9.7 Hadamard Matrices

A Hadamard matrix of order $n$ is an $n\times n$ matrix $H$ with entries $+1,-1$ such that

In other words, the rows are orthogonal and hence so are the columns. These are named so as they arose in the following question of Hadamard : Let $A$ be an $n\times n$ matrix whose entries have absolute value at most $1$. How large can $det(A)$ be ? Each row is a vector of length at most $\sqrt{n}$ and so the volume of parallelogram (also the determinant) spanned by the row vectors is at most $n^{n/2}$. When is the equality achieved ? For all row vectors to have length $\sqrt{n}$, the entries have to be $-1,+1$. For the volume of parallelogram to be equal to product of the length of row vectors, the vectors have to be orthogonal i.e., the matrix $A$ is Hadamard !

Permutation by rows or columns preserves Hadamard property. Also multiplying a row or column by $-1$ preserves Hadamard property. Hadamard matrices obtained this way are called equivalent. Trivially, by multiplying each row and column suitably for any Hadamard matrix, one can find an equivalent Hadamard matrix with first row and column being all $+1$’s. Such Hadamard matrices are called normalized. Since the remaining rows of a normalized matrix are orthogonal to the first row, the number of $+1$’s and $-1$’s must be equal. Thus of $n\ne 1$, it must be even ! In fact, we can say more.