The Pauli matrices and the identity matrix form a basis of $M_2(\mathbb{C})$, and it is easy to show that any matrix in $M_2(\mathbb{C})$ can be written like $$ M = \frac{\text{Tr}(MI)}{2}I + \frac{\text{Tr}(M\sigma_x)}{2}\sigma_x + \frac{\text{Tr}(M\sigma_y)}{2}\sigma_y + \frac{\text{Tr}(M\sigma_z)}{2}\sigma_z . $$ This led me to suspect that $\langle A, B\,\rangle\,:= \frac{\text{Tr}(AB)}{2}$ is an inner product in this space and that the Pauli matrices and identity form an orthonormal basis with regard to this inner product. However it is not positive definite and does not have conjugate symmetry.
I have looked around a bit and found mention of it being an inner product in the real vector space of hermitian matrices called the "Hilbert-Schmidt inner product", but searching for that did not help.
So, my question is, is there any special reason why this identity holds? Am I missing something and is this really an inner product?
The Hilbert-Schmidt inner product is defined by $$ \langle A,B \rangle = \operatorname{Tr}(AB^*) $$ where $B^*$ denotes the adjoint (conjugate-transpose) of $B$. Notably, if $B$ is Hermitian, then $B = B^*$.
The Pauli matrices indeed form an orthonormal basis with respect to the Hilbert-Schmidt inner product, which is the usual sesquilinear inner product over the complex vector space $M_n(\Bbb C)$.