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IMC / 1994 / Problems / Day 1, P1

IMC 1994 · Day 1 · P1

linear algebraworth 13 pts

a) Let AA be a n×nn \times n, n≥2n \ge 2, symmetric, invertible matrix with real positive elements. Show that zn≤n2−2nz_n \le n^2 - 2n, where znz_n is the number of zero elements in A−1A^{-1}.

b) How many zero elements are there in the inverse of the n×nn \times n matrix A=(1111…11222…21211…11212…2………………1212……)?A = \begin{pmatrix} 1 & 1 & 1 & 1 & \dots & 1 \\ 1 & 2 & 2 & 2 & \dots & 2 \\ 1 & 2 & 1 & 1 & \dots & 1 \\ 1 & 2 & 1 & 2 & \dots & 2 \\ \dots & \dots & \dots & \dots & \dots & \dots \\ 1 & 2 & 1 & 2 & \dots & \dots \end{pmatrix}?

Solution (official)

Denote by aija_{ij} and bijb_{ij} the elements of AA and A−1A^{-1}, respectively. Then for k≠mk \ne m we have ∑i=0nakibim=0\sum\limits_{i=0}^{n} a_{ki} b_{im} = 0 and from the positivity of aija_{ij} we conclude that at least one of {bim:i=1,2,…,n}\{ b_{im} : i = 1,2,\dots,n \} is positive and at least one is negative. Hence we have at least two non-zero elements in every column of A−1A^{-1}. This proves part a). For part b) all bijb_{ij} are zero except b1,1=2b_{1,1} = 2, bn,n=(−1)nb_{n,n} = (-1)^n, bi,i+1=bi+1,i=(−1)ib_{i,i+1} = b_{i+1,i} = (-1)^i for i=1,2,…,n−1i = 1,2,\dots,n-1.

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