# Smallest subgroup of unitary group, containing diagonal matrices and a fixed unitary matrix is the whole group

Suppose that $$U_n(\mathbb{C})$$ is the group of unitary matrices of dimension $$n$$ over complex numbers. Fix a unitary matrix $$A \in U_n(\mathbb{C})$$ and consider the smallest closed subgroup $$K \subseteq U_n(\mathbb{C})$$, that contains all diagonal matrices (maximal torus group) and also $$A$$. It seems that apart from some exceptions for $$A$$ (like when $$A$$ is a diagonal matrix), $$K = U_n(\mathbb{C})$$. Do you have any idea, if this is true, how to prove it and how to derive the exception cases?

Particularly, I'm interested in the case where $$A$$ is a Circulant matrix, i.e. it has the following form: $$A=F^{-1}\cdot L\cdot F,$$ where $$L$$ is a diagonal matrix and $$F$$ is the DFT matrix. I'm not sure if this restriction simplifies the problem or not.

P.S. I have asked the question here in math.stackexchange group, but I guess, the question should be more relevant to this group.

• For large families of familiar counterexamples to the first paragraph, take $A$ in the normalizer of the diagonal torus $T$, or in the centralizer of any subtorus $S\subset T$. – Francois Ziegler May 27 '20 at 10:16

## 1 Answer

Here is a suggestion: If you want $$A$$ and the diagonal matrices to generate the full unitary matrix, then, in particular, the only matrices which commute with both $$A$$ and all the diagonal matrices should be scalar matrices, using Schur's Lemma. Furthermore, it is easy to see that the only matrices which commute with all diagonal matrices are themselves diagonal. Hence a necessary condition for $$A$$, together with diagonal matrices to generate the full unitary group is that $$A$$ commutes with no non-scalar diagonal matrix.

It is true that $$A$$ has this property if $$A$$ is the permutation matrix associated to the $$n$$-cycle $$(12 \ldots n)$$ (or any other $$n$$-cycle would do). However, in this case $$A$$, together with diagonals, does not generate the full unitary group, but generates a group of "monomial matrices" (matrices with one non-zero entry in each row and column).

• Thanks. It can be verified that this necessary condition is satisfied for Circulant matrices iff it has at least one non-zero non-diagonal element. – Mini May 27 '20 at 11:43