Finding the number of Jordan Form Matrices

Question

I'm asked to find the number of Jordan form matrices $A$ with $A^n = O$ for $n \in \mathbb{Z}^+$ such that $\dim (\text{Null} A^2) = 7, \dim(\text{Null} A^5) = 14$ and $\dim (\text{Null} A^5 \cap \text{Range} A^5) = 1$ with all the ways the Jordan blocks can be ordered.

I been at this question for majority of the day now. It can be noted that $A^n = O$ means the only eigenvalues of $A$ is 0, so I think I only need to somehow determine how many blocks there are of each size, but I'm struggling with it. So far, I was able to count 36 of such matrices. Any help would be appreciated!

Answer 1

Suppose that $A$ has size $m \times m$ and $m_j$ Jordan blocks of size $j$ for $j = 1,2,3,\dots$. The fact that $\dim [\operatorname{null}(A^5) \cap \operatorname{range}(A^5)] = 1$ tells us that $A$ has exactly one Jordan block with size greater than $5$. In general, $$ \dim [\operatorname{null}(A^5) \cap \operatorname{range}(A^5)] \geq \dim [\operatorname{null}(A) \cap \operatorname{range}(A^5)] \\ = m_6 + m_7 + \cdots = m - (m_1 + \cdots + m_5), $$ but the fact that $A$ is nilpotent implies that $\dim \operatorname{range}(A^5) > 0 \implies \dim [\operatorname{null}(A) \cap \operatorname{range}(A^5)] > 0$.

On the other hand, we have $$ m = m_1 + 2m_2 + 3m_3 + \cdots\\ \dim \operatorname{null}(A^2) = m_1 + 2m_2 + 2m_3 + \cdots \\ = m - m_3 - 2m_4 - 3m_5 - \cdots, \\ \dim \operatorname{null}(A^5) = m_1 + 2m_2 + 3m_3 + 4m_4 + 5m_5 + 5m_6 + \cdots \\ = m - m_6 - 2m_7 - \cdots.\\ $$ This is all the information that we have about $A$.

Answer 2

Since $A^n = O$ for some $n\in \mathbb{Z}^+$, then for any eigenvector $v$, $$ 0 = A^nv = A^{n - 1} \lambda v = \dots = \lambda^n v = 0.$$ Hence, $\lambda = 0$ is the only suitable eigenvalue of $A$. This means, \begin{align*} A = \begin{pmatrix} J_0^{m_1} & & \\ & \ddots & \\ & & J_0^{m_l} \end{pmatrix} \end{align*} Now, for any single Jordan block $J = J_0^m$. $J^k = 0$ for $k = m$ since $J$ has $1$ above the diagonal. For $0 \le k \le m$, we have the $1$'s shifted towards the upper right corner. So then for $0 \le k \le m$, $\text{Null}(J^k) = \{e_1, \dots, e_k\}$ and Range$(J^k) = \{e_1, \dots, e_{m-k}\}$. So we have nullity$(J^k) = k$ and rank$(J^k) = m - k$ and $\dim(\text{Null}(J_k) \cap \text{Range}(J^k)) = \min(k, m - k)$. Noting that for $k > m$, $\dim(\text{Null}(J_k) \cap \text{Range}(J^k)) = 0$. Let $a_j$ be the number of Jordan blocks of size $m = j$ and $b_j$ be the number of Jordan blocks size $m \ge j$. Then, $$\text{nullity}(A^k) = a_1 + 2a_2 + 3a_3 + \dots + (k-1)a_{k-1} + kb_k = b_1 + \dots + b_k.$$ From the premise, we know $\text{nullity}(A^2) = 7$ and $\text{nullity}(A^5) = 14$, and so, $b_1 + b_2 = 7$ and $b_1 + \dots + b_5 = 14\iff b_3 + b_4 + b_5 = 7$. Let $B = \text{Null}(A^5) \cap \text{Range}(A^5)$ and $B_i = \text{Null}(J_0^{m_i})^5 \cap \text{Range}(J_0^{m_i})^5$. Since $\dim(B) = 1$, exactly one of the Jordan blocks $J'$ must have $\dim(B_i) = 1$ and the rest must have $\dim(B_i) = 0$ and so for those Jordan blocks with dimension 0, $m_i \le 5$ so $\text{rank}(J_0^{m_i})^5 = 0$. For the jordan block with dimension 1, we must have $m = 6$ and thus $\text{rank}(A^5) = \sum^l_{i = 1}\text{rank}(J_0^{m_i})^5 = 1$. By the rank nullity theorem, the size of $A$ is therefore $\text{rank}(A^5) + \text{nullity}(A^5) = 1 + 14 = 15$. Since exactly one Jordan block has $m = 6$ and $m \le 5$ for all others, we must have $a^6 = 1$ and $a_j = 0$ for all $j > 6$. Given the above and $b_1 + b_2 = 7 = b_3 + b_4 + b_5$, and that $b_1 \ge b_2$, we have three options, $(b_1, b_2) \in \{(6,1), (5, 2), (4, 3)\}$. We can see the only viable choice is $(b_1, b_2) = (4, 3)$ as all other choices results in a invalid partition of 15. Consider $(6, 1)$, by definition of $a_j$, $b_j$, we have $a_1 = b_1 - b_2$ and so it requires, $a_1 = 5, a_6 = 1$ and all other $a_j$ to be $0$ and corresponds to the partition $(1, 1, 1, 1, 1, 6)$. Similarly, for $(b_1, b_2) = (5, 2)$ results in another invalid partition $(1, 1, 1, 6, 6)$ because it would make $a_6 = 2\not = 1$.

Now, given $b_1 = 4$ and $b_2 = 3$. We have $a_1 = a_6 = 1$, so there remain two blocks we can use to partition 15. Hence, the size of these two blocks sums to $15 - 6 - 1 = 8$ and the size of these two blocks satisfies the condition $b_3 + b_4 + b_5 = 7$. It can be seen that the following block sizes, $(1, 3, 5, 6)$ and $(1, 4, 4, 6)$ satisfy these requirements. With the first block size combo having $4!/2$ permutations and $4!$ permutations for the second choice for a total of 36 matrices.