3: Constrained Extrema of Quadratic Forms

In this section it is convenient to write \[{\bf X}= \left[\begin{array}{ccccccc} x_{1}\\x_{2}\\\vdots\\x_{n} \end{array}\right]. \nonumber \]

An eigenvalue of a square matrix \(\mathbf{A} = [a_{ij}]_{i,j=1}^{n}\) is a number \(\lambda\) such that the system

\[\mathbf{A}\mathbf{X} = \lambda \mathbf{X}, \nonumber \] or, equivalently, \[(\mathbf{A}-\lambda \mathbf{I})\mathbf{X}=\mathbf{0}, \nonumber \]

has a solution \(\mathbf{X} \ne \mathbf{0}\). Such a solution is called an eigenvector of \(\mathbf{A}\). You probably know from linear algebra that \(\lambda\) is an eigenvalue of \(\mathbf{A}\) if and only if

\[\det(\mathbf{A} -\lambda \mathbf{I}) = {0}. \nonumber \]

Henceforth we assume that \(\mathbf{A}\) is symmetric \((a_{ij} = a_{ji}, 1 \le i, j \le n)\). In this case,

\[\det(\mathbf{A}-\lambda \mathbf{I}) = (-1)^{n}(\lambda-\lambda_{1})(\lambda-\lambda_2) \cdots (\lambda-\lambda_{n}), \nonumber \]

where \(\lambda_{1},\lambda_2,\dots,\lambda_{n}\) are real numbers.

The function \[Q(\mathbf{X}) = \sum^{n}_{i,j=1} a_{ij} x_{i}x_{j} \nonumber \] is a quadratic form. To find its maximum or minimum subject to \(\displaystyle{\sum^{n}_{i=1} x^{2}_{i}=1}\), we form the Lagrangian

\[L=Q(\mathbf{X}) - \lambda \sum^{n}_{i=1}x^{2}_{i}. \nonumber \]

Then

\[L_{x_{i}}= 2 \sum^{n}_{j=1} a_{ij}x_{j} - 2\lambda x_{i}=0, \quad 1 \le i \le n, \nonumber \]

so

\[\sum_{j=1}^{n}a_{ij}x_{j0}=\lambda x_{i0},\quad 1\le i\le n. \nonumber \]

Therefore, \(\mathbf{X_{0}}\) is a constrained critical point of \(Q\) subject to \(\displaystyle{\sum^{n}_{i=1} x^{2}_{i}=1}\) if and only if  \({\mathbf A}{\mathbf X}_{0}=\lambda{\mathbf X}_{0}\) for some \(\lambda\); that is, if and only if \(\lambda\) is an eigenvalue and \(\mathbf{X}_{0}\) is an associated unit eigenvector of \(\mathbf{A}\). If \({\mathbf A}\mathbf{X}_{0}={\bf X}_{0}\) and \(\displaystyle{\sum_{i}^{n}x_{i0}^{2}}=1\), then

\[\begin{aligned} Q(\mathbf{X}_{0}) & =& \sum^{n}_{i=1} \left(\sum^{n}_{j=1} a_{ij}x_{j0} \right) x_{i0} = \sum^{n}_{i=1} (\lambda x_{i0})x_{i0} \\ & =& \lambda \sum^{n}_{i=1} x^{2}_{i0} = \lambda;\end{aligned} \nonumber \]

therefore, the largest and smallest eigenvalues of \({\bf A}\) are the maximum and minimum values of \(Q\) subject to \(\displaystyle{\sum_{i=1}^{n}x_{i}^{2}}=1\).

Example \(\PageIndex{1}\)

Nov 2, 2019, 9:12 PM

[example:6]

Find the maximum and minimum values

\[Q(\mathbf{X}) = x^{2}+y^{2}+2z^{2}-2xy + 4xz + 4yz \nonumber \] subject to the constraint \[\label{eq:18} x^{2}+y^{2}+z^{2}=1. \]

Solution

The matrix of \(Q\) is

\[\mathbf{A} = \left[\begin{array}{rrrrr} 1 & -1 & 2 \\ -1 & 1 & 2 \\ 2&2&2 \end{array}\right] \nonumber \]

and

\[\begin{aligned} \det(\mathbf{A} - \lambda \mathbf{I}) & =& \left|\begin{array}{ccccccc} 1-\lambda &-1 & 2 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \end{array}\right| \\ & =& -(\lambda+2)(\lambda-2)(\lambda-4),\end{aligned} \nonumber \]

so

\[\lambda_{1}=4, \quad \lambda_2=2, \quad \lambda_3=-2 \nonumber \]

are the eigenvalues of \(\mathbf{A}\). Hence, \(\lambda_{1}=4\) and \(\lambda_3 = -2\) are the maximum and minimum values of \(Q\) subject to Equation \ref{eq:18}.

To find the points \((x_{1},y_{1},z_{1})\) where \(Q\) attains its constrained maximum, we first find an eigenvector of \({\bf A}\) corresponding to \(\lambda_{1}=4\). To do this, we find a nontrivial solution of the system

\[(\mathbf{A}-4\mathbf{I}) \left[\begin{array}{ccccccc} x_{1}\\ y_{1}\\ z_{1} \end{array}\right]= \left[\begin{array}{ccccccc} -3 & -1 & \phantom{-}2 \\ -1 & -3 & \phantom{-}2 \\ \phantom{-}2 & \phantom{-}2 & -2 \end{array}\right] \left[\begin{array}{ccccccc} x_{1}\\y_{1}\\z_{1} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0 \end{array}\right]. \nonumber \]

All such solutions are multiples of \(\left[\begin{array}{ccccccc} 1\\1\\2 \end{array}\right].\) Normalizing this to satisfy Equation \ref{eq:18} yields

\[{\bf X}_{1}=\frac{1}{\sqrt6} \left[\begin{array}{ccccccc} x_{1}\\y_{1}\\z_{1} \end{array}\right]=\pm \left[\begin{array}{ccccccc} 1\\ 1\\1 \end{array}\right]. \nonumber \]

To find the points \((x_{3},y_{3},z_{3})\) where \(Q\) attains its constrained minimum, we first find an eigenvector of \({\bf A}\) corresponding to \(\lambda_{3}=-2\). To do this, we find a nontrivial solution of the system

\[(\mathbf{A}+2\mathbf{I}) \left[\begin{array}{ccccccc} x_{3}\\ y_{3}\\ z_{3} \end{array}\right]= \left[\begin{array}{rrrcccc} 3 & -1 & \phantom{-}2 \\ -1 & 3 & \phantom{-}2 \\ \phantom{-}2 & \phantom{-}2 & 4 \end{array}\right] \left[\begin{array}{ccccccc} x_{3}\\y_{3}\\z_{3} \end{array}\right]= \left[\begin{array}{ccccccc} 0\\0\\0 \end{array}\right]. \nonumber \]

All such solutions are multiples of \(\left[\begin{array}{rcccccc} 1\\1\\-1 \end{array}\right].\) Normalizing this to satisfy Equation \ref{eq:18} yields

\[{\bf X}_{3}= \left[\begin{array}{ccccccc} x_{2}\\y_{2}\\z_{2} \end{array}\right]=\pm \frac{1}{\sqrt{3}} \left[\begin{array}{rcccccc} 1\\ 1\\-1 \end{array}\right]. \nonumber \]

As for the eigenvalue \(\lambda_{2}=2\), we leave it you to verify that the only unit vectors that satisfy \({\bf A}{\bf X}_{2}=2{\bf X}_{2}\) are

\[{\bf X}_{2}=\pm \frac{1}{\sqrt{2}} \left[\begin{array}{rcccccc} 1\\ 1\\-1 \end{array}\right]. \nonumber \]

For more on this subject, see Theorem [theorem:4].