In David Tong's Quantum field theory lecture notes, page 101, line 5, he shows that:
$$(p.\sigma)(p.\overline{\sigma})=m^2.I_2$$
(I have placed the identity matrix $I_2$ for clarity)
Then I don't know how in line 10 of the same page he can take:
$$(p.\sigma)(\sqrt{p.\overline{\sigma}})\xi=m.(\sqrt{p.\sigma})\xi$$
As I see, it comes from the formula: (which is used again in the notes)
$$(\sqrt{p.\sigma})(\sqrt{p.\overline{\sigma}})=m.I_2$$
I have $(\sqrt{p.\sigma})(\sqrt{p.\sigma})=p.\sigma$ and $(\sqrt{p.\overline{\sigma}})(\sqrt{p.\overline{\sigma}})=p.\overline{\sigma}$ and $(p.\sigma)(p.\overline{\sigma})=m^2.I_2$ but I don't know how to prove that $(\sqrt{p.\sigma})(\sqrt{p.\overline{\sigma}})=m.I_2$ since to the extend I know, matrix multiplication is not commute.
Note: $p.\sigma=\sum_{\mu=0}^4 p_\mu\sigma^\mu$ where $\sigma^\mu=(I,\sigma^i)$ and $\overline{\sigma}^\mu=(I,-\sigma^i)$. $\sigma^i$ is Pauli matrices with $i=1,2,3$