Question

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

3.\(Q\left( x \right) = 5x_1^2 + 6x_2^2 + 7x_3^2 + 4x_1^{}x_2^{} - 4x_2^{}x_3^{}\).

Short answer

1. The maximum value of \(Q\left( {\rm{x}} \right)\) the subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\) is \(9\).
2. A unit vector \({\rm{u}}\) where this maximum is attained is \({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{{ - 1} \mathord{\left/
3. {\vphantom {{ - 1} 3}} \right.
4. \kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/
5. {\vphantom {{ - 2} 3}} \right.
6. \kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/
7. {\vphantom {2 3}} \right.
8. \kern-\nulldelimiterspace} 3}}\end{array}} \right]\).
9. 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 \({\lambda _2} = 6\).

Step-by-step solution

Find the greatest eigenvalue

As per the question, we have:

\(Q\left( {\rm{x}} \right) = 5x_1^2 + 6x_2^2 + 7x_3^2 + 4x_1^{}x_2^{} - 4x_2^{}x_3^{}\)

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

So, from exercise 1, the eigenvalues are 9, 6, and 3. This implies that the greatest eigenvalue is 9.

\({\lambda _1} = 9\)

Hence, the maximum value of \(Q\left( {\rm{x}} \right)\) the subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\) is \(9\).

Find the vector for this greatest eigenvalue

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 that corresponds to this eigenvalue is:

\(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,a unit vector \({\rm{u}}\) where this maximum is attained 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]\).

Find the second greatest eigenvalue. 

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

\({\lambda _2} = 6\)

Hence, the maximum of \(Q\left( {\rm{x}} \right)\) the subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1\) and \({{\rm{x}}^T}{\rm{u}} = 0\) is \({\lambda _2} = 6\).