Question

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

16. \( - 6{\bf{x}}_1^2 - 10{\bf{x}}_2^2 - 13{\bf{x}}_3^2 - 13{\bf{x}}_4^2 - 4{{\bf{x}}_1}{{\bf{x}}_2} - 4{{\bf{x}}_1}{{\bf{x}}_3} - 4{{\bf{x}}_1}{{\bf{x}}_4} + 6{{\bf{x}}_3}{{\bf{x}}_4}\)

Short answer

1. \({\lambda _1} = - 4\)is the maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}A{\bf{x}} = 1\).
2. The unit vector in which the maximum is attained is \({\bf{u}} = \pm \left[ {\begin{array}{*{20}{c}}{\frac{{ - 3}}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\end{array}} \right]\).
3. \({\lambda _2} = - 10\)isthe maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}{\bf{x}} = 1\) and \({{\bf{x}}^T}{\bf{u}} = 0\).

Step-by-step solution

Determine the eigenvalues of the matrix

Obtain the quadratic form of the matrix as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}{ - 6}&{ - 2}&{ - 2}&{ - 2}\\{ - 2}&{ - 10}&0&0\\{ - 2}&0&{ - 13}&3\\{ - 2}&0&3&{ - 13}\end{array}} \right]\)

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

\(\begin{array}{l} > > {\mathop{\rm A}\nolimits} = \left[ { - 6\,\,\, - 2\,\,\,\, - 2\,\,\,\, - 2;\,\, - 2\,\,\,\, - 10\,\,\,\,0\,\,\,\,0;\, - 2\,\,\,\,0\,\,\, - 13\,\,\,\,3;\, - 2\,\,\,\,0\,\,\,\,3\,\,\,\, - 13} \right];\\ > > {\mathop{\rm E}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right);\end{array}\)

\({\mathop{\rm E}\nolimits} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{ - 10}\\{ - 12}\\{ - 16}\end{array}} \right]\)

Therefore, the eigenvalues of the matrix \(A\) are \({\lambda _1} = - 4,{\lambda _2} = - 10,{\lambda _3} = - 12,\) and \({\lambda _4} = - 16\).

Determine the maximum value of \(Q\left( x \right)\) subject to the constraint \({{\bf{x}}^T}{\mathop{\rm x}\nolimits}  = 1\) 

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 \({\lambda _1} = - 4\).

According to theorem 6, the greatest eigenvalues \({\lambda _1} = - 4\).is the maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}A{\bf{x}} = 1\).

Determine a unit vector u where this maximum is attained

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:

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

\({\mathop{\rm V}\nolimits} = \left[ {\begin{array}{*{20}{c}}{ - 3}&0&1&0\\1&{ - 2}&1&0\\1&1&1&{ - 1}\\1&1&1&1\end{array}} \right]\)

Therefore, the eigenvector corresponds to the greatest eigenvalues \({\lambda _1} = - 4\) is \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\1\\1\end{array}} \right]\).

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

\(\begin{array}{c}{\bf{u}} = \frac{1}{{\sqrt {12} }}\left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\1\\1\end{array}} \right]\\ = \pm \left[ {\begin{array}{*{20}{c}}{\frac{{ - 3}}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\end{array}} \right]\end{array}\)

Therefore, the unit vector in which the maximum is attained is \({\bf{u}} = \pm \left[ {\begin{array}{*{20}{c}}{\frac{{ - 3}}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\\{\frac{1}{{\sqrt {12} }}}\end{array}} \right]\).

Determine a maximum value of \(Q\left( x \right)\) subject to the constraints \({{\bf{x}}^T}{\mathop{\rm x}\nolimits}  = 1\) and \({{\bf{x}}^T}{\bf{u}} = 0\)

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 \({\lambda _2} = - 10\).

According to theorem 7, the second greatest eigenvalue \({\lambda _2} = - 10\) isthe maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}{\bf{x}} = 1\) and \({{\bf{x}}^T}{\bf{u}} = 0\).

Therefore, \({\lambda _2} = - 10\) is the maximum value of \({{\bf{x}}^T}A{\bf{x}}\) subject to the constraint \({{\bf{x}}^T}{\bf{x}} = 1\) and \({{\bf{x}}^T}{\bf{u}} = 0\).