olymmath

Putnam 1988 B5

  • Session B

For positive integers nn, let MnM_n be the 2n+12n+1 by 2n+12n+1 skew-symmetric matrix for which each entry in the first nn subdiagonals below the main diagonal is 1 and each of the remaining entries below the main diagonal is -1. Find, with proof, the rank of MnM_n. (According to one definition, the rank of a matrix is the largest kk such that there is a k×kk \times k submatrix with nonzero determinant.)

One may note that

M1=(0−1110−1−110)M2=(0−1−11110−1−11110−1−1−1110−1−1−1110).\begin{aligned}M_1 &= \left( \begin{array}{ccc} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{array}\right) \\ M_2 &= \left( \begin{array}{ccccc} 0 & -1 & -1 & 1 & 1 \\ 1 & 0 & -1 & -1 & 1 \\ 1 & 1 & 0 & -1 & -1 \\ -1 & 1 & 1 & 0 & -1 \\ -1 & -1 & 1 & 1 & 0 \end{array} \right).\end{aligned}

Curated text Kedlaya Putnam archive. Quoted with source.

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