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A Matrix Eigenvalue Problem: Problem 79-2Author(s): G. Efroymson, A. Steger and S. SteinbergSource: SIAM Review, Vol. 22, No. 1 (Jan., 1980), pp. 99-100Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2029883 .
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PROBLEMS AND SOLUTIONS 99
A Dice Problem
Problem 80-5* bv M. S. KIAMKIN and A. flu (University of Alberta). Given tn identical polyhedral dice whose corresponding faces are numbered
identically with arbitrary integers; (a) Prove or disprove that if the dice are tossed at random, the probability that the
sum of the bottom n face numbers is divisible by n is at least 1/2"'. (b) Determine the maximum probability for the previous sum being equal to
k(mod n) for k = 1, 2, n - 1. In (a), the special case for n = 3 was set by the first author as a problem in the 1979
U.S.A. Mathematical Olympiad.
SOLUTIONS
A Matrix Eigenvalue Problem
Problem 79-2 by G. EFROYMSON, A. STEGER and S. STEINBERG (University of New Mexico). Let M, denote the n x n matrix whose (j, k) entry M,A (j, k) is given by
j-1)(k --1)/ln 1 j, k n,
where Cl = e2r/ n. Determine all the eigenvalues of Mn. The matrix Mn arises in some work on finite Fourier transforms.
Solution by T. SEKIGUCHI and N. KIMURA (University of Arkansas). Let c1, c2, , cn be the eigenvalues of the matrix Ml,. Then cl, C2, ,c, are the
eigenvalues of M2. A simple calculation gives M2 = (a1k ), where a,k = 1 if j + k = 2 or j + k = n +2, and a,k = 0 otherwise. The eigenvalues of M2 are 1 and -1, with multiplicities [(n + 2)/2] and [(n - 1)/2] respectively. Thus each Ck is one of 1, -1, i, -i. Let Ln (c) be the multiplicity of eigenvalue c of Mn. Then
Ln ()+L, 1-1)=[(n +2)/2]
and
L, (i) +L(-i) = [(n - 1)/21.
On the other hand, the trace T,, of Mn is
T, = L, (1) -L, (- 1) + (L,, (i)- Ln W )i,
and it is computed directly from M,, as follows:
! I n T,,= Y to (k-- 1 )2
k-1
However, this is the so-called Gaussian sum, and it is known that
l +i if n-0 1 if n 1
T,, = (1 + i)(1 + i-F)/2 } if-2 (mod 4).
i if n 3
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100 PROBLEMS AND SOLUTIONS
So we can count Ln(c) for these four c's as the following table shows:
nl 4k+ 1 4k+2 4k+3 4k+4
Ln(1) k +1 k+1 k+1 k+2
Ln(-1) k k + 1 k + 1 k + 1
Ln(i) k k k+1 k+1
Ln(-i) k k k k
Also solved by C. GIVENs and 0. RUEHR (Michigan Technological University), M. H. GUTKNECHT (Eidgen6ssische Technische Hochschule, Zurich, Switzerland), W. B. JORDAN (Scotia, NY), 0. P. LoSSERS (Eindhoven University, Eindhoven, the Netherlands), P. J. NIKOLAI (U.S. Air Force Flight Dynamics Laboratory, Wright- Patterson AFB, Ohio), K. WANG (SUNY at Buffalo), and the proposers.
A Determinant
Problem 79-3* by A. E. BARKAUSKAS and D. W. BANGE (University of Wisconsin- La Crosse). Find either a closed form solution or a simple recurrence to evaluate the n x n
determinant la1,, where aj = aj, a, = c + 1 (c an integer> 1), a12 = 1, a,C+k =1 for k = 1 to c and ci + k - n; all other a,, =0.
The problem arose in counting the spanning trees of a certain class of outerplanar graphs.
A composite of solutions by WILLIAM B. JORDAN (Scotia, N.Y.) and E. L. ALLGOWER (Colorado State University).
For notational convenience, we generalize the problem slightly to evaluating the determinant of the symmetric n x n matrix Anc) (a1,. , an) whose entries are zero except for a,, a, 12 =1 and a, Cl k= 1 for all those i1, ,n and k = 1, , c,that satisfy ci + k ? n. Solution of this problem rests upon the following observations.
1. The determinant is easy for n < c since A"1(aj, . , an) is then block diagonal, with blocks
[a, 61 and diag (a3, an
2. For n -> c, the division algorithm gives n = cn1 + r1, with 0- r1 < c. Thus, ci + k > n for i > n 1 allows the partitioned form
(*) Anc(a, , agn) = [1 l () A*c) ( a, an,) E]
with
D diag (aEnl,1 + an
3. The staggered positioning of ones implied by the defining relation a,cl+k = 1 makes the rows of E mutually orthogonal. Thus, ED- 1E' is diagonal and consequently
nA nlC) - ED,-'Et = A
( C)e in) {
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