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# 3: Constrained Extrema of Quadratic Forms

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

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

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

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

$\det(\mathbf{A} -\lambda \mathbf{I}) = {0}.$

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

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

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

The function $Q(\mathbf{X}) = \sum^{n}_{i,j=1} a_{ij} x_{i}x_{j}$ 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

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

Then

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

so

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

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}

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 $$\PageIndex{1}$$

Find the maximum and minimum values

$Q(\mathbf{X}) = x^{2}+y^{2}+2z^{2}-2xy + 4xz + 4yz$ 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}

so

$\lambda_{1}=4, \quad \lambda_2=2, \quad \lambda_3=-2$

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].