Problem 82-6, A Matrix ProblemAuthor(s): C. G. BroydenSource: SIAM Review, Vol. 24, No. 2 (Apr., 1982), p. 223Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2029361 .

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SIAM REVIEW ? 1982 Society for Industrial and Applied Mathematics Vol. 24, No. 2, April 1982 0036-1445/82/2402-0005 $01.00/0

PROBLEMS AND SOLUTIONS

EDITED BY MURRAY S. KLAMKIN

COLLABORATING EDITORS: HENRY E. FETTIS CECIL C. ROUSSEAU YUDELL L. LUKE OTTO G. RUEHR

All problems and solution should be sent, typewritten in duplicate, to Murray S. Klamkin, Department of Mathematics, University of Alberta, Edmonton, Alberta, Canada T6G 2G1. An asterisk placed beside a problem number indicates that the problem was submitted without solution. Proposers and solvers whose solutions are published will receive 10 reprints of the corresponding problem section. Other solvers will receive just one reprint provided a self-addressed stamped envelope is enclosed. Proposers and solvers desiring acknowledgment of their contributions should include a self-addressed stamped postcard (no stamps necessary for outside the U.S.A. and Canada). Solutions should be received by July 15, 1982.

PROBLEMS

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-lXiS

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

A Triple Sum

Problem 82-7, by P. E. MERILEES (International Meteorological Institute, Stockholm). Given that Imnp = 1 if m = n + p, = 0 otherwise, show that

M M M

TIjk- Z Z Z exp {i(mnXI-nXj-pXk)}Imnp m=-M n=-M p=-M

sin (2M?+ )'Osin (2M?+ 1)0" sin 0' sin 0"

where

-j 1 Jsin [(2M +? )0+ MO'] sin (M + 1)0'

Qrjk - sin 0 sin G' sin [(2M + 1)0 + MO"] sin (M + 1)0"

sin 0" J and 20 =X - Xk, 20' = ,- Xk, 20" = A,- X. Show further that

2sin (2M ?1) (- X)sin (2M + 1) (X ; j)sin (2M + 1) (X -Xk>a

2Ijk r sin XI) sin ( 2 ) sin (X Xk)

The problem arose in the theory of spectral models of atmospheric flow. 223

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