Let $A, B$ be two $n \times n$ matrices over a field $F$ and $A,B$ have the same characteristic and minimal polynomials. If no eigenvalue has algebraic multiplicity greater than $3,$ then $A$ and $B$ are similar (characteristic polynomial of $A,B$ is $f = (x - c_1)^{d_1} \cdots (x - c_k)^{d_k}$).
I have to use the following result.
If $A, B$ are two $3 \times 3$ nilpotent matrices. Then $A, B$ are similar if and only if they have same minimal polynomial.
My attempt has been to consider the following cases $ d_i = 1,2,3$ and then the corresponding jordan blocks which becomes same in both the cases. Since two matrices having the same rational form are similar, matrices $A,B$ are similar. My doubt here lies in the fact that where do I need to use the theorem that two $3 \times 3$ nilpotent matrices are similar iff they have the same minimal polynomial.
Hint: First you can show that there is a decomposition of $F^n$ such that
$$F^n = \bigoplus_i \ker (A-\lambda_iI)^{d_i}\,.$$
What can you say about $\dim\ker (A-\lambda_iI)^3$ for a given $i$ ? How does $A-\lambda_i I$ act on this subspace?
Edit: You are right that using the Jordan form completely trivializes this result. However this result is much easier than Jordan form. In fact you could use a slight generalization of it to prove the existence and unicity of the Jordan form. The above hint gives clues for using the theorem about nilpotent matrices to prove the result.