Are $n \times n$ special orthogonal matrices, all the entries of which have the same absolute value, possible for $n \neq 4$?

Question

As I noted in my preceding question https://math.stackexchange.com/questions/3510189/give-a-general-class-to-which-a-specific-4-times-4-special-orthogonal-matrix in equation (62) of their recent publication https://arxiv.org/abs/1708.05336, "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}

For $n=3, 5, 6$, I have tried (via direct enumeration) unsuccessfully to construct analogous $n \times n$ special orthogonal matrices, in which all the (equal) entries of the last column and row are (for probabilistic reasons) positive, and the remaining $n^2- 2n +1$ entries are all equal in absolute value. Such matrices might be helpful in extending the Li-Qiao framework to the construction of separable decompositions of length $n \neq 4$. (It is not clear, however, that matrices must be of the specific requested form to so extend their framework. Perhaps, other than having the last row and columns positive, all remaining entries could be unrestricted, other than for the orthogonality requirement.)

It was observed by Robert Israel in the noted preceding question that Q is proportional to a Hadamard matrix. However, the next larger-sized ($8 \times 8$) Hadamard matrices are not orthogonal in character, so this does not seem to be a productive direction to take. (But as the comments below of others and mine indicate I was in error in making this claim.)

Answer 1

Well, if $Q\in\mathrm{SO}(n)$ satisfies your (original) conditions then so does $ \frac1{\sqrt2} \begin{pmatrix} -{}^tQ&Q\\ \phantom{-}{}^tQ&Q \end{pmatrix} \in\mathrm{SO}(2n). $

Added: As S. Stadnicki since commented, your desired set $\mathrm S$ of possible orders $n$ is contained in and conjecturally equal to $\{1,2\}\cup4\mathbf N$ (Hadamard conjecture);* the above construction ($\cong$ Sylvester’s) just shows $\smash{2^{\mathbf N}\subset\mathrm S}$. (Voting to close, as R. Israel had really said all this at the mis-linked question.)

* Paley (1933, front page) proved $\mathrm S\subset\{1,2\}\cup4\mathbf N$ thus: Assume w.l.o.g. that all entries are $\pm1$ and $Q$ has 3 distinct columns $u,v,w$. Then their orthogonality gives $$ n=\|u\|^2=\langle u+v,u+w\rangle=\sum\nolimits_i(u_i+v_i)(u_i+w_i), $$ a sum all of whose terms are $0$ or $\pm 4$.

Answer 2

In reading the ArXiv paper referenced in the post, the authors are using matrices with complex entries. (Near the beginning they talk about generators of SO(2), one of them being a square root of -I.) If one is looking for complex matrices in an orthogonal group SO(n), one can choose a scaled version of a complex Hadamard matrix of order n, where n is any positive integer, to get a matrix with entries having the same norm (absolute value), as well as having a row and a column having all entries equal to 1/(scale value, which may be n or square root of n).

(For the section being considered having display (62), it is unclear to me if the authors restrict themselves to matrices with real entries. For the purposes of the question, it seems to me that using matrices with complex entries is appropriate for carrying out their analysis, and that restricting the order to accommodate real Hadamard matrices is unnecessary.)

Gerhard "Complex Numbers Are Numbers Two" Paseman, 2020.02.03.