3: Constrained Extrema of Quadratic Forms
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- Oct 27, 2024
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( \newcommand{\kernel}{\mathrm{null}\,}\)
In this section it is convenient to write X=[x1x2⋮xn].
An eigenvalue of a square matrix A=[aij]ni,j=1 is a number λ such that the system
AX=λX, or, equivalently, (A−λI)X=0,
has a solution X≠0. Such a solution is called an eigenvector of A. You probably know from linear algebra that λ is an eigenvalue of A if and only if
det(A−λI)=0.
Henceforth we assume that A is symmetric (aij=aji,1≤i,j≤n). In this case,
det(A−λI)=(−1)n(λ−λ1)(λ−λ2)⋯(λ−λn),
where λ1,λ2,…,λn are real numbers.
The function Q(X)=n∑i,j=1aijxixj is a quadratic form. To find its maximum or minimum subject to n∑i=1x2i=1, we form the Lagrangian
L=Q(X)−λn∑i=1x2i.
Then
Lxi=2n∑j=1aijxj−2λxi=0,1≤i≤n,
so
n∑j=1aijxj0=λxi0,1≤i≤n.
Therefore, X0 is a constrained critical point of Q subject to n∑i=1x2i=1 if and only if AX0=λX0 for some λ; that is, if and only if λ is an eigenvalue and X0 is an associated unit eigenvector of A. If AX0=X0 and n∑ix2i0=1, then
Q(X0)=n∑i=1(n∑j=1aijxj0)xi0=n∑i=1(λxi0)xi0=λn∑i=1x2i0=λ;
therefore, the largest and smallest eigenvalues of A are the maximum and minimum values of Q subject to n∑i=1x2i=1.
Find the maximum and minimum values
Q(X)=x2+y2+2z2−2xy+4xz+4yz subject to the constraint x2+y2+z2=1.
Solution
The matrix of Q is
A=[1−12−112222]
and
det(A−λI)=|1−λ−12−11−λ2222−λ|=−(λ+2)(λ−2)(λ−4),
so
λ1=4,λ2=2,λ3=−2
are the eigenvalues of A. Hence, λ1=4 and λ3=−2 are the maximum and minimum values of Q subject to Equation ???.
To find the points (x1,y1,z1) where Q attains its constrained maximum, we first find an eigenvector of A corresponding to λ1=4. To do this, we find a nontrivial solution of the system
(A−4I)[x1y1z1]=[−3−1−2−1−3−2−2−2−2][x1y1z1]=[000].
All such solutions are multiples of [112]. Normalizing this to satisfy Equation ??? yields
X1=1√6[x1y1z1]=±[111].
To find the points (x3,y3,z3) where Q attains its constrained minimum, we first find an eigenvector of A corresponding to λ3=−2. To do this, we find a nontrivial solution of the system
(A+2I)[x3y3z3]=[3−1−2−13−2−2−24][x3y3z3]=[000].
All such solutions are multiples of [11−1]. Normalizing this to satisfy Equation ??? yields
X3=[x2y2z2]=±1√3[11−1].
As for the eigenvalue λ2=2, we leave it you to verify that the only unit vectors that satisfy AX2=2X2 are
X2=±1√2[11−1].
For more on this subject, see Theorem [theorem:4].
[example:6]