# Unitary matrix with elements of equal modulus

It is a well-known fact that there are unitary matrices $U$, whose elements have equal modulus ($|u_{ij}|=c, \forall i,j$). The most known example is the DFT Matrix.

My question is, are the DFT Matrices and their permutations the only matrices that satisfy this relation? A similar question has been asked in this forum before (alas, with no answer), but with the constraint of symmetry. In my case, I do not care if the matrix is symmetric or not.

## 2 Answers

Your conjecture is false because you can multiply each column of a solution by any complex of modulus $1$.

When $n=2$, the set of solutions is $V(2)=\{\dfrac{1}{\sqrt{2}}\begin{pmatrix}e^{iu}&e^{iv}\\e^{i(u+\alpha)}&-e^{i(v+\alpha)}\end{pmatrix};u,v,\alpha\in\mathbb{R}\}$.

Note that $V(2)$ is a real variety of dimension $3$ and that $U(2)$ is a real variety of dimension $4$.

EDIT. When $n=3$, we obtain $V(3)=\{\dfrac{1}{\sqrt{3}}\begin{pmatrix}e^{iu_1}&e^{iu_2}&e^{iu_3}\\e^{iu_4}&j^2e^{i(u_4-u_1+u_2)}&je^{i(u_4-u_1+u_3)}\\e^{iu_7}&je^{i(u_7-u_1+u_2)}&j^2e^{i(u_7-u_1+u_3)}\end{pmatrix};u_1,u_2,u_3,u_4,u_7\in\mathbb{R};1+j+j^2=0\}$.

Thus $dim(V(3))=5$ while $dim(U(3))=9$.

EDIT. Answer to @Bryson of Heraclea. Yes, all the possible cases. Yet, beware, the case $n=3$ is special; indeed, when the sum of $3$ complex numbers $a,b,c$ of modulus $1$ is $0$, then $b=ja,c=j^2a$; this result doesn't generalize to higher dimensions;

• The examples you have provided so far are based on the DFT Matrix, which has been modified by multiplying each row or/and column with a different number of unit modulus. Would you say that this exhausts all the possible cases? – Bryson of Heraclea Jan 18 '18 at 17:03
• Which result does not generalize to higher dimensions? Furthermore, what do you make of user1551's answer (Hadamard matrix). Wouldn't fall out of the category of permuted and column/row-multiplied DFT matrices? – Bryson of Heraclea Jan 19 '18 at 9:31
• You did not understand my edit. For $n>3$ a sum of $n$ complex numbers $a_1,\cdots,a_n$ of modulus $1$ cannot be entirely solved by relations of the type $a_2=\omega a_1,a_3=\omega ^2 a_1,\cdots$ where $\omega$ is $n^{th}$ root of unity (the previous solution is only a particular solution). For example, if $n=4$, then the solutions depend also on an arbitrary parameter. Thus, the user1551's instance is not a scoop. – loup blanc Jan 19 '18 at 9:42

The answer is no. E.g. by halving a Hadamard matrix $H_4$ of order $4$, you get a real orthogonal matrix $$\frac12H_4=\pmatrix{ 1& 1& 1& 1\\ 1&-1& 1&-1\\ 1& 1&-1&-1\\ 1&-1&-1& 1},$$ but it is neither a permutation matrix nor a DFT matrix.