Say $ A\in\mathbb{C}^{6\times6} $ and has eigenvalues $\lambda_1$ and $\lambda_2$ of multiplicity $ 3$ both of them. And for $\kappa=1,2,3$ the echelon form of the matrix
$$ (A-\lambda_1I)^\kappa $$
has exactly $\kappa$ zero lines and for $\kappa=1,2$
$$ (A-\lambda_2I)^{\kappa} $$
has exactly $\kappa+1$ zero lines.
Find the Jordan of the matrix subject to $\lambda_1$, $\lambda_2$. What I did is for $\kappa=1$
$$ (A-\lambda_1I) $$
has $1$ zero line so the rank of the the matrix is $6-1=5$. And so the nullspace has rank of $1$. Thus the geometric multiplicity of $\lambda_1$ is $1$, so I have $1$ Jordan block with $\lambda_1$ same logic for $\lambda_2$ found $2$ Jordan blocks. So I have 1 Jordan block of $3\times3$ with $\lambda_1$ and one $1\times 1 $,$2\times 2$ for $\lambda_2$.
Question: what am I supposed to do with the rest information for the different values of $\kappa$?