Prove that $A$ is similar to $B$

Question

Let $A, B \in M_n(\mathbb{F})$ such that $m_A(x) = m_B(x)$ and $$f_A(x)=f_B(x)=(x-\lambda_1)^{d_1}\cdots (x-\lambda_k)^{d_k}$$ for different $\lambda_1, \ldots, \lambda_k$ such that $1 \le d_l \le 3$ for all $1 \le l \le k$. Prove that $A$ and $B$ are similar matrices.

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1. We can assume that $A,B$ are in Jordan form because $A_J \sim A \sim B \sim B_J$.
2. We also may assume $A,B$ are Jordan blocks of some $\lambda_i$, because if each $J_{A_i} \sim J_{B_i}$ (order doesn't matter) then $A\sim B$. (Right? I am not sure about it)
3. So now we're looking at $A, B$ two jordan blocks. We know that $$m_A(x) = m_B(x) = (x-\lambda_i)^{d'_i}$$ and $1 \le d'_i \le d_i \le 3$.

What do I do now?

Answer 1

When $d_i \leq 3$, then $d_i'$, the degree of $(x - \lambda_i)$ in the minimal polynomial, which is the size of the largest Jordan block associated with $\lambda$, completely determines the Jordan form of the $\lambda$-blocks up to similarity.

If $d_i = 1$, there is one Jordan block of size $1$.

If $d_i = 2$, then if $d_i' = 2$, there is one Jordan block (of size 2). If $d_i' = 1$, there are two (of size 1).

If $d_i = 3$, then if $d_i' = 3$, there is one Jordan block (of size 3). If $d_i' = 2$, then there is one Jordan block of size 2, and the other Jordan block must then have size $1$. If $d_i' = 1$, then all Jordan blocks are size $1$.