Bsrhrki

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 $langle P, Q rangle = operatornametrace(PQ^*)$.

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

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.