# Is there always a complete, orthogonal set of unitary matrices?

The set of size-$$n$$ unitary matrices span $$\Bbb C^{n \times n}$$ (this can be proven nicely using polar decomposition). If we select a maximal linear subset of unitary matrices, then we have a basis of $$\Bbb C^{n \times n}$$ consisting of $$n^2$$ unitary matrices. My question is whether we can find a basis that satisfies the additional constraint of orthogonality. That is:

Does there exist a basis $$\mathcal B$$ of $$\Bbb C^{n \times n}$$ such that every $$P \in \mathcal B$$ is unitary (that is, $$P^*P = I$$) and for all distinct $$P,Q \in \mathcal B$$, we have $$\langle P,Q \rangle = 0$$?

Here, $$\langle \cdot , \cdot \rangle$$ refers to the Frobenius (AKA Hilbert-Schmidt) inner product, namely $$\langle P, Q \rangle = \operatorname{trace}(PQ^*)$$.

When $$n = 2$$, the Pauli matrices provide a convenient solution. That is, we can take $$\mathcal B = \{I,\sigma_1,\sigma_2,\sigma_3\} \subset \Bbb C^{2 \times 2}.$$ We can use this to produce a solution whenever $$n = 2^k$$. In particular, if we define $$\sigma_0 = I$$ for convenience, we can take $$\mathcal B = \{\sigma_{m_1} \otimes \cdots \otimes \sigma_{m_k} : 0 \leq m_j \leq 3\} \subset \Bbb C^{2^k \times 2^k}.$$ Could we come up with a basis for any other $$n$$? Could we do so for every $$n$$?

Some observations so far:

• Without loss of generality, we can assume that $$\mathcal B$$ contains the $$n \times n$$ identity matrix $$I$$. If $$I$$ is an element of the basis, it follows that the remaining matrices form a basis for the subspace of all trace-$$0$$ matrices.
• A commuting set of matrices spans at most an $$n$$-dimensional subset, so there must be elements of $$\mathcal B$$ that fail to commute

## 1 Answer

Yes, consider the group of $$n^2$$ matrices generated by the shift $$e_i \mapsto e_{i+1}$$ and the diagonal matrix with entries $$(1,\omega,\omega^2,\cdots,\omega^{n-1})$$ where $$\omega$$ is a primitive $$n$$th root of unit.

• Very elegant! Thank you – Ben Grossmann Mar 30 '19 at 20:26
• For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices. – Nathaniel Johnston Mar 31 '19 at 17:08
• Let $g,h$ be these matrices. Do you rather mean the set of those $n^2$ matrices of the form $\{g^kh^\ell\}$? The group $G_n$ generated by these two matrices is larger for $n\ge 2$. It is a (non-abelian for $n\ge 2$) nilpotent group order $n^3$. Indeed the commutator $z=hgh^{-1}g^{-1}$ is the scalar multiplication by $\omega$, and this group consists of the $n^3$ matrices of the form $g^kh^\ell z^m$. – YCor Apr 20 '20 at 12:52
• Good point, indeed the formulation was wrong. – Guillaume Aubrun Apr 20 '20 at 14:44