Hadamard matrix

Remarks

The condition says that any two distinct rows have dot product $0$, so they agree in exactly $n/2$ places. Since $H^{-1} = H^T / n$, also $H^T H = n I_n$, so the columns are pairwise orthogonal as well. Example: $H_2 = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ satisfies $H_2 H_2^T = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2 I_2$. The Kronecker product of two Hadamard matrices is a Hadamard matrix, so the Kronecker powers of $H_2$ give Hadamard matrices of every order $2^k$. Non-example: no $3 \times 3$ matrix with entries $\pm 1$ is Hadamard, since the dot product of two rows is a sum of three terms $\pm 1$, which is odd and so not $0$. A Hadamard matrix of order $n$ can exist only if $n = 1$, $n = 2$ or $n$ is a multiple of $4$; the Hadamard conjecture asks whether all multiples of $4$ occur. For such $H$, $\abs{\det H} = n^{n/2}$.