{aligned}\)

Thus, the new quadratic form is \(Q\left( {\rm{y}} \right) = 9y_1^2 + 9y_2^2 - y_3^2 - y_4^2\).