Write J - 2I and solve the homogeneous system directly, or note that J splits into 2x2 diagonal blocks and count how many of those blocks belong to eigenvalue 2.
Step 1: Set up the system $(J - 2I)v = 0$ directly.
Let $v = (v_1, v_2, v_3, v_4, v_5, v_6)^T$. Subtracting $2$ from each diagonal entry of $J$ gives:
$$J - 2I = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$
Step 2: Write out the equations from $(J-2I)v = 0$ row by row.
Row 1: $v_2 = 0$.
Row 2: $0 = 0$ (no restriction from this row).
Row 3: $v_4 = 0$.
Row 4: $0 = 0$ (no restriction).
Row 5: $v_5 + v_6 = 0$.
Row 6: $v_6 = 0$.
Step 3: Solve the system for the free variables.
From row 6, $v_6 = 0$. Putting this in row 5 gives $v_5 = 0$ as well.
From row 1, $v_2 = 0$, and from row 3, $v_4 = 0$.
The variables $v_1$ and $v_3$ never appear in any equation, so they stay completely free.
So the general solution is $v = (v_1, 0, v_3, 0, 0, 0)^T$, with $v_1$ and $v_3$ arbitrary.
Step 4: Count the dimension of the solution space and conclude.
The solution space is spanned by the two independent vectors $(1,0,0,0,0,0)^T$ and $(0,0,1,0,0,0)^T$, since $v_1$ and $v_3$ are free parameters.
So the null space of $J - 2I$ has dimension $2$, which means the geometric multiplicity of the eigenvalue $2$ is $2$.
$$\boxed{2}$$
