 Solution

Yes. In ${\mathbb{F}}_7$, the minimal polynomial splits asThe three factors are distinct, so the minimal polynomial criterion for diagonalizability applies. For example the eigenvectors $v_{\lambda }=(1,{\lambda }^2,\lambda {)}^{\mathsf{T}}$ for $\lambda =1,2,4$ give an eigenbasis over this finite field. The contrast with characteristic three is that the shift polynomial now splits without repeated roots.