[M] In Exercise 14-17, follow the instructions given for Exercise 3-6.

14. \(3{{\bf{x}}_1}{{\bf{x}}_2} + 5{{\bf{x}}_1}{{\bf{x}}_3} + 7{{\bf{x}}_1}{{\bf{x}}_4} + 7{{\bf{x}}_2}{{\bf{x}}_3} + 5{{\bf{x}}_2}{{\bf{x}}_4} + 3{{\bf{x}}_3}{{\bf{x}}_4}\)

Obtain the quadratic form of the matrix as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}0&{\frac{3}{2}}&{\frac{5}{2}}&{\frac{7}{2}}\\{\frac{3}{2}}&0&{\frac{7}{2}}&{\frac{5}{2}}\\{\frac{5}{2}}&{\frac{7}{2}}&0&{\frac{3}{2}}\\{\frac{7}{2}}&{\frac{5}{2}}&{\frac{3}{2}}&0\end{array}} \right]\)

Use MATLAB code to obtain the eigenvalues of the matrix as shown below:

\(\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left[ {0\,\,\,\frac{3}{2}\,\,\,\,\,\frac{5}{2}\,\,\,\,\frac{7}{2};\,\frac{3}{2}\,\,\,\,0\,\,\,\,\frac{7}{2}\,\,\,\,\,\frac{5}{2};\,\frac{5}{2}\,\,\,\,\frac{7}{2}\,\,\,\,0\,\,\,\frac{3}{2};\,\frac{7}{2}\,\,\,\frac{5}{2}\,\,\,\,\frac{3}{2}\,\,\,\,\,0} \right];\\ > > {\mathop{\rm E}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right);\end{array}\)

\({\mathop{\rm E}\nolimits} = \left[ {\begin{array}{*{20}{c}}{\frac{{15}}{2}}\\{ - \frac{1}{2}}\\{ - \frac{5}{2}}\\{ - \frac{5}{2}}\end{array}} \right]\)

Therefore, the eigenvalues of the matrix \(A\) are \({\lambda _1} = \frac{{15}}{2},{\lambda _2} = - \frac{1}{2},{\lambda _3} = - \frac{5}{2},\) and \({\lambda _4} = - \frac{9}{2}\).

Theorem 6states that consider \(A\) as a symmetric matrixand \(m = \min \left\{ {{{\bf{x}}^T}A{\bf{x}}:\left\| {\bf{x}} \right\| = 1} \right\},M = \max \left\{ {{{\bf{x}}^T}A{\bf{x}}:\left\| {\bf{x}} \right\| = 1} \right\}\). Then the greatest eigenvalue \({\lambda _1}\) of A is \(M\) and the least eigenvalue of \(A\) is \(m\). \(M\) is the value of \({{\bf{x}}^T}A{\bf{x}}\) if the unit eigenvalue of \({{\bf{u}}_1}\) is \({\bf{x}}\) that corresponds to \(M\). \(m\) is the value of \({{\bf{x}}^T}A{\bf{x}}\) if the unit eigenvalue of \({{\bf{u}}_1}\) is \({\bf{x}}\) that corresponds to \(m\).

It is observed that the greatest eigenvalue of \(A\) is \({\lambda _1} = \frac{{15}}{2}\).

According to theorem 6, the greatest eigenvalues \({\lambda _1} = \frac{{15}}{2}\) is the maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}A{\bf{x}} = 1\).

According to theorem 6, the maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}A{\bf{x}} = 1\) happens at a unit eigenvector that corresponds to the greatest eigenvalue \({\lambda _1}\).

Use the MATLAB code to compute the eigenvector of the matrix A as shown below:

\( > > \left[ {{\mathop{\rm V}\nolimits} \,\,{\mathop{\rm D}\nolimits} } \right] = {\mathop{\rm eigs}\nolimits} \left( A \right);\)

\({\mathop{\rm V}\nolimits} = \left[ {\begin{array}{*{20}{c}}1&1&{ - 1}&{ - 1}\\1&{ - 1}&1&{ - 1}\\1&{ - 1}&{ - 1}&1\\1&1&1&1\end{array}} \right]\)

Therefore, the eigenvector corresponds to the greatest eigenvalues \({\lambda _1} = \frac{{15}}{2}\) is \(\left[ {\begin{array}{*{20}{c}}1\\1\\1\\1\end{array}} \right]\).

Normalize the vector to obtain the unit vector as shown below:

\(\begin{array}{c}{\bf{u}} = \frac{1}{{\sqrt 4 }}\left[ {\begin{array}{*{20}{c}}1\\1\\1\\1\end{array}} \right]\\ = \pm \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\end{array}} \right]\end{array}\)

Thus, the unit vector in which the maximum is attained is \({\bf{u}} = \pm \left[ {\begin{array}{*{20}{c}}{\frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\\{\frac{1}{2}}\end{array}} \right]\).

Theorem 7states that consider that \(A,{\lambda _1}\), and \({{\bf{u}}_1}\) as in theorem 6. The maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint

\({{\bf{x}}^T}{\bf{x}} = 1\)and \({{\bf{x}}^T}{\bf{u}} = 0\)

Is equal to the second greatest eigenvalue, \({\lambda _2}\) and the maximum is attained if the eigenvector \({{\bf{u}}_2}\) that corresponds to \({\lambda _2}\) is x.

It is observed that the second greatest eigenvalue of A is \({\lambda _2} = - \frac{1}{2}\).

According to theorem 7, the second greatest eigenvalue \({\lambda _2} = - \frac{1}{2}\) isthe maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}{\bf{x}} = 1\) and \({{\bf{x}}^T}{\bf{u}} = 0\).

Thus, \({\lambda _2} = - \frac{1}{2}\) is the maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}{\bf{x}} = 1\) and \({{\bf{x}}^T}{\bf{u}} = 0\).