49th Putnam 1988

Problem B5

Find the rank of the 2n+1 x 2n+1 skew-symmetric matrix with entries given by aij = 1 for (i - j) = -2n, -(2n-1), ... , -(n+1); -1 for (i - j) = -n, -(n-1), ... , -1; 1 for (i - j) = 1, 2, ... , n; -1 for (i - j) = n+1, n+2, ... , 2n+1. In other words, the main diagonal is 0s, the n diagonals immediately below the main diagonal are 1s, the n diagonals below that are -1s, the n diagonals immediately above the main diagonal are -1s, and the n diagonals above that are 1s.



Answer: 2n.

The determinant is zero, since each column sums to zero. Hence the rank is not 2n+1. However, we will show that the matrix D2n formed by deleting the first row and the first column has determinant ±1, which is non-zero. Hence the rank is 2n.

We proceed by induction. The result is clearly true for n = 1, since D2 = 1. Suppose it is true for n-1. We carry out row and column operations on D2n to show that its determinant is ±D2n-2.

Adding the row n+2 to row 1 gives 0s except in for the -1 in col 1 and the 1 in col n+2. Adding or subracting the new row 1 from all others except row n+2 gives 0s in col 1 except for -1 in rows 1 and n+2. It also gives 0s in col n+2 except for 1 in row 1. Note that the submatrix formed by deleting rows 1 and n+2 and cols 1 and n+2 has not been changed by these operations and is D2n-2.

We may now expand by col n+2. This gives a single term ± the determinant formed by deleting row 1 and col n+2. This has only one non-zero entry in col 1, that in the old row n+2. So expanding gives ±D2n-2 as required.



49th Putnam 1988

© John Scholes
1 Jan 2001