2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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;

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Bryson of Heraclea Jan 18 '18 at 17:03
  • $\begingroup$ 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? $\endgroup$ – Bryson of Heraclea Jan 19 '18 at 9:31
  • $\begingroup$ 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. $\endgroup$ – loup blanc Jan 19 '18 at 9:42
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.