Existence of orthogonal basis of symmetric $n\times n$ matrices, where each matrix is unitary?

For a positive integer $$n$$, let $$S_n$$ denote the set of $$n\times n$$ symmetric matrices over $$\mathbb{C}$$. As a complex vector space, this set has dimension $$\mathrm{dim}(S_n)=\binom{n+1}{2}$$. The standard Hilbert-Schmidt inner product $$\langle A,B\rangle = \mathrm{Tr}(AB^*)$$ for $$A,B\in S_n$$ turns this space into an inner product space.

Question. Does there always exist an orthogonal basis $$\{U_1,U_2,\dots,U_{\frac{n(n+1)}{2}}\} \subseteq S_n$$ of symmetric matrices such that each $$U_i$$ is unitary?

Case for $$n=2$$

The case $$n=2$$ is simple, since we may take the matrices $$U_1 = \begin{pmatrix} 1&0\\0&1\end{pmatrix},\quad U_2 = \begin{pmatrix} 1&0\\0&-1\end{pmatrix},\quad\text{and}\quad U_3 = \begin{pmatrix} 0&1\\1&0\end{pmatrix}.$$ These matrices are all symmetric and unitary, and they satisfy $$\mathrm{Tr}(U_iU_j^*) = 0$$ whenenver $$i\neq j$$. Moreover, these matrices span the space of $$2\times 2$$ symmetric matrices.

Case for $$n$$ even

I've come up with a way to construct such a basis for any even $$n$$.

Consider the matrices $$\{E_{i,j}\,:\, i,j\in\{1,\dots,n\}\}$$, where $$E_{i,j}$$ is the $$n\times n$$ matrix that has a 1 in the $$i$$th row and $$j$$th column with zeros elsewhere. One orthogonal basis for the set of $$n\times n$$ symmetric matrices is the set $$\{H_{i,j}: i,j\in\{1,\dots,n\},\, i\leq j\}$$ where we define $$H_{i,j} = \left\{\begin{array}{ll}E_{i,i} & \text{if }i=j\\ \frac{1}{\sqrt{2}}(E_{i,j}+E_{j,i}) & \text{if }i\neq j\end{array}\right.$$ Consider the complete graph $$K_n$$ with $$n$$ vertices. We may identify the edges of this graph with the set $$\mathcal{E} = \{H_{i,j} : i,j\in\{1,\dots,n\},\, i where the edge connecting vertices $$i$$ and $$j$$ (with $$i) is denoted $$H_{i,j}$$. Note that $$\mathcal{E}$$ has $$\binom{n}{2}=\frac{n(n-1)}{2}$$ elements.

If $$n$$ is even, there is a 1-factorization of this graph (see here). A 1-factorization corresponds to a partitioning the edges $$\mathcal{E}$$ into $$n-1$$ subsets, $$\mathcal{F}_1,\dots,\mathcal{F_{n-1}}$$, each with $$n/2$$ edges, such that, for each $$k$$, no two edges in $$\mathcal{F_k}$$ are adjacent and $$\mathcal{F_1}\cup\cdots\cup\mathcal{F_{n-1}} = \mathcal{E}.$$ For each $$k\in\{1,\dots,n-1\}$$, we can label the elements of $$\mathcal{F_k}$$ as $$\mathcal{F_k} = \{F_{k,1},\dots,F_{k,n/2}\}.$$ Let $$\omega = e^{i2\pi/n}$$ denote the $$n$$th root of unity, and for each $$k\in\{1,\dots,n-1\}$$ and $$\ell\in\{1,\dots,\frac{n}{2}\}$$ we define the matrix $$G_{k,\ell} = 2\sum_{a=1}^{n/2} \omega^{2\ell a} F_{k,a}.$$ It can be verified that $$G_{k,\ell}$$ is symmetric and unitary. Finally, define symmetric unitary matrices $$A_1,\dots,A_n$$ by $$A_j = \sum_{a=1}^n \omega^{ja} E_{a,a}$$ for each $$j\in\{1,\dots n\}$$. It can be shown that the set $$\{A_1,\dots,A_n\}\cup\{G_{k,\ell}\, :\, k\in\{1,\dots,n-1\},\, \ell\in\{1,\dots,n/2\}\}$$ is an orthogonal basis of $$S_n$$.

Case when $$n=3$$

The case for $$n$$ odd seems to be a bit trickier. I've at least been able to construct a basis of the desired form when $$n=3$$ as follows: \begin{align*} U_1 &= \frac{\sqrt{3}}{2}\begin{pmatrix} \frac{2}{\sqrt{3}} & 0 & 0\\ 0 &\frac{1}{\sqrt{3}} & 1\\ 0 & 1 & -\frac{1}{\sqrt{3}}\end{pmatrix} & U_2&=\frac{\sqrt{3}}{2}\begin{pmatrix} \frac{2}{\sqrt{3}} & 0 & 0\\ 0 &\frac{1}{\sqrt{3}} & -1\\ 0 & -1 & -\frac{1}{\sqrt{3}}\end{pmatrix}\\ U_3 &= \frac{\sqrt{3}}{2}\begin{pmatrix} \frac{-\overline{\beta}}{\sqrt{3}} & 0 & 1\\ 0 &\frac{\alpha}{\sqrt{3}} & 0\\ 1 & 0 & \frac{\beta}{\sqrt{3}}\end{pmatrix} & U_4&=\frac{\sqrt{3}}{2}\begin{pmatrix} \frac{-\overline{\beta}}{\sqrt{3}} & 0 & -1\\ 0 &\frac{\alpha}{\sqrt{3}} & 0\\ -1 & 0 & \frac{\beta}{\sqrt{3}}\end{pmatrix} \\ U_5 &= \frac{\sqrt{3}}{2}\begin{pmatrix} \frac{i\overline{\beta}}{\sqrt{3}} & 1 & 0\\ 1 &\frac{i\beta}{\sqrt{3}} & 0\\ 0 & 0 & \frac{i\alpha}{\sqrt{3}}\end{pmatrix} & U_6&=\frac{\sqrt{3}}{2}\begin{pmatrix} \frac{i\overline{\beta}}{\sqrt{3}} & -1 & 0\\ -1 &\frac{i\beta}{\sqrt{3}} & 0\\ 0 & 0 & \frac{i\alpha}{\sqrt{3}}\end{pmatrix} \end{align*} where $$\alpha = \sqrt{\frac{27}{8}} - i\sqrt{\frac{5}{8}}$$ and $$\beta=\sqrt{\frac{3}{8}} + i\sqrt{\frac{5}{8}}$$. It can be verified that the matrices $$\{U_1,\dots,U_6\}$$ are symmetric, unitary, and pairwise orthogonal.

Existence for odd $$n$$ greater than $$3$$?

Numerically, I've been able to find an orthogonal basis of symmetric unitary matrices for odd dimensions up to $$n=11$$. But I found no discernible pattern that allowed me to construct an exact solution.

I'd like to know if there exists a collection of $$\binom{n+1}{2}$$ orthogonal symmetric $$n\times n$$ unitary matrices for any odd $$n>3$$.

• Just a trivial comment: you normalize your matrices to give them (Hilbert Schmidt) norm $1$, but of course they won't be unitary after that (none of your $U$'s is). – Christian Remling Dec 6 '18 at 0:03
• Thanks, you're right! I've un-normalized all the matrices such that they are in fact unitary. – luftbahnfahrer Dec 6 '18 at 4:04
• This will sound dumb, but did you by chance try to construct an example with $n=7$ after $n=3$? – Josiah Park Dec 6 '18 at 4:39
• @JosiahPark: according to the last part the OP has examples up to $n=11$. – Nik Weaver Dec 6 '18 at 4:46
• @NikWeaver Sure, but these are not exact, correct? I read that they were numeric. The dimension count for $n=7$ is even like $n=3$ so it makes sense to look there possibly before $n=5$. – Josiah Park Dec 6 '18 at 4:48