ordered partition, block matrix given by $r_j \times r_j$ nilpotent Jordan blocks is nilpotent, rational canonical form, jordan canonical form

Question

Let $F$ be a field. For an integer $n \ge 1$, and ordered partition of $n$ is a sequence $\underline{r} = \{r_1, \dots, r_m\}$ of positive integers such that $r_1 \le \dots \le r_m$ and $\sum r_j = n$. For each such $\underline{r}$, define $N(\underline{R}) \in \text{Mat}_{n \times n}(F)$ to be the block matrix given by $r_j \times r_j$ nilpotent Jordan blocks $($i.e., $1$'s just below main diagonal, $0$'s elsewhere$)$, with blocks stacked along the main diagonal in order of increasing size.

1. Prove that $N(\underline{r})$ is nilpotent and compute its rational canonical form.
2. Use Jordan canonical form to prove that every nilpotent $N \in \text{Mat}_{n \times n}(F)$ is $\text{GL}_n(F)$-conjugate to $N(\underline{r})$ for a unique $\underline{R}$.

I understand that for the second part we want to show $N(\underline{r}) = MNM^{-1}$ for some $M \in \text{GL}_n(F)$. I also know that Jordan canonical form only requires the minimal polynomial to split into linears in $F[t]$, which is an automatic condition when $F$ is algebraically closed, and also when the linear map is nilpotent.

Answer 1

1.

Raising $N(\underline{r})$ to the $m$th power has the effect of raising each of its Jordan blocks to the $m$th power, so it is enough to show that the given Jordan blocks are nilpotent. These matrices are lower triangular with $0$'s on the diagonal, so their characteristic polynomial is clearly a power of $t$: that is, the only eigenvalue in any extension is $0$, so we have nilpotence. $($This can also be proved by inductive matrix calculation with each Jordan block.$)$ In terms of the structure theorem, $V = F^n$ as an $F[t]$-module with $t$ acting as $N(\underline{r})$ is $\bigoplus(F[t]/t^{r_i})$ with $r_1 \le \dots \le r_m$. This is the Jordan canonical form. Since $t^r|t^{r'}$ whenever $r \le r'$, this is also the rational canonical form.

2.

Any nilpotent endomorphism on an $n$-dimensional vector space has characteristic polynomial $t^n$ that is a product of linears, so the theorem on Jordan canonical form may be applied to such a map without restriction on $F$. The Jordan blocks have along their diagonals exactly the roots of the linear factors of the characteristic $($or equivalently, minimal$)$ polynomial, which in this case is just the root $0$. Hence, the Jordan canonical form for $N$ is exactly given by some $N(\underline{R})$; the arrangement of the $r_j$'s in monotonically increasing order is the translation of the condition on exponents in Jordan canonical form to ensure the uniqueness aspect of this form. Hence, $N$ is $GL_n(F)$-conjugate to a unique $N(\underline{R})$.

In this way, we see that the set of $GL_n(F)$-conjugacy classes of nilpotent elements in $\text{Mat}_{n \times n}(F)$ $($$``$nilpotent orbits$"$ for the conjugation action of $\text{GL}_n(F)$ on the vector space $\text{Mat}_{n\times n}(F)$$)$ is parameterized by the finite set of $\underline{r}$'s.