If a <basis> consists of <eigenvectors>, let $\lambda_1,\ldots,\lambda_r\in K$ 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 \b[(i) implies (ii)].

Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2004 / 2 / i / Solution

 Solution (source code)