Sprinkling signs in unitary matrices

Question

Let $A$, $B$ be $n\times n$ unitary complex matrices, such that for all indices $i,j$ we have $|a_{ij}|=|b_{ij}|$. Does there then exist diagonal unitary matrices $D,D’$ such that $DAD’=B$?

This can also be phrased as, take a unitary matrix, and sprinkle norm one signs on all of its entries. If the resulting matrix is unitary, then your sign at $(i,j)$ is equal to $z_i\cdot w_j$ for some lists of norm one complex numbers $(z_i)_{i=1}^n$, $(w_j)_{j=1}^n$.

The unitary case implies the analogous result for orthogonal matrices, and the positivity here makes the problem feel much easier. I would also be interested in a proof for this orthogonal case.

I have checked this for some small $n$, and the orthogonal case has the feeling of a combinatorial problem on intersecting families of sets, but I wasn’t able to make precise this intuition.

Answer 1

There are 5 inequivalent Hadamard matrices of order 16; if I understand correctly that's a counterexample.

Answer 2

I am able to quickly and easily compute counterexamples in both the unitary case and the orthogonal case numerically.

To do this, one should have access to automatic differentiation because I we not want to compute all the gradients by hand.

Consider the loss function: $$L(A,B)=\|AA^*-I\|_2^2+\|BB^*-I\|_2^2+\sum_{i=1}^n\sum_{j=1}^n(|a_{i,j}|-|b_{i,j}|)^2$$ where $\|C\|_2$ denotes the Frobenius norm.

We minimize the loss using gradient descent until we get a loss of about zero. We then normalize $A,B$ by multiplying $A,B$ on the left and the right by diagonal unitary matrices so that the first row of $A$ and the first column of $B$ consist of non-negative real numbers. If $A\neq B$, then you have obtained your desired counterexample.

Conjugation also gives counterexamples. If $A$ is any unitary matrix and $B=\overline{A}=(A^T)^*=(A^*)^T$, then we typically can't find diagonal unitary $D,D'$ with $DAD'=B$, but the entries in $A,B$ will still all have the same absolute value.

To produce more counterexamples, you might want to look into complex Hadamard matrices (and maybe quaternionic Hadamard matrices) which are just complex matrices where each entry has the same absolute value (for example, the Fourier matrix is a complex Hadamard matrix). If $X,Y$ are complex Hadamard matrices, then we say that $X,Y$ are equivalent if $Y=D_0P_0XP_1D_1$ for diagonal unitary $D_0,D_1$ matrices and permutation matrices $P_0,P_1$. There are generally many equivalence classes of $N\times N$-complex Hadamard matrices.