Possible Jordan Canonical Forms Given Minimal Polynomial

Solution 1:

Yes, you did.

Based on the minimal polynomial, you must have a two-by-two Jordan block for eigenvalue 2 and a one-by-one block for eigenvalue 1. You can fill in the five-by-five matrix with more of those blocks or with one-by-one blocks for eigenvalue 2. Using those rules yields precisely your first four matrices. Your fifth matrix is not correct.

Furthermore, you can permute the blocks. Thus,

• your first matrix yields 3!/2! = 3 Jordan forms,
• your second and third matrices yield 4!/2! = 12 forms each, and
• your four matrix yields 4!/3! = 4 forms

for a total of 3 + 2 · 12 + 4 = 31 forms.