# Give a general class to which a specific $4 \times 4$ special orthogonal matrix belongs

In equation (62) of their recent publication, "Separable Decompositions of Bipartite Mixed States", Li and Qiao present the matrix $$Q \in \mbox{SO}(4)$$,

$$\begin{equation} Q=\frac{1}{2}\left( \begin{array}{cccc} 1 & -1 & -1 & 1 \\ -1 & -1 & 1 & 1 \\ -1 & 1 & -1 & 1 \\ 1 & 1 & 1 & 1 \\ \end{array} \right). \end{equation}$$

We see that all the entries of the matrix have the same absolute value ($$\frac{1}{2}$$), and in the last row and column, all the entries (for probabilistic reasons) are positive. I would be interested in other $$4 \times 4$$ special orthogonal matrices with these properties, and more broadly still in $$n \times n$$ special orthogonal counterparts. (I have an application in mind with $$n=11$$.)

• See Hadamard matrix. But the order of a Hadamard matrix must be $1$, $2$ or a multiple of $4$, so $n=11$ won't work. – Robert Israel Jan 15 at 16:32
• So, $Q$ is itself a Hadamard matrix? Is it some (harmless) permutation of the $H_4$ given in the Wikipedia reference. The positivity of the last (alternatively, first) row and column is important for probabilistic reasons in the Li-Qiao scheme. – Paul B. Slater Jan 15 at 17:15
• $2Q$ is Hadamard. Interchange the first and last columns, and the first and last rows. There is, up to row and column permutations and multiplying rows and columns by $-1$, only one $4 \times 4$ Hadamard matrix. – Robert Israel Jan 15 at 18:49
• I generated an $8 \times 8$ Hadamard matrix, but its product with its transpose is not proportional to the identity matrix, so my quest for orthogonality is not fulfilled. – Paul B. Slater Jan 28 at 15:11
• Perhaps, Paul, you would like to share your remarkable $8\times8$matrix with us. – Gerry Myerson Feb 2 at 21:27