Let \(Q\left( x \right) = 7x_1^2 + x_2^2 + 7x_3^2 - 8x_1^{}x_2^{} - 4x_1^{}x_3^{} - 8x_2^{}x_3^{}\). Find a unit vector \(x{\rm{ in }}{\mathbb{R}^3}\) at which \(Q\left( x \right)\) is maximized, subject to \({x^T}x = 1\).

[Hint: The eigenvalues of the matrix of the quadratic form \(Q\) are \(9,{\rm{ and }} - 3\).]

As per the question, we have:

\(Q\left( {\rm{x}} \right) = 7x_1^2 + x_2^2 + 7x_3^2 - 8x_1^{}x_2^{} - 4x_1^{}x_3^{} - 8x_2^{}x_3^{}\)

And the eigenvalues as:

\(\begin{array}{l}{\lambda _1} = 9\\{\lambda _2} = - 3\end{array}\)

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

So, we have:

\({\lambda _1} = 9\)

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 \({\lambda _1}\).

The eigenvector corresponding to \({\lambda _1} = 9\)are a linear combination of\(\left[ {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right]\)and \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right]\)

The unit eigenvector that corresponds to the eigenvalue\({\lambda _1} = 9\)is:

\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right]\).

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 \({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right]\).