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Let $S$ send successive standard basis vectors of $K^n$ to the next one, wrapping the last to the first. Its minimal polynomial is $X^n-1$: this polynomial annihilates it, and the cyclic vectors $e_1,S{e}_1,\dots ,S^{n-1}{e}_1$ form a basis, excluding any annihilator of smaller degree. The diagonalizability criterion using the minimal polynomial gives the displayed equivalence. A splitting polynomial is square-free exactly when the characteristic of a field does not divide $n$, as follows by taking its derivative $n{X}^{n-1}$ and checking common roots. Splitting over the specified field is a separate requirement. For $n=3$, the shift is diagonalizable over the complex numbers and ${\mathbb{F}}_7$, but not over ${\mathbb{F}}_2$ or ${\mathbb{F}}_3$.

The minimal polynomial factors over $\mathbb{C}$ as $(t-3)(t-3i)(t+3i)$ with distinct roots. The diagonalizability criterion using the minimal polynomial therefore makes $\alpha$ diagonalizable. Its characteristic polynomial gives multiplicities two, one and one, so its Jordan normal form isup to reordering the one-dimensional Jordan blocks.