A Matrix Problem: Problem 82-6Author(s): C. G. BroydenSource: SIAM Review, Vol. 25, No. 3 (Jul., 1983), p. 405Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2029394 .

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PROBLEMS AND SOLUTIONS 405

A Matrix Problem

Problem 82-6, by C. G. BROYDEN (University of Essex, Colchester, England). Let X be an m x n matrix with no null columns and with qi _ 1 nonzero elements in

its ith row, 1 _ i _ m. Let Q = diag (qi), and D be the diagonal matrix whose diagonal elements are equal to the corresponding diagonal elements of XTX. Determine the conditions under which the matrix D - XTQ- 'Xis

(a) positive definite, (b) positive semidefinite. This problem arose in connection with an algorithm for scaling examination marks.

Solution by the proposer. Denote the ith row of X by xT and define the row vector sT to be that vector for which

si1 = 0 if xi1 = 0 and si1 = 1 if xij = 0. Let Y be a diagonal matrix of order n and define the row vector uT by

( 1 ) ~~~~~~~uT = xT Y(I - q-'S Sj T)-

Note that I - q- I SiST is idempotent. It follows that

(2) uT u1 = xT Y(I - qy's1siS) Yxi = x,Y2xi - qy1 (x Ysi) 2.

Since xi1 = 0 whenever si] = 0, it follows that

(3) x,iYs1 = x[ Ye, where e = (1, 1, . . ., 1 )T. Letting y = Ye, a straightforward calculation yields

m in

(4) Z uT'ui = E {xqY2xi - q7'(xTy)2} = yT (D - XTQX)y. i=l ~ ~ ~ i=l Consequently, D - XTQ-' Xis positive semidefinite for all X.

Let (a,, a2, . . *, am)" and (bl, b2, . . ., bn be vectors having no zero components and choose each xij to be either 0 or else aibj. Then setting Y = diag (b,-' ),we find

(5) x,Y= aiS T

so that (I) yields

(6) u'= 0T, i= 1,* ,m.

Thus, in this case D - XTQ-IX is not positive definite. In general, D - XTQ-IX is positive definite unless there exists a nonzero diagonal matrix Yand a sequence of scalars al, a2, . . ., am such that (5) holds for i = 1, 2, * *, m.

Also solved by B. SCHAFFRIN (Institut fur Theoretische Geodasie, Universitat Bonn, Germany).

On the Expected Value of Two Statistics

Problem 82-13, by K. A. RAO (State University of New York at Binghamton). If

R2= max (Y,, Y2) - min (Y,, Y2),

R3= max (Y,, Y2, Y3) - min (Y,, Y2, Y3),

where Y1, Y2, and Y3 are cyclically exchangeable random variables, prove that

2E(R3) = 3E(R2).

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