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The question is on the title ($n \in \mathbb{N}^*$). To be clear, unitary basis here means basis consisting of (complex) unitary matrices.

I wonder this question because recently I've read about Pauli matrices, which are all unitary and, together with the identity matrix $I_2$, make up a basis of the $\mathbb{R}$-vector space of $2 \times 2$ Hermitian matrices. There are 4 elements in that basis, suitably as the vector space has $\dim = 2 \times 2 = 4$. In the general case, this would be $n^2$.

Disclaimer: I should prove the existence of such basis first, but I haven't done it. Still I think there exists at least one $\forall n$.

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    $\begingroup$ See this wiki page and this post on MO. $\endgroup$ – Omnomnomnom Apr 12 at 18:33
  • $\begingroup$ @Omnomnomnom The basis is very elegant! However, it's not Hermitian (though it spans Hermitian matrices). Is there a way to change this? $\endgroup$ – Vincent J. Ruan Apr 12 at 19:50
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    $\begingroup$ The Gell-Mann matrices from that same wiki page are Hermitian, but they're not generally unitary. $\endgroup$ – Omnomnomnom Apr 12 at 21:24
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    $\begingroup$ It’s notable that when $n$ is odd, there are no trace-zero unitary and Hermitian matrices $\endgroup$ – Omnomnomnom Apr 12 at 22:07
  • $\begingroup$ @Omnomnomnom So we have 2 bases, one is unitary but not Hermitian, one is traceless Hermitian but not generally unitary (definitely not for odd $n$). Still, the desired basis hasn't been reached yet. Also, for $n = 2^k$, we can tensor product to create the desired basis. $\endgroup$ – Vincent J. Ruan Apr 13 at 11:42

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