Find maximum number of Jordan Canonical Form given the minimal polynomial and geometric multiplicity of one eigenvalue

Question

Let $(x-2)^2 (x-3)$ be minimal polynomial of a $5$ by $5$ matrix s.t geometric multiplicity of $k=2$ is $2$. Find maximum number of Jordan Canonical Form.

Options are 2,3,1. Gm of k=2 is 2 means no. Of linearly independent vectors corresponding to k=2 is 2. How to proceed??

Answer 1

Since the minimal polynomial is $(x-2)^2(x-3)$ so the characteristic polynomial will be $(x-2)^2(x-3)^3$.

Hence the possible options are $$\begin{align}J_2(2)\\J_2(3)\\J_1(3) \end{align}$$

or

$$\begin{align}J_2(2)\\J_1(3)\\J_1(3)\\J_1(3) \end{align}$$

where $J_2(2)$=\begin{bmatrix}2 &1\\0&2\end{bmatrix}