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Lemma: Let $N_1$ and $N_2$ be $3 \times 3$ nilpotent matrices over field $\mathbb F$. Then, $N_1$ and $N_2$ are similar if and only if they have the same minimal polynomial.

Use the result above and the Jordan form to prove the following:

Theorem: Let $A$ and $B$ be $n \times n$ matrices over the field $F$ which have the same characteristic polynomial $$f = (x - C_1)^{d_1} \cdots (x - C_k)^{d_k}$$ and the same minimal polynomial. If no $d_i$ is greater than $3$, then $A$ and $B$ are similar.

My idea is to leave $A$ and $B$ in the form of Jordan and separate $A = D + N$ and $B = D_0 + N_0$ with $D, D_0$ diagonal and $N, N_0$ nilpotent matrices. Hence we have that $D$ is similar to $D_0$ because both are diagonal matrices with the same eigenvalues ​​less than order. For $N$ and $N_0$ I would break both matrices in direct sum with blocks of sizes $d_i \times d_i$. He would use the minimal polynomials of each piece of the direct sum of $N$ and $N_0$ to coincide and use the Lemma of the beginning. Then each matrix of the direct sum of one would be similar to the other. I think this argument is not right because even if they are similar $D$ and $D_0, N$ and $N_0$, what guarantees that the sum is similar? Any suggestion?

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Your argument basically works. Choosing and concatenating bases of each of the generalised eigenspaces, we obtain changes of basis that respectively make $A$ and $B$ block diagonal matrices $A'$ respectively $B'$ with square blocks of sizes $d_1,d_2,\ldots,d_k$. One gets characteristic and minimal polynomials of each block$~i$ by taking the power of $(x-C_i)$ in the common characteristic respectively minimal polynomial of $A$ and $B$, and then as you argued $d_i\leq3$ implies that each block of $A'$ is conjugate to the corresponding block of $B'$.

Finally if one takes the change of basis matrices $P_i$ that perform these individual conjugations, then the block diagonal matrix with blocks $P_1,\ldots,P_k$ will conjugate $A'$ to $B'$.

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