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

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

Short answer

The required values are:

1. The maximum value of\(Q\left( {\rm{x}} \right)\)the subject to the constraint\({{\rm{x}}^T}{\rm{x}} = 1\)is\({\lambda _1} = 7\).
2. A unit vector\({\rm{u}}\)where this maximum is attained is\({\rm{u}} = \pm \left[ {\begin{array}{*{20}{c}}{{1 \mathord{\left/
3. {\vphantom {1 {\sqrt 3 }}} \right.
4. \kern-\nulldelimiterspace} {\sqrt 3 }}}\\{{1 \mathord{\left/
5. {\vphantom {1 {\sqrt 3 }}} \right.
6. \kern-\nulldelimiterspace} {\sqrt 3 }}}\\{{1 \mathord{\left/
7. {\vphantom {1 {\sqrt 3 }}} \right.
8. \kern-\nulldelimiterspace} {\sqrt 3 }}}\end{array}} \right]\).
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\({\lambda _2} = 4\).

Step-by-step solution

Find the greatest eigenvalue

As per the question, we have:

\(Q\left( {\rm{x}} \right) = 3x_1^2 + 3x_2^2 + 5x_3^2 + 6x_1^{}x_2^{} + 2x_1^{}x_3^{} + 2x_2^{}x_3^{}\)

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

So, from exercise 2, the eigenvalues are 7, 4, and 0. This implies that the greatest eigenvalue is 7.

\({\lambda _1} = 7\)

Hence, the maximum value of \(Q\left( {\rm{x}} \right)\) the subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\) is \({\lambda _1} = 7\).

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:

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

{\vphantom {1 {\sqrt 3 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 3 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 3 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 3 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 3 }}} \right.

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

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

{\vphantom {1 {\sqrt 3 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 3 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 3 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 3 }}}\\{{1 \mathord{\left/

{\vphantom {1 {\sqrt 3 }}} \right.

\kern-\nulldelimiterspace} {\sqrt 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:

\({\lambda _2} = 4\)

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