Matrix anticommuting with four or five Dirac $\Gamma$-matrices

Consider the following four or five Dirac $$\Gamma$$-matrices$$\begin{gather}σ_{x,y}\otimes τ_0,σ_{x,z}\otimes\tau_z,\tag{*1}\\σ_{x,y}\otimes τ_0,σ_{x,y,z}\otimes τ_z,\tag{*2}\end{gather}$$ where $$\sigma$$ and $$\tau$$ are conventional Pauli matrices. Only three of them, $$\sigma_{x,y}\otimes\tau_0,\sigma_{z}\otimes\tau_z$$, satisfy the standard anticommutation relation. (The linked page and some papers call all the 16 basis matrices as Dirac $$\Gamma$$-matrices. Perhaps a bit misleading.)

I found $$\sigma_z\otimes\tau_x$$ anticommuting with those three. But is it possible to find one matrix that anticommutes with all the four or even five? If not, how to show? I feel that there must be some not complex way and some basic thing I'm not aware of.

Let's first denote $$\alpha_1=\sigma_x\otimes\tau_0,\alpha_2=\sigma_y\otimes\tau_0,\alpha_0=\sigma_z\otimes\tau_z,\beta_1=\sigma_x\otimes\tau_z,\beta_2=\sigma_y\otimes\tau_z$$ for the five matrices in the question. We also define $$\alpha_3=\sigma_z\otimes\tau_x,\alpha_4=\sigma_z\otimes\tau_y.$$
The first step is to learn the fact that a maximal mutually anticommuting set consists of five $$\Gamma$$-matrices $$\alpha_{0,1,2,3,4}$$. (This is is not unique. Six different sets share all the same proof, actually.)
Secondly, note that $$\{\alpha_{3},\beta_1\}\neq0$$ and $$\{\alpha_{4},\beta_1\}\neq0$$ are linearly independent and so is for $$\beta_2$$.
Thus, any linear combination of $$\alpha_{3,4}$$, the sole possibility anticommuting with $$\alpha_{0,1,2}$$, will not anticommute with $$\beta_{1,2}$$. This proves no matrix anticommuting with all the four or five in the question.
• Can you direct me to a proof of the claim about a maximal anticommuting set of five Dirac matrices? How do you find the six different such sets? Arfken (Mathematical Methods for Physicists, 3rd edition, pp. 213--214) lists six such sets, but they don't seem to be the only possible ones. For example, in his notation, an unlisted set would be $\{\sigma_1, \sigma_2, \sigma_3, \alpha_1, \rho_2 = \alpha_5\}$. Oct 8, 2019 at 11:17
• @splitcomplexes It's combinatorics. There are 16 matrices in total. One can try to select out each one in the five one by one. Multiplying the number of choices (requiring mutual anticommutation) at each step gives us $15\times 8\times 3\times 2\times 1=720$. Then, the fact that ordering doesn't matter leads exactly to $720/5!=6$ such maximal sets. Oct 8, 2019 at 21:16