 Solution

No. The same cyclic-vector argument gives minimal polynomial $X^3-1=(X-1)(X^2+X+1)$ over the finite field ${\mathbb{F}}_2$. The quadratic has no root in ${\mathbb{F}}_2$: its value is one at both zero and one. Thus the minimal polynomial does not split into linear factors over the specified field, violating the minimal polynomial criterion for diagonalizability. It becomes diagonalizable over a suitable larger finite field, but that is a different question.