# Find a unitary basis of the $\mathbb{R}$-vector space of $n \times n$ (complex) Hermitian matrices

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 $$\forall n$$.

• – Omnomnomnom Apr 12 at 18:33
• @Omnomnomnom The basis is very elegant! However, it's not Hermitian (though it spans Hermitian matrices). Is there a way to change this? – Vincent J. Ruan Apr 12 at 19:50
• The Gell-Mann matrices from that same wiki page are Hermitian, but they're not generally unitary. – Omnomnomnom Apr 12 at 21:24
• It’s notable that when $n$ is odd, there are no trace-zero unitary and Hermitian matrices – Omnomnomnom Apr 12 at 22:07
• @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. – Vincent J. Ruan Apr 13 at 11:42