In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

5. \(Q\left( x \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\).

As per the question, we have:

\(Q\left( {\rm{x}} \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\)

The matrix with its eigenvalues has the form:

\(\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&{ - 5}\\{ - 5}&1\end{array}} \right]\\ \Rightarrow {\lambda _1} = 6\;\;\& \,\,\,{\lambda _2} = - 4\end{array}\)

The maximum value of the given function subjected to constraints\({{\rm{x}}^T}{\rm{x}} = 1\)will be the greatest eigenvalue.

So, we have:

\({\lambda _1} = 6\)

Hence,the maximum value of\(Q\left( {\rm{x}} \right)\)the subject to the constraint\({{\rm{x}}^T}{\rm{x}} = 1\)is\({\lambda _1} = 6\).

Apply the theorem which states that the value of\({{\rm{x}}^T}A{\rm{x}}\) is maximum when \({\rm{x}}\) is a unit eigenvector \({{\rm{u}}_1}\) corresponding to the greatest eigenvalue \({\lambda _1}\).

Find the eigenvector corresponding to \({\lambda _1} = 6\).

\(\begin{array}{c}A - 6I = \left[ {\begin{array}{*{20}{c}}{1 - 6}&{ - 5}\\{ - 5}&{1 - 6}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 5}&{ - 5}\\{ - 5}&{ - 5}\end{array}} \right]\end{array}\)

Theorem the null space is \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right]\).

The eigenvector that corresponds to eigenvalue\({\lambda _1} = 6\)is:

\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{{ - 1} \mathord{\left/

{\vphantom {{ - 1} {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\end{array}} \right]\).

Hence,a unit vector\({\rm{u}}\)where this maximum is attained is\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{{ - 1} \mathord{\left/

{\vphantom {{ - 1} {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 2 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 2 }}}\end{array}} \right]\).

The maximum value of the given function is subjected to constraints\({{\rm{x}}^T}{\rm{x}} = 1\)and\({{\rm{x}}^T}{\rm{u}} = 0\)will be the second-largest eigenvalue. That is:

\({\lambda _2} = - 4\)

Hence,the maximum of\(Q\left( {\rm{x}} \right)\)subject to the constraints\({{\rm{x}}^T}{\rm{x}} = 1\)and \({{\rm{x}}^T}{\rm{u}} = 0\) is\({\lambda _2} = - 4\).