Hoffman Kunze linear algebra section 7.4

Question

How many possible Jordan forms are there for a $ 6 \times 6$ matrix with characteristics polynomial $(x+2)^4 (x-1)^2$ ?

My answer is $P(4) \times P(2) = 5 \times 2 = 10$, where $P(n)$ is number of partitions.

Is my approach correct?

Answer 1

No, because Jordan form doesn't have a canonical ordering of either the eigenvalues (because algebraically closed fields aren't ordered fields!) or the blocks (because, well, if you can't even order the eigenvalues what would be the point?). There are in fact 10 fundamentally different transformations here, but each of them has more than one possible Jordan form. Consider the transformation with 2 blocks of dimension 1 for the eigenvalue 1, and with blocks of dimensions 1 and 3 for the eigenvalue -2. There are four blocks that can be put in any order; there would be 24 ways to do this, but two of the blocks are identical, so there are just 12 ways. Carry this out for each of the 10 essentially distinct transformations, and sum that up to get your answer.