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.
Best Answer
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.