Find all possible Jordan Canonical form for nilpotent matrices with the characteristic polynomial $\lambda ^6$

Question

As far as I understand the way to solve it is to go from $(m_A=\lambda^k)_{k=1}^6$, the largest block will be of size $k$ and find the sizes that the rest of the blocks can be by finding ways to add Jordan blocks of smaller or equal size to the previous one and have their sizes sum to $6$.

Am I missing something? Is there a better way to do it with nilpotent matrices?

Answer 1

You are correct. In other words, finding the possible Jordan Canonical forms for a $6 \times 6$ nilpotent matrix is the same as finding all the sequences of natural numbers $b_1 \geq b_2 \geq \dots \geq b_i$ such that $b_1 + \dots + b_i = 6$. Each such sequence describes a block matrix with $i$ blocks, $b_1$ will be the size of the largest block (and so the minimal polynomial will be $x^{b_1}$), $b_2$ will be the size of the second largest block and so on.

Explicitly, you have the following options:

1. Minimal polynomial $x^6$: One option, corresponding to the sequence $6$ (so $i=1$ and $b_1 = 6$).
2. Minimal polynomial $x^5$: One option, corresponding to the sequence $5 \geq 1$.
3. Minimal polynomial $x^4$: Two options, corresponding to the sequences $4 \geq 2$ and $4 \geq 1 \geq 1$.
4. Minimal polynomial $x^3$: Three options, corresponding to the sequences $3 \geq 3$, $3 \geq 2 \geq 1$ and $3 \geq 1 \geq 1 \geq 1$.
5. Minimal polynomial $x^2$: Three options, corresponding to the sequences $2 \geq 2 \geq 2$, $2 \geq 2 \geq 1 \geq 1$, $2 \geq 1 \geq 1 \geq 1 \geq 1$.
6. Minimal polynomial $x^1$: One option (the zero matrix), corresponding to the sequence $1 \geq 1 \geq 1 \geq 1 \geq 1 \geq 1$.