Suppose you are given n linearly independent vectors in n-dimensional Euclidean space. You move the vectors so that each vector becomes longer, but their inner products remain the same. What happens to the volume of the parallelepiped they span?
This month I am in Cambridge, at the Isaac Newton Institute pretending to be a statistician. (The picture shows one of the striking John Robinson artworks outside the Institute.) Dennis Lin, from Penn State, asked me a very interesting question, which so far I can’t solve.
In a post on team games, I discussed conference matrices. Let me repeat the facts here. A conference matrix is a square matrix C with entries 0 on the diagonal and ±1 elsewhere, which satisfies CCT = (n−1)I. Conference matrices only exist for even order n; they are equivalent to symmetric (resp. skew-symmetric) matrices if n is congruent to 2 (resp. 0) mod 4.
If C is a skew-symmetric conference matrix, and H = C+I, then H is a Hadamard matrix, that is, it has all entries ±1 and satisfies HHT = nI. Such a matrix is called a skew-Hadamard matrix (it is skew-symmetric apart from the +1s on the diagonal). Conversely, given a skew-Hadamard matrix H, the matrix C = H−I is a skew-symmetric conference matrix.
Let us widen the enquiry to deal with the case where no conference or Hadamard matrix exists. The letters C and H suggest a fanciful terminology which I will use. We define a hot matrix to be a square matrix with entries ±1, with +1 on the diagonal and off-diagonal entries skew-symmetic; a cold matrix is a skew-symmetric matrix with 0 on the diagonal and all off-diagonal elements ±1. There is an obvious bijection between cold and hot matrices, given by H = C+I.
Now a theorem of Hadamard implies that a hot matrix has determinant at most nn/2, with equality if and only if it is a skew-Hadamard matrix. For consider the rows of such a matrix as vectors in n-dimensional space. Each vector has length n1/2, so the volume of the parallelepiped they span is at most nn/2, with equality if and only if the rows are pairwise orthogonal. Similarly, a cold matrix has determinant at most (n−1)n/2, with equality if and only if it is a skew conference matrix.
Thus, if a skew-Hadamard matrix exists (as is conjectured to be the case for all multiples of 4), then it has maximum the determinant among all hot matrices, and the corresponding skew conference matrix has maximum determinant among all cold matrices.
What happens if n is not a multiple of 4? This was Lin’s question.
If n is odd, then the determinant of a cold matrix is zero. (More generally, any skew-symmetric matrix of odd order has determinant zero.) So there is nothing to say in this case.
Dennis Lin asked whether it is true that, for n congruent to 2 (mod 4), if H is a hot matrix with maximum determinant, then H−I is a cold matrix with maximum determinant. Perhaps it goes the other way as well.
Computation shows that the assertion is correct for n = 6: the largest possible determinants for hot and cold matrices are 160 and 81 respectively, and the matrices representing the maxima correspond under the natural bijection. There are too many matrices for n = 10 for an exhaustive search; and in lieu of doing anything clever, I simply looked at a few million random matrices, and found that the matrices with maximum determinant which showed up did indeed correspond. (The best I found were 64000 for hot and 33489 for cold.)
The connection with my introductory remarks simply comes from the observation that, if hot and cold matrices correspond under our bijection (that is, H = C+I), then the rows of H, regarded as vectors, are longer than those of C (length n1/2 as opposed to (n−1)1/2), but the inner products of corresponding pairs of distinct rows in the two matrices are the same. However, I haven’t been able to use this observation to get anywhere.
Let me end on a cautionary note. The correspondence between hot and cold matrices does not induce a monotone map on their determinants. In other words, the orderings might agree at the top end, but definitely differ lower down.