 Solution

The nonconstant invariant factors of a linear operator satisfy $d_1\mid \cdots \mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an irreducible polynomial, so every $d_j$ must equal $X^2+1$. Their polynomial degrees sum to $n$, giving $n=2k$.

The companion matrix of $X^2+1$, in the basis $(1,X)$ of $\mathbb{Q}[X]/(X^2+1)$, isTherefore, for each fixed even $n$, every such $A$ has rational canonical formThere are no such matrices when $n$ is odd. This is an instance of the rational canonical form with irreducible minimal polynomial principle; equality of minimal polynomials alone does not determine rational canonical form in general.