Recipe

Find the partition of a nilpotent operator

Use when: you have a nilpotent operator or matrix $N$ and need the exact sizes of its Jordan chains, the data that fixes its similarity class completely

Goal: the full partition $(k_1,k_2,\dots)$ of chain lengths, with the lengths summing to $\dim V$

Taught in Nilpotent operators and their structure

Quick recipe

Find the smallest $k$ with $N^k=\mathbf 0$
Compute $d_j=\dim\ker(N^j)$ for $j=0,\dots,k+1$
Count chains of each length with $c_j=2d_j-d_{j-1}-d_{j+1}$
Assemble the partition and check the dimension count

1
Compute $N^2, N^3, \dots$ in turn and stop at the first power that comes out to the zero matrix; call that exponent $k$ .

Example

For $N=\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$ , $N^2$ still sends $e_3\mapsto e_1$ (not zero), but $N^3$ is the zero matrix, so $k=3$ .
- $N^j$ is still nonzero — raise the power by one and test again.
- $N^j$ first comes out to the zero matrix — stop; that exponent is $k$ .
⚡ Shortcut

If $N$ is already written as block-diagonal Jordan blocks (1's on the superdiagonal, 0 everywhere else), read $k$ directly as the size of its largest block, skipping the power test entirely.

Why: every later step compares kernel dimensions only up through this exponent, so the recipe cannot proceed without knowing where they stop growing.
2
Solve $N^j\mathbf v=\mathbf 0$ for each power from $j=0$ up through $j=k+1$ , and record the dimension of each solution space as $d_j$ .

Example

For the same $N$ : $d_0=0$ (the identity's kernel is always trivial), $d_1=2$ , $d_2=3$ , $d_3=4$ , and $d_4=4$ .

Why: the chain-counting formula in the next step needs the dimension immediately below and above each $j$ , not just the value at $j$ itself.
3
For every length $j$ from $1$ to $k$ , compute $c_j=2d_j-d_{j-1}-d_{j+1}$ . This number is exactly how many chains have length $j$ , not length at least $j$ .

Example

$c_1=2(2)-0-3=1$ , $c_2=2(3)-2-4=0$ , and $c_3=2(4)-3-4=1$ .

Why: this double difference isolates the chains that stop at exactly $j$ , since a longer chain would still be contributing to $d_{j+1}$ too.
4
List each length $j$ with $c_j>0$ , repeated $c_j$ times, largest first, then add up length times count over every $j$ .

Example

Here $c_3=1$ and $c_1=1$ (with $c_2=0$ contributing nothing), so the partition is $(3,1)$ , and $3+1=4=\dim V$ .

Why: a partition whose lengths fail to add up to $\dim V$ signals an arithmetic error in an earlier step, not a valid answer.

Common mistakes

Treating $d_0$ as $\dim V$ instead of $0$ ; $N^0$ is the identity, which has a trivial kernel.
Stopping the power test one exponent too early, before $N^j$ actually reaches the zero matrix.
Reading $c_j$ as the number of chains of length at least $j$ rather than exactly $j$ .
Dropping a length whose count $c_j$ came out to $0$ , instead of simply leaving it out of the partition.
Listing the partition out of order or skipping the final check that the lengths add up to $\dim V$ .