I'm studying for my qualifying exam. A question asks to find the RCF of each $4 \times 4$ matrix over $\mathbb{R}$ having minimal polynomial $(x^2+1)^2$.
There is only one option: $$ R = \begin{bmatrix} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 0 \end{bmatrix} $$ Now we are asked to find the RCF over $\mathbb{C}$. Since the minimal polynomial splits over $\mathbb{C}$, then the RCF is $$ P = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 2i & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -2i \end{bmatrix} $$
Let $A$ and $B$ be the matrices of transition for $R$ and $P$ respectively. Then $ARA^{-1} = R = BPB^{-1}$. It follows that $R = (A^{-1}B)R(A^{-1}B)^{-1}$.
Finally, we must repeat the first two steps for $5 \times 5$ matrices.
Claim: This is impossible over $\mathbb{R}$. For the only possible invariant factors are $(x^2+1)$ and $(x^2+1)^2$. $5 = 4+1$. There is no invariant factor having degree 1, so no such matrix exists.
Since $(x^2+1)^2$ splits over $\mathbb{C}$, then there are possibilities. They are diag$(i,P)$ and diag$(-i,P)$. These last parts are where I am uncertain.