Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be an $n\times n$ symmetric unitary matrix with elements of equal magnitude and the elements of the first row (and the first column, of course) are $1/\sqrt{n}$, i.e. $X_{j,k} = e^{i \phi_{j,k}}/\sqrt{n}$ with $\phi_{j,k}=\phi_{k,j} \in \mathbb{R}$ and $\phi_{1,k}=0$. Prove (or disprove) that such matrix is unique up to permutations.

The background here is to get understanding of the set of bases in $\mathbb{C}^n$ with coordinates differ by phase factors only. The orthogonality relation $(v_j, v_k) = n^{-1}\sum_l \exp(i(\phi_{j,l}-\phi_{k,l})) = 0$ for $j \ne k$ is invariant with respect to $\phi_{j,k} \to \phi_{j,k}+\omega_j + \omega_k$. This would be great if the set could be obtained from a particular choice

$$X_{j,k} = \frac{1}{\sqrt{n}}\exp\left(\frac{2\pi i}{n} (j-1)(k-1)\right)$$

by permutations and the symmetry transformation.

The statement seems plausible since the matrix $X$ is determined by $n(n-1)/2$ phases with $n(n-1)/2$ constraints imposed by orthogonality relations. The proof, however, is lacking. I would greatly appreciate help in a form of proof/disproof or a strong hint.

share|cite|improve this question
If the elements of the first row and column are all ones, then the upper left entry of $XX^*$ is $n$, so $XX^*\ne I$, so $X$ is not unitary. – Gerry Myerson Mar 8 '13 at 0:36
@Gerry: I think that was just a somewhat sloppy formulation; the formula right after it does include the factor $\sqrt n$. – joriki Mar 8 '13 at 7:45
Yes, sorry for inaccuracies. The important property of the matrix in question is that its elements have the same magnitude and therefore the whole matrix was thought with a common normalization factor. – Misha E Mar 8 '13 at 20:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.