 Solution

If a basis consists of eigenvectors, let ${\lambda }_1,\dots ,{\lambda }_r\in K$ be the distinct eigenvalues. A polynomial annihilates the diagonal action exactly when it vanishes at each ${\lambda }_j$. Consequently the minimal polynomial is ${\prod }_j(X-{\lambda }_j)$, a product of distinct linear factors. It therefore has no repeated irreducible factor, proving (i) implies (ii).