$A$ is a complex nilpotent $15X15$ matrix of order 5 ($A^5=0$), $\operatorname{rank}(A)=10$, $\operatorname{rank}(A^3)=4$. Find all possible JCFs

Question

I know how to use every detail except for the rank of $A^3$: A is nilpotent so all the eigenvalues are zero. I know that because the minimal polynomial is $\lambda^5$, the largest block is of size 5

$n-r=5$ so there are 5 Jordan Blocks.

I'm left with many different options. Some of them intuitively don't make sense (for example, $J= \operatorname{diag}(J_5(0),J_4(0),J_4(0),J_1(0),J_1(0)$) but I don't know any theorem to contradict them.

Furthermore, I know that the difference between the nullity of $A^j$ and $A^{j-1}$ is the number of jordan blocks of size that is at least $j$ x $j$, but because I don't know the ranks of $A^2$ and $A^4$ I can't deal with it.

Answer 1

A (nilpotent) Jordan block of size $k$ contributes $k-n$ to the rank of $A^n$ if $k\ge n$, and contributes $0$ if $k\le n$. Let $n_k$ denote that number of Jordan blocks of size $k$.

• From $A^5=0$, we known that all blocks have size $\le 5$, i.e., $n_k=0$ for $k>5$.
• From $\operatorname{rank}(A)=10$, we known that there are five Jordan blocks, i.e., $$\tag1n_1+n_2+n_3+n_4+n_5=5.$$
• We note that $\operatorname{rank}(A^2)=10-(n_2+n_3+n_4+n_5)$ and $\operatorname{rank}(A^3)=10-(n_2+n_3+n_4+n_5)-(n_3+n_4+n_5)$. Hence $$\tag2n_2+2n_3+2n_4+2n_5=6.$$
• And finally, the matrix size is $15$, so $$\tag3n_1+2n_2+3n_3+4n_4+5n_5=15$$

From $(3)-(2)-(1)$, we have $$\tag4n_4+2n_5=4,$$ in particular $n_5\le 2$.

• If $n_5=0$, we find $n_4=4$, which make the left hand side of $(3)$ at least $16$, so leads to no solution.
• If $n_5=1$, then $n_4=2$. Then $(1)$ and $(2)$ become $n_1+n_2+n_3=2$, $n_2+2n_3=0$, so $n_2=n_3=0$ and $n_1=2$.
• If $n_5=2$, then $n_4=0$. Then $(1)$ and $(2)$ become $n_1+n_2+n_3=3$, $n_2+2n_3=2$, so either $n_3=0, n_2=2, n_1=1$, or $n_3=1, n_2=0, n_1=2$.

In summary, we find three different possible Jordan forms.

Answer 2

Denote the sizes of the blocks by $x_1 \geq x_2 \geq \dots \geq x_5 \geq 1$ with $x_1 = 5$. We must have $x_2 + \dots + x_5 = 10$. The rank of $J_k(0)^l$ is $\max(k-l,0)$. If $A$ is similar to $\operatorname{diag}(J_5(0), J_{x_2}(0), \dots, J_{x_5}(0))$ then $A^3$ is similar to $\operatorname{diag}(J_5^3(0), J_{x_2}^3(0), \dots, J_{x_5}^3(0))$ whose rank is $2 + \max(x_2 - 3, 0) + \dots + \max(x_5 - 3,0) = 4$. Now, we have two options:

1. If $\max(x_i - 3, 0) = 2$ then $x_i = 5$ and $x_j \leq 3$ for all $i \neq j$ (with $2 \leq i,j \leq 4$). This gives the following options: $$ x_1 = x_2 = 5, x_3 = 3, x_4 = x_5 = 1, \\ x_1 = x_2 = 5, x_3 = x_4 = 2, x_5 = 1.$$
2. If $\max(x_i - 3, 0) = \max(x_j - 3, 0) = 1$ for $2 \leq i < j \leq 5$ then $x_i = x_j = 4$ and $x_k \leq 3$ for $2 \leq k \leq 5$ and $k \neq i,j$. This gives the following option: $$ x_1 = 5, x_2 = x_3 = 4, x_4 = x_5 = 1. $$