234 V Hadamard’s Determinant Problem
by measuring each separately. Suppose, for definiteness, that we havemobjects
whose weights are to be determined and we performn≥mweighings. The whole
experiment may be represented by ann×mmatrixA=(αjk).Ifthek-th object is
not involved in thej-th weighing, thenαjk =0; if it is involved, thenαjk=+ 1
or−1 according as it is placed in the left-hand or right-hand pan of the balance. The
individual weightsξ 1 ,...,ξmare connected with the observed resultsη 1 ,...,ηnof
the weighings by the system of linear equations
y=Ax, (1)wherex=(ξ 1 ,...,ξm)t∈Rmandy=(η 1 ,...,ηn)t∈Rn.
We will again denote by‖y‖the Euclidean norm(|η 1 |^2 +···+|ηn|^2 )^1 /^2 of the
vectory.Letx ̄∈Rmhave as its coordinates the correct weights and lety ̄=Ax ̄. If,
because of errors of measurement,yranges over the ball‖y− ̄y‖≤ρinRn,then
xranges over the ellipsoid(x− ̄x)tAtA(x− ̄x)≤ρ^2 inRm. Since the volume of
the ellipsoid is [det(AtA)]−^1 /^2 times the volume of the ball, we may regard the best
choice of the design matrixAto be that for which the ellipsoid has minimum volume.
Thus we are led to the problem of maximizing det(AtA)among alln×mmatrices
A=(αjk)withαjk∈{ 0 ,− 1 , 1 }.
A different approach to the best choice of design matrix leads (by§2) to a similar
result. Ifn>mthe linear system (1) is overdetermined. However, the least squares
estimate for the solution of (1) is
x=Cy,whereC=(AtA)−^1 At.Letak∈Rnbe thek-th column ofAand letck∈Rnbe the
k-th row ofC.SinceCA=Im,wehaveckak=1. Ifyranges over the ball‖y− ̄y‖≤ρ
inRn,thenξkranges over the real interval|ξk−ξ ̄k|≤ρ‖ck‖. Thus we may regard the
optimal choice of the design matrixAfor measuringξkto be that for which‖ck‖is a
minimum.
By Schwarz’s inequality (Chapter I,§4),
‖ck‖‖ak‖≥ 1 ,with equality only ifctkis a scalar multiple ofak.Also‖ak‖≤n^1 /^2 , since all elements
ofAhave absolute value at most 1. Hence‖ck‖≥n−^1 /^2 , with equality if and only if all
elements ofakhave absolute value 1 andckt=ak/n. It follows that the design matrixA
is optimal for measuringeachofξ 1 ,...,ξmif all elements ofAhave absolute value 1
andAtA=nIm. Moreover, in this case the least squares estimate for the solution
of (1) is simplyx=Aty/n. Thus the individual weights are easily determined from
the observed measurements by additions and subtractions, followed by a division byn.
Suppose, for example, thatm=3andn=4. If we take