For $A$ in rational canonical form, find $B$ with $AB=BA$ but for which $B \notin Span\{ A^i \mid 0\leq i \leq n-1\}$

Question

As part of a larger question, I've reached the point where I need to prove the following claim:

Let $\mathbb{F}$ be a field. Let $A \in M_n(\mathbb{F})$ be in rational canonical form. Prove that if $A$ is not a one-block matrix (one companion matrix), then there exists a $B\in M_n(\mathbb{F})$, with $AB=BA$ but for which $B \notin Span\{ A^i \mid 0\leq i \leq n-1\}$.

Besides playing around with specific examples and specific $n$, I'm unable to come up with a general way of constructing such a $B$. My question is twofold, with the second one being even more important to me:

1. What could be such a $B$?
2. How does one tackle such a question? I used to think that playing around with examples should give sufficient intuition, but I'm often encountering these situations where I'm still none the wiser after doing so (as in this case). I would be extremely grateful if the suggestion for a solution would offer the thought process of how you came up with it too.

Answer 1

Let $A=C_1\oplus C_2\oplus\cdots\oplus C_r$ where each $\chi_i$ (the characteristic polynomial of $C_i$) divides $\chi_{i+1}$. Let $B=I\oplus0\oplus\cdots\oplus0$. If $B=p(A)$ for some polynomial $p$, then $\chi_2|p$. In turn, $\chi_1|p$ and $p(C_1)=0$. Now we arrive at a contradiction because the first diagonal sub-block of $B$ is $I$.