Let $\mathbb{F}$ be a field, $\lambda \in \mathbb{F}$ and $A,B \in M_n(\mathbb{F})$ such that $m_A(x)=m_B(x)=(x-\lambda)^k$ and such that the geometric multiplicity of $\lambda$ in $A$ equals to the geometric multiplicity of $\lambda$ in $B$.
a. Prove that for $n=5$ or $n=6$- $A$ and $B$ are similar
b. Find an example for $n=7$ for which $A$ and $B$ are NOT similar
My main question here is whether I should try proving it for $n=5$ and $n=6$ seperately? Or should I try proving it for any $n$ and then observe that it holds only for $n=5$ or $n=6$?
Plus, any hints or suggestion regarding my main question or the question above will be welcome
For n = 7 consider jordan blocks $(3,2,2)$ and $(3,3,1)$ both these have algebraic and geometric multiplicity $3$, but are not similar. That is, consider matrices $A, B$ such that,
$\chi_A = (x- \lambda)^3(x- \lambda)^3(x- \lambda)^1$ and $\chi_B = (x- \lambda)^3(x- \lambda)^2(x- \lambda)^2$
For $n > 7$ append rest to these jordan blocks as i.e. blocks $(3,2,2,1, \dots ,1)$ and $(3,3,1,1, \dots ,1)$ to get examples of non-similar matrices. For the case of $n = 5$ and $n = 6$, we can list out the possible Jordan blocks and show that this kind of a configurations are not possible. This is not an elegent proof.