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 1


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.

  • $\begingroup$ Very elegant! Thank you $\endgroup$ Mar 30, 2019 at 20:26
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    $\begingroup$ For what it's worth, these matrices are described in some detail on Wikipedia. They are sometimes called the "shift" and "clock" matrices. $\endgroup$ Mar 31, 2019 at 17:08
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    $\begingroup$ 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 $g^kh^\ell z^m$. $\endgroup$
    – YCor
    Apr 20, 2020 at 12:52
  • $\begingroup$ Good point, indeed the formulation was wrong. $\endgroup$ Apr 20, 2020 at 14:44

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