91Ó°ÊÓ

As per the question, we have:

\(Q\left( {\rm{x}} \right) = - 2x_1^2 - x_2^2 + 4x_1^{}x_2^{} + 4x_2^{}x_3^{}\)

And the eigenvalues as:

\(\begin{array}{l}{\lambda _1} = 2\\{\lambda _2} = - 1\\{\lambda _3} = - 4\end{array}\)

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

So, we have:

\({\lambda _1} = 2\)

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 \({\lambda _1} = 2\).

\({\rm{v}} = \left[ {\begin{array}{*{20}{c}}{1/2}\\1\\1\end{array}} \right]\)

The unit eigenvector that corresponds to the eigenvalue\({\lambda _1} = 2\)is:

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

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{array}} \right]\).

Hence, aunit vector \({\rm{x}}\) in \({\mathbb{R}^3}\) at which \(Q\left( {\rm{x}} \right)\) is maximized, subject to \({{\rm{x}}^T}{\rm{x}} = 1\) is \({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{array}} \right]\).