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

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 $$Q \in \mbox{SO}(4)$$,

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

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.)

• How is an 8 by 8 Hadamard matrix "not orthogonal in character"? Feb 2, 2020 at 20:56
• Echoing @user44191: Hadamard matrices are precisely the $\pm 1$-valued matrices such that $A^\top A$ is equal to a scalar multiple of the identity, so rescaling one of these always gives an orthogonal matrix whose entries share the same absolute value. The determinant 1 condition should be easy to check in various examples. Feb 2, 2020 at 22:10
• Also, there are matrices called complex Hadamard matrices of all orders which can supply such orthogonal matrices with entries from the complex numbers. Gerhard "Going Off In Another Dimension" Paseman, 2020.02.02. Feb 2, 2020 at 22:36
• Thanks for the comments! Well, I had taken the Kronecker product of a $2 \times 2$ and a $4 \times 4$ Hadamard matrix to presumably get an $8 \times 8$ one, which proved to be not orthogonal, that is its matrix product with its transpose was not proportional to the identity. But it looks like I should probably recheck my computations/references--in view of the comments---just rechecked. My $H_2$ was miscoded--so mea culpa and the $H_8$ is orthogonal. But what about $n$ not a mulitple of 4. Feb 2, 2020 at 23:45
• Note that the construction works both ways; the existence of matrices of the form you're looking for is precisely equivalent to the existence of a Hadamard matrix, by scaling of the entries. The Wikipedia page on Hadamard matrices notes that they can only exist in dimensions $1$, $2$, and $4n$, so you won't find any other examples. Feb 3, 2020 at 0:29

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 $$\pm 4$$.

• Unfortunately it's only in the comments, but @PaulB.Slater has revised the question to require that $n$ not be a multiple of $4$. Feb 3, 2020 at 0:11
• OK--thanks for the answer! So, how about a conjecture on my part that for $m$ not a multiple of 4, one can not have an SO(m) matrix, with the entries of the $m$-th row and column all equal and positive, and all the remaining entries of the same absolute value as the entries of the last row and column. So, Hadamard matrices would be special in this regard. Feb 3, 2020 at 0:25
• Hasn't the $n \in 4 \mathbf{Z}$ case been settled by the answer of Ziegler, given that the orginal $4 \times 4$ $Q$ is in SO(n)? Feb 3, 2020 at 0:29
• @PaulB.Slater I double the size each time, so I think it solves only powers of 2 — your own arises that way starting from the 1$\times$1 identity. But yes, if you now want to exclude those it actually shouldn’t be by changing the question but by asking another. (Else no one knows who’s been answering what.) Feb 3, 2020 at 0:38
• OK, point well-taken, FZ. So, my conjecture should be for integers that are not powers of 2, rather than integers that are not multiples of 4. Feb 3, 2020 at 0:46

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.