Nilpotent Operators

Properties

Suppose $ T \in \mathcal{L}(V) $, then

$ T $ is nilpotent $\iff \dim G(0,T) = \dim V $.

Jordan Basis

Every nilpotent operator has a Jordan basis.

Let $N \in \mathcal{L}(V)$ be nilpotent. Fix a Jordan basis: there are vectors $v_1,\dots,v_k$ (the tops) and lengths $m_1,\dots,m_k$ with $m_1+\dots+m_k = \dim V$ such that

\[\big\{\, N^j v_i \;:\; i = 1,\dots,k,\;\; j = 0,\dots,m_i-1 \,\big\}\]

is a basis of $V$, and $N^{m_i}v_i = 0$ for each $i$.

Sort the lengths so $m_1 \ge m_2 \ge \dots \ge m_k$ and picture the chains as columns, aligned at the bottom:

\[\begin{array}{cccc} v_1 & & & \\ Nv_1 & v_2 & & \\ \vdots & \vdots & \ddots & \\ N^{m_1-2}v_1 & N^{m_2-2}v_2 & & v_k \\ N^{m_1-1}v_1 & N^{m_2-1}v_2 & \cdots & N^{m_k-1}v_k \end{array}\]

$N$ moves each entry one step down its column, and off the bottom to $0$. The bottom level is exactly the chain bottoms, which is a basis of $\operatorname{null}N$.

Sorted into decreasing order, the chain lengths form a partition of $\dim V$:

\[\mu = (m_1 \ge m_2 \ge \dots \ge m_k), \qquad m_1 + \dots + m_k = \dim V.\]

The individual $m_i$ are its parts. In the array above, each part is the height of one column, because one column is one chain.

Now read the same boxes across instead of down. Level $j$ contains one box from every chain tall enough to reach it, so its size is

\[\mu'_j \;:=\; \#\{\,i : m_i \ge j\,\}.\]

The sequence $\mu’ = (\mu’_1, \mu’_2, \dots)$ is again a partition of $\dim V$, called the conjugate of $\mu$. It has $m_1$ parts, and conjugating twice returns $\mu$.

One set of boxes, two partitions: $\mu$ counts down the columns, $\mu’$ counts across the levels.

Example. Two chains, of lengths $3$ and $2$ — the running example of this post:

\[\begin{array}{cc} x^2 & \\ 2x & xy \\ 2 & y \end{array}\]

The columns have heights $3$ and $2$, so $\mu = (3,2)$. The levels, read from the bottom up, have sizes $2$, $2$, $1$, so $\mu’ = (2,2,1)$. Both sum to $5 = \dim V$, as they must — they count the same five boxes.

Most references draw a partition with its parts as rows; the conventional Young diagram of $\mu$ is therefore this array reflected across a diagonal. Nothing below depends on the choice. Parts are columns here because $N$ acts down them.

Level $j$ means the vectors killed by $N^j$ but not by $N^{j-1}$ — an algebraic condition that makes no reference to the picture. A chain of length $m_i$ contributes its entry $N^{m_i - j}v_i$ to level $j$ whenever $m_i \ge j$, and nothing beyond level $m_i$. In the bottom-aligned array the levels are therefore the horizontal slices, and $\mu’_j$ is the size of level $j$.

That is what bottom alignment buys, and it is why $\mu’$, not $\mu$, is the sequence the rest of this post counts. Cumulatively, the boxes in the bottom $j$ levels are a basis of $\operatorname{null}N^j$:

\[d_j \;:=\; \dim\operatorname{null}N^j \;=\; \mu'_1 + \dots + \mu'_j .\]

So $d$ is the partial-sum sequence of $\mu’$, and differencing recovers it:

\[\mu'_j = d_j - d_{j-1}.\]

On the example: $\mu’ = (2,2,1)$ gives $(d_1,d_2,d_3) = (2,4,5)$, matching $\operatorname{null}N = \operatorname{span}(1,y)$, $\operatorname{null}N^2 = \operatorname{span}(1,y,x,xy)$, and $\operatorname{null}N^3 = V$.

The span of one Jordan chain is one Jordan block. The whole space is the direct sum of the chain-spans.

Let $U_i = \operatorname{span}(v_i, Nv_i, \dots, N^{m_i-1}v_i)$ be the span of one chain. It’s invariant under $N$, so $\left. N \right\rvert_{U_i}$ makes sense. Order the chain bottom to top:

\[u_1 = N^{m_i-1}v_i,\quad u_2 = N^{m_i-2}v_i,\quad \dots,\quad u_{m_i} = v_i.\]

Then $Nu_1 = N^{m_i}v_i = 0$ and $Nu_j = u_{j-1}$ for $j \ge 2$. Since column $j$ of a matrix records the image of the $j$-th basis vector, the $1$ from $Nu_j = u_{j-1}$ sits in row $j-1$, column $j$ — the superdiagonal:

\[\mathcal{M}(N|_{U_i}) = \begin{pmatrix} 0&1& & \\ &0&\ddots& \\ & &\ddots&1\\ & & &0\end{pmatrix} = J_{m_i}(0).\]

And since $V = U_1\oplus\dots\oplus U_k$ with each $U_i$ invariant, the matrix of $N$ with respect to the concatenated basis is block diagonal:

\[\mathcal{M}(N) = J_{m_1}(0) \oplus \dots \oplus J_{m_k}(0).\]

$\dim\operatorname{null}N^m = \sum_j \min(m, m_j)$.

A $k \times m$ rectangle, bottom-aligned, and $d_m$ is the number of diagram boxes it covers.

Column $i$ contributes $\min(m, m_i)$ boxes to it. Whichever runs out first wins:

• Short column ($m_i \le m$): the reach of $N^m$ exceeds the column, so the whole column dies — $m_i$ boxes.
• Tall column ($m_i > m$): only the bottom $m$ die — $m$ boxes.

Summing over columns gives $\dim\operatorname{null}N^m = \sum_i \min(m,m_i)$. $\blacksquare$

Let $N$ be nilpotent on $V$. The partition $\mu$ does not depend on which Jordan basis produced it, and two nilpotent operators are similar if and only if their partitions agree.

Independence of the basis. A Jordan basis fills in the diagram, and the diagram determines $\mu’$ as its level sizes. But those level sizes are $d_j - d_{j-1}$, and $d_j = \dim\operatorname{null}N^j$ makes no reference to a basis. So $\mu’$ is forced by $N$ alone, and therefore so is $\mu$. Different Jordan bases genuinely differ — the vectors are choices, and Step 1 of the algorithm below makes them — but the shape they fill is fixed before any choice is made.

Similar operators agree. If $N’ = ANA^{-1}$ then $(N’)^j = AN^jA^{-1}$, so $A$ restricts to an isomorphism $\operatorname{null}N^j \to \operatorname{null}(N’)^j$. The two towers have equal dimensions, hence equal $\mu’$, hence equal $\mu$.

Agreeing operators are similar. Each of $N$ and $N’$ admits a Jordan basis in which its matrix is $J_{m_1}(0)\oplus\dots\oplus J_{m_k}(0)$. Equal $\mu$ means this is literally the same matrix, so each is similar to it and hence to the other. $\blacksquare$

A column ends at level $j$ exactly when it reaches level $j$ and fails to reach level $j+1$, so

\[\#\{\,i : m_i = j\,\} \;=\; \mu'_j - \mu'_{j+1} \;=\; 2d_j - d_{j-1} - d_{j+1},\]

with the convention $\mu’_{p+1} = 0$. Together with $k = \mu’_1 = d_1 = \dim\operatorname{null}N$ for the number of chains, this recovers $\mu$ from the null-space tower by pure arithmetic.

Since $\#\{i : m_i = j\} \ge 0$, the identity above says the sequence $(d_j)$ is concave:

\[2d_j \ge d_{j-1} + d_{j+1}.\]

Equivalently, its increments $\mu’_j$ are non-increasing — the picture’s “levels widen as you descend”. A hand-computed tower violating this has an error in the tower itself, and it is worth checking before running the algorithm rather than after.

Any two Jordan bases of $N$ give matrices that differ only by a permutation of the diagonal blocks.

Two Extremes

Everything so far has been about a general shape. Two degenerate shapes are worth naming, because the whole framework collapses in opposite directions: a diagram that is a single column, and a diagram that is a single level. Each is worth working out on its own before comparing them.

One quantity below needs saying carefully first.

Free parameters in a Jordan basis. The shape of the diagram is forced by $N$, but the basis filling it is not. To compare how much choice is left in different cases, fix the following count.

A Jordan basis is a list of $n$ vectors, so in coordinates it is $n^2$ scalars. Usually those scalars are not independent of each other: fixing some of them forces the rest. Write $r$ for how many can be chosen freely.

A few choices always have to be thrown out. In the cases below it is the vectors $v$ with $N^{n-1}v = 0$, and the lists of $n$ vectors that fail to be a basis. Such choices are degenerate rather than numerous, and discarding them does not reduce $r$.

The Jordan bases do not form a subspace of anything, so $r$ is not the dimension of a subspace. It is a count of independent choices.

That $r$ is well defined — that no cleverer parametrization uses fewer scalars — is not proved here, and is not obvious from either computation below. Commuting Operators settles it: the invertible operators commuting with $N$ act simply transitively on the Jordan bases of a given shape, so the Jordan bases are always in bijection with the invertible part of $\mathcal{C}(N)$, and $r = \dim\mathcal{C}(N)$ in every case.

Until then, read $r$ as the number of scalars the parametrization at hand happens to use.

One Column: a Single Chain

$\mu = (n)$ and $\mu’ = (1,\dots,1)$. One chain of length $n$: a single top, a single bottom, and every level of size $1$.

\[k = \mu'_1 = 1, \qquad p = m_1 = n, \qquad d_j = \min(j,n), \qquad \operatorname{rank}N^j = \max(0, n-j),\]

and $\mathcal{M}(N) = J_n(0)$ is a single Jordan block.

This is as far from the zero operator as a nilpotent operator gets: $N^{n-1} \neq 0$, so the tower

\[\\{0\\} \subsetneq \operatorname{null}N \subsetneq \operatorname{null}N^2 \subsetneq \dots \subsetneq \operatorname{null}N^n = V\]

climbs one dimension at a time, with no repeats and no jumps. Equivalently $d_j = j$ until it saturates, which is the only way a concave sequence of $n$ steps can rise as slowly as possible while still reaching $n$.

Free parameters. A Jordan basis here is determined by its top $v$ alone: the rest of the chain is $Nv, N^2v, \dots$, forced. So the $n$ coordinates of $v$ specify the whole basis, subject to the single non-degeneracy condition $N^{n-1}v \neq 0$. That gives $r = n$: of the $n^2$ scalars a basis nominally carries, only $n$ are free.

Invariant subspaces. The only subspaces invariant under $N$ are the $n+1$ subspaces $\operatorname{null}N^j$ for $0 \le j \le n$ — a single chain of them, totally ordered by inclusion. This is the most rigid an invariant-subspace lattice can be.

The proof uses the geometric series from the Invertibility section. Let $W \ne \{0\}$ be invariant and let $s$ be smallest such that some $w \in W$ has a nonzero coefficient on $N^sv$; write $w = q(N)N^sv$ with $q(N) = c_sI + c_{s+1}N + \dots$ and $c_s \ne 0$. Then $q(N)$ is a nonzero scalar plus a nilpotent, hence invertible with inverse again a polynomial in $N$. Since $W$ is invariant it is closed under polynomials in $N$, so $N^sv = q(N)^{-1}w \in W$, and applying $N$ repeatedly gives $W \supseteq \operatorname{null}N^{\,n-s}$. Minimality of $s$ gives the reverse inclusion.

nonderogatory/cyclic operator. The minimal polynomial equals the characteristic polynomial.

One Level: the Zero Operator

$\mu = (1,\dots,1)$ and $\mu’ = (n)$. Now there are $n$ chains, each of length $1$, so every box is simultaneously a top and a bottom.

\[k = \mu'_1 = n, \qquad p = m_1 = 1, \qquad d_j = n \ \text{for all } j \ge 1, \qquad \operatorname{rank}N^j = 0,\]

and $\mathcal{M}(N)$ is the zero matrix.

Here $p = 1$ says $N^1 = 0$, so this shape is the zero operator. Every null space is already $V$ and the tower stops before it starts.

Free parameters. All of them. Since $N = 0$, the chain condition is vacuous and every basis of $V$ is a Jordan basis. The $n^2$ coordinates of the $n$ vectors range freely, subject only to the non-degeneracy condition that they be independent, so $r = n^2$ — the largest a Jordan basis count can be, since $n^2$ scalars is all a basis has.

Invariant subspaces. Every subspace of $V$ is invariant, again since $N = 0$. This is the least rigid an invariant-subspace lattice can be, and the exact opposite of the totally ordered chain above.

The nilpotent shadow of diagonalizability. If $T$ has eigenvalue $\lambda$ and $N = \left. (T - \lambda I) \right\rvert_{G(\lambda,T)}$, then

\[G(\lambda, T) = E(\lambda, T) \iff N = 0 \iff \mu = (1,\dots,1),\]

and the block for $\lambda$ is $\lambda I$ rather than a nontrivial Jordan form.

Comparing the Ends

The two shapes are conjugate to each other. Conjugation reflects the diagram, and a single column reflects to a single level, so the two extremes of the framework are one partition and its transpose. Anything proved by counting columns in one case is proved by counting levels in the other.

They are also the only two shapes that are extreme in this way: every other $\mu$ has both a column of height $\ge 2$ and a level of size $\ge 2$.

Side by side, they bracket every invariant in this post.

	one column	one level
$\mu$	$(n)$	$(1^n)$
$\mu’$	$(1^n)$	$(n)$
chains $k = \mu’_1$	$1$	$n$
index of nilpotency $p = m_1$	$n$	$1$
$d_j$	$\min(j, n)$	$n$ for all $j \ge 1$
$\operatorname{rank}N^j$	$\max(0, n-j)$	$0$ for all $j \ge 1$
$\mathcal{M}(N)$	$J_n(0)$	the zero matrix
invariant subspaces	$n+1$, a chain	all of them
free parameters $r$	$n$	$n^2$

The last two rows are the ones that do not obviously belong to the same story as the others, and they are the reason for the sequel. Commuting Operators shows that the operators commuting with $N$ form a space of dimension $\sum_j (\mu’_j)^2$, which is $n$ for one column and $n^2$ for one level — the smallest and largest values it can take, and exactly the two entries in the last row.

That is what makes the free-parameter count above legitimate rather than ad hoc. The two cases were counted by unrelated arguments here — one by parametrizing a single top vector, the other by observing that no condition applies at all — and they nonetheless landed on $\dim\mathcal{C}(N)$ both times. The sequel explains why they had to.

Jordan Block

$\dim E(0, J_s(\lambda)) = 1$

Nilpotent Block

Let $N = J_m(0)$. The band of $1$s migrates one step toward the upper right with each power, and loses one entry each time. With $m = 5$:

\[N = \begin{pmatrix}0&1&0&0&0\\ &0&1&0&0\\ & &0&1&0\\ & & &0&1\\ & & & &0\end{pmatrix},\quad N^2 = \begin{pmatrix}0&0&1&0&0\\ &0&0&1&0\\ & &0&0&1\\ & & &0&0\\ & & & &0\end{pmatrix},\quad N^3 = \begin{pmatrix}0&0&0&1&0\\ &0&0&0&1\\ & &0&0&0\\ & & &0&0\\ & & & &0\end{pmatrix}\]

Then $N^4$ has a single $1$ in the top-right corner, and $N^5 = 0$.

In the bottom-to-top ordering $u_1 = N^{m-1}v, \dots, u_m = v$, we have $Nu_j = u_{j-1}$, hence $N^ru_j = u_{j-r}$, with $u_i := 0$ for $i \le 0$. Column $j$ therefore has its $1$ in row $j - r$, which is the $r$-th superdiagonal. The entries with $j - r \le 0$ fall off the top, leaving $\max(0, m - r)$ ones.

The columns that vanish are $j \le r$, so

\[\dim\operatorname{null}N^r = \min(r, m), \qquad \operatorname{rank}N^r = \max(0, m-r).\]

That’s the single-chain case of $\dim\operatorname{null}N^r = \sum_j\min(r,m_j)$, and the “band exits the corner” picture is the same statement as “$N^r$ annihilates the bottom $\min(r,m)$ entries of a column of height $m$.”

Non-nilpotent Block

If $\lambda \neq 0$, with $A = J_m(\lambda) = \lambda I + N$, the two terms commute, so

\[A^r = \sum_{t=0}^{\min(r,\,m-1)} \binom{r}{t}\lambda^{r-t}N^t.\]

Rather than shifting a single band, this fills the entire upper triangle, with $\lambda^r$ on the diagonal, $r\lambda^{r-1}$ on the first superdiagonal, $\binom{r}{2}\lambda^{r-2}$ on the second, and so on — constant along each diagonal, i.e. upper triangular Toeplitz. For $m = 4$:

\[A^r = \begin{pmatrix} \lambda^r & r\lambda^{r-1} & \binom{r}{2}\lambda^{r-2} & \binom{r}{3}\lambda^{r-3}\\ & \lambda^r & r\lambda^{r-1} & \binom{r}{2}\lambda^{r-2}\\ & & \lambda^r & r\lambda^{r-1}\\ & & & \lambda^r \end{pmatrix}\]

Two structural differences worth noting: $A^r$ is never $0$, since $\det A^r = \lambda^{rm} \neq 0$, so $J_m(\lambda)$ is invertible rather than nilpotent; and the shifting behaviour is recovered only through $A - \lambda I = N$, which is precisely why the general Jordan form is proved by subtracting $\lambda$ on each generalized eigenspace and working with the nilpotent part.

Find a Jordan Basis

Input. A nilpotent $N \in \mathcal{L}(V)$.

Precompute. The increasing chain of null spaces and their dimensions,

\[\{0\} = K_0 \subsetneq K_1 \subsetneq \dots \subsetneq K_p = V, \qquad K_j := \operatorname{null}N^j, \quad d_j := \dim K_j,\]

where $p$ is the smallest exponent with $N^p = 0$. Keep an explicit basis of each $K_j$.

Recall \(\mu'_j = d_j - d_{j-1}\), the size of level $j$ from the Jordan Basis section. The algorithm reconstructs the diagram level by level from the bottom up, so this is the number of vectors it must produce at level $j$.

State carried between levels. As the loop runs it carries a list $H_j$ of vectors sitting in $K_j$, pushed down from the level above. It also outputs a list $T_j$ of new chain tops at each level.

---

Step 0 (initialize). Set $j := p$ and $H_p := \varnothing$.

Step 1 (choose new tops at level $j$). Form one list, in this order:

1. a basis of $K_{j-1}$,
2. the vectors of $H_j$,
3. any spanning set of $K_j$.

Extend the first two groups to a basis of $K_j$ using vectors from the third. The vectors taken from group 3 are $T_j$. (In practice: put the list in as columns of a matrix and row reduce; the pivot columns from group 3 are $T_j$.)

Each $v \in T_j$ is the top of a new chain of length exactly $j$:

\[v,\; Nv,\; \dots,\; N^{j-1}v.\]

Count check: $\lvert H_j \rvert + \lvert T_j \rvert = \mu’_j$. Every vector of $H_j$ should survive as a pivot; if one doesn’t, there is an arithmetic error upstream.

Step 2 (push down). Set

\[H_{j-1} := \{\,Nw \;:\; w \in H_j \cup T_j\,\}.\]

Step 3 (loop). Set $j := j-1$. If $j \ge 1$, go to Step 1. If $j = 0$, stop.

At $j = 1$, group 1 is empty ($K_0 = {0}$), so Step 1 is ordinary extension of $H_1$ to a basis of $\operatorname{null}N$.

Output. For each $v \in T_j$ (over all $j$), the chain $v, Nv, \dots, N^{j-1}v$. Concatenating all chains gives a Jordan basis.

Verification. You should have $\lvert T_j \rvert = 2d_j - d_{j-1} - d_{j+1}$ chains of length $j$, and $\sum_j j\,\lvert T_j \rvert = \dim V$ vectors in total.

Why it works. Let $\pi_j : K_j \to K_j/K_{j-1}$ be the quotient map. Step 1 is really “extend $\pi_j(H_j)$ to a basis of $K_j/K_{j-1}$, then lift the added cosets to vectors”; the append-and-extend recipe is exactly that computation done without forming the quotient. Different lifts give different, equally valid Jordan bases.

The engine. For $j \ge 2$ the map

\[\bar N_j : K_j/K_{j-1} \longrightarrow K_{j-1}/K_{j-2}, \qquad v + K_{j-1} \longmapsto Nv + K_{j-2},\]

is well defined ($v \in K_j \Rightarrow Nv \in K_{j-1}$, and $v \in K_{j-1} \Rightarrow Nv \in K_{j-2}$) and injective ($Nv \in K_{j-2}$ means $N^{j-1}v = 0$, i.e. $v \in K_{j-1}$). The restriction $j \ge 2$ is just so that $K_{j-2}$ exists; the smallest case is $\bar N_2 : K_2/K_1 \to K_1$.

The square commutes.

\[\begin{array}{ccc} K_j & \xrightarrow{\ \ N\ \ } & K_{j-1} \\[2pt] \big\downarrow{\scriptstyle \pi_j} & & \big\downarrow{\scriptstyle \pi_{j-1}} \\[2pt] K_j/K_{j-1} & \xrightarrow{\ \bar N_j\ } & K_{j-1}/K_{j-2} \end{array}\]

For $w \in K_j$:

\[\pi_{j-1}(Nw) = Nw + K_{j-2} = \bar N_j\big(w + K_{j-1}\big) = \bar N_j\big(\pi_j(w)\big).\]

Loop invariant. Step 1 at level $j$ can only run if $H_j \subseteq K_j$ and $\pi_j(H_j)$ is independent in $K_j/K_{j-1}$, since a dependent list cannot be extended to a basis. Step 2 is what re-establishes this one level down. Applying the commuting square elementwise to the list $H_j \cup T_j \subseteq K_j$,

\[\pi_{j-1}(H_{j-1}) = \pi_{j-1}\big(N(H_j \cup T_j)\big) = \bar N_j\big(\pi_j(H_j \cup T_j)\big),\]

where $\pi_j(H_j \cup T_j)$ is a basis of $K_j/K_{j-1}$ by Step 1 and $\bar N_j$ is injective. So:

• $H_{j-1} \subseteq K_{j-1}$, since $w \in K_j \Rightarrow N^{j-1}(Nw) = N^jw = 0$. This clause is what makes $\pi_{j-1}(H_{j-1})$ a well-formed expression at all.
• $\pi_{j-1}(H_{j-1})$ is independent in $K_{j-1}/K_{j-2}$, since injective maps preserve independence.

That is precisely the precondition Step 1 needs at level $j-1$, so the loop hands itself a valid input each time and nothing carried is ever discarded. The argument runs for $j \ge 2$, which is all that is needed. At $j = p$ both clauses hold vacuously, since $H_p = \varnothing$. At $j = 1$ every $w \in H_1 \cup T_1$ lies in $K_1 = \operatorname{null}N$, so Step 2 produces only $H_0 = {0,\dots,0}$, but Step 3 stops there and $H_0$ is never used.

Free consequence, useful for hand-checking: $\pi_j(H_j \cup T_j)$ is a basis and $\bar N_j$ is injective, so pushing down neither merges two chains nor loses one:

\[\lvert H_{j-1}\rvert = \lvert H_j\rvert + \lvert T_j\rvert = \mu'_j.\]

Chains have the advertised length. Each $v \in T_j$ has $\pi_j(v) \neq 0$, i.e. $v \notin K_{j-1}$, so $N^{j-1}v \neq 0$ while $N^jv = 0$. Passing to the quotient is what rules out a chain dying early; plain linear independence would not.

The chains form a basis. By induction on $j$: the chain vectors at levels $\le j-1$ form a basis of $K_{j-1}$, and $H_j \cup T_j$ lifts a basis of $K_j/K_{j-1}$, so together they give a basis of $K_j$. At $j = p$ this is a basis of $V$, and each chain-span has matrix $J_j(0)$.

The counts. Two facts combine. First, Step 1 makes $\pi_j(H_j \cup T_j)$ a basis of $K_j/K_{j-1}$, so the level is exactly as wide as the dimension jump:

\[\lvert H_j \cup T_j\rvert = \dim\left(K_j/K_{j-1}\right) = \mu'_j.\]

Second, each chain visits each level at most once. Let a chain have top $v \in T_i$, so $N^{i-1}v \neq 0$ and $N^iv = 0$. Its member $N^tv$ then satisfies

\[N^{i-t-1}\left(N^tv\right) = N^{i-1}v \neq 0, \qquad N^{i-t}\left(N^tv\right) = N^iv = 0,\]

so $N^tv$ lies in $K_{i-t}$ but not in $K_{i-t-1}$, putting it at level $i-t$. As $t$ runs from $0$ to $i-1$ the level runs from $i$ down to $1$: one member at each level from $1$ to $i$, and nothing above level $i$. Step 2 files that member into $T_j$ if $j = i$ and into $H_j$ if $j < i$. Hence $H_j \cup T_j$ is in bijection with the chains of length $\ge j$, and

\[\#\{\text{chains of length} \ge j\} = \mu'_j.\]

Subtracting consecutive levels isolates the chains that stop at level $j$, which are exactly the ones born there, namely $T_j$:

\[\lvert T_j\rvert = \#\{\text{chains of length } = j\} = \mu'_j - \mu'_{j+1} = 2d_j - d_{j-1} - d_{j+1},\]

with the convention $\mu’_{p+1} = 0$.

The same subtraction reads as a statement about $\bar N_{j+1}$. Start from Step 2 one level up, which sets $H_j = N(H_{j+1} \cup T_{j+1})$. Pushing that definition through the commuting square gives

\[\pi_j(H_j) = \bar N_{j+1}\big(\pi_{j+1}(H_{j+1} \cup T_{j+1})\big),\]

and by Step 1 at level $j+1$ the list inside the parentheses is a basis of $K_{j+1}/K_j$. An injective map carries a basis of its domain to a basis of its range, so

\[\pi_j(H_j) \ \text{ is a basis of } \ \operatorname{range}\bar N_{j+1}, \qquad \dim\operatorname{range}\bar N_{j+1} = \mu'_{j+1}.\]

Now recall the general principle: if a basis of a subspace $W’ \subseteq W$ is extended to a basis of $W$, the added vectors represent a basis of $W/W’$. Step 1 at level $j$ performs exactly such an extension, with

\[W' = \operatorname{range}\bar N_{j+1}, \qquad W = K_j/K_{j-1},\]

and the added vectors are $\pi_j(T_j)$. So $\pi_j(T_j)$ represents a basis of

\[W/W' = \big(K_j/K_{j-1}\big)\big/\operatorname{range}\bar N_{j+1} = \operatorname{coker}\bar N_{j+1},\]

of dimension \(\mu'_j - \mu'_{j+1} = \lvert T_j\rvert\). Level $j+1$ sends \(\mu'_{j+1}\) independent directions down into level $j$, level $j$ has \(\mu'_j\) directions to fill, and the new tops fill the shortfall. The cokernel is the part of level $j$ that nothing above it reaches. At $j = p$ there is no level above, so read $\operatorname{range}\bar N_{p+1} = {0}$, consistent with \(\mu'_{p+1} = 0\).

Do not run this bottom-up. Most vectors of $\operatorname{null}N$ are not in $\operatorname{range}N$, so choosing a basis of $\operatorname{null}N$ first and hunting for preimages typically fails. Top-down, low-level vectors are manufactured by pushing down, never guessed.

The algorithm’s count is the diagram’s count.

Level $j$ of the algorithm carries $\lvert H_j \cup T_j\rvert = d_j - d_{j-1}$ vectors, and level $j$ of the diagram has size $\mu’j = d_j - d{j-1}$. They agree because both are the same dimension jump, computed once from the null-space tower and once from the chain lengths. Neither derivation assumed the other.

The monotonicity \(\mu'_p \le \dots \le \mu'_1\) then has two independent proofs: levels widen as you descend, and $\bar N_{j}$ is injective.

Worked example

Let $V = \operatorname{span}(1,\, x,\, y,\, x^2,\, xy)$ and $N = \partial/\partial x$. Then $N$ sends

\[1 \mapsto 0,\quad x \mapsto 1,\quad y \mapsto 0,\quad x^2 \mapsto 2x,\quad xy \mapsto y.\]

Precompute: $\operatorname{null}N = \operatorname{span}(1, y)$, $\operatorname{null}N^2 = \operatorname{span}(1,y,x,xy)$, $\operatorname{null}N^3 = V$. So $p = 3$ and $(d_0,d_1,d_2,d_3) = (0,2,4,5)$.

So $\mu’ = (d_1 - d_0,\; d_2 - d_1,\; d_3 - d_2) = (2, 2, 1)$, whose conjugate is $\mu = (3,2)$: one chain of length $3$ and one of length $2$. The algorithm above recovers exactly this, but the shape is already determined by the null-space tower alone.

$j$	$H_j$ (pushed down)	need $d_j - d_{j-1}$	new tops $T_j$
$3$	$\varnothing$	$5-4 = 1$	${x^2}$
$2$	${2x}$	$4-2 = 2$	${xy}$
$1$	${2,\; y}$	$2-0 = 2$	$\varnothing$

Reading the levels:

• $j=3$: nothing carried in. Need $1$ vector of $\operatorname{null}N^3 = V$ outside $\operatorname{null}N^2$; take $x^2$. Push down: $H_2 = {2x}$.
• $j=2$: carrying $2x$. Need $2$ vectors total independent modulo $\operatorname{null}N = \operatorname{span}(1,y)$; $2x$ supplies one, so add one more from $\operatorname{null}N^2 = \operatorname{span}(1,y,x,xy)$ — take $xy$. Push down: $H_1 = {N(2x), N(xy)} = {2, y}$.
• $j=1$: carrying ${2,y}$, which is already a basis of the $2$-dimensional $\operatorname{null}N$. Nothing to add.

Chains: $x^2 \to 2x \to 2 \to 0$ and $xy \to y \to 0$. Jordan basis (each chain bottom-to-top)

\[(2,\; 2x,\; x^2,\; y,\; xy),\]

giving $\mathcal{M}(N) = J_3(0)\oplus J_2(0)$. Checks: $\lvert T_3 \rvert = 1 = 2d_3-d_2-d_4 = 10-4-5$, $\lvert T_2 \rvert = 1 = 2(4)-2-5$, $\lvert T_1 \rvert = 0 = 2(2)-0-4$, and $3 + 2 = 5 = \dim V$.

Row reduction (Gaussian elimination) example at level $j = 2$.

Coordinates are with respect to the ordered basis $(1, x, y, x^2, xy)$.

The list is: group 1 is $1, y$; group 2 is $2x$; group 3 is $1, y, x, xy$. As columns,

\[\left(\begin{array}{ccc|cccc} 1 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 2 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)\]

Reducing (here only row swaps are needed) puts pivots in columns $1, 2, 3, 7$. Columns $4$ and $5$ repeat group 1, and column $6$ is a scalar multiple of column $3$, so all three are non-pivots. The single pivot inside group 3 is column $7$, giving

\[T_2 = \{xy\},\]

which matches what was found by hand. The count check holds: $\lvert H_2\rvert + \lvert T_2\rvert = 1 + 1 = 2 = \mu’_2$.

• Full reduced echelon form is unnecessary. Forward elimination to echelon form already reveals the pivot positions, which is all you need.
• Group 3 can be any spanning set of $\operatorname{null}N^j$; a basis is the convenient choice, and duplicates cost nothing since they simply fail to be pivots.
• The same elimination produces the null spaces in the first place, by solving $N^j x = 0$, so you can compute the entire tower $\operatorname{null}N, \dots, \operatorname{null}N^p$ with the same tool before the loop starts.

Invertibility

Let $W$ be a vector space over $\mathbf{F}$, let $N \in \mathcal{L}(W)$ be nilpotent with $N^p = 0$, and let $\lambda \in \mathbf{F}$ with $\lambda \neq 0$. Then $\lambda I + N$ is invertible, and

\[(\lambda I+N)^{-1} \;=\; \sum_{t=0}^{p-1} \frac{(-1)^t}{\lambda^{\,t+1}}\,N^t .\]

Set $A = -\lambda^{-1}N$. Since scalars commute with everything, $A^p = (-\lambda^{-1})^pN^p = 0$, and

\[\lambda I + N = \lambda\left(I - A\right).\]

Let $G = \sum_{t=0}^{p-1}A^t$, a finite sum. Both products telescope:

\[(I-A)G \;=\; \sum_{t=0}^{p-1}A^t \;-\; \sum_{t=0}^{p-1}A^{t+1} \;=\; \sum_{t=0}^{p-1}A^t \;-\; \sum_{t=1}^{p}A^{t} \;=\; A^0 - A^p \;=\; I,\]

and the same computation with the factors reversed gives $G(I-A) = I$, since $A$ commutes with its own powers. So $I - A$ is invertible with inverse $G$, hence $\lambda I + N = \lambda(I-A)$ is invertible with inverse $\lambda^{-1}G$. Expanding $A$:

\[\lambda^{-1}G \;=\; \lambda^{-1}\sum_{t=0}^{p-1}\left(-\lambda^{-1}\right)^tN^t \;=\; \sum_{t=0}^{p-1}\frac{(-1)^t}{\lambda^{\,t+1}}N^t. \qquad \blacksquare\]

Worth noting what the argument does not use: no finite-dimensionality, no eigenvalues, no assumption on $\mathbf{F}$, and no separate case for $W = \{0\}$. The single input is that the series terminates, which is exactly what nilpotency provides. Over $\mathbb{R}$ or $\mathbb{C}$ this is the Neumann series for $(I+A)^{-1}$, with the convergence hypothesis replaced by the stronger fact that all but finitely many terms are $0$.

Any $p$ with $N^p = 0$ works; the index of nilpotency is simply the smallest choice, and larger $p$ only appends zero terms.

The inverse is a polynomial in $N$ of degree $< p$, so

The inverse is a polynomial in $N$ of degree $< p$. In particular any operator commuting with $N$ also commutes with $(\lambda I+N)^{-1}$, and the inverse of a Jordan block is again a polynomial in the nilpotent part.

Commuting Operators takes this further: for a single Jordan block, every operator commuting with $N$ is a polynomial in $N$, so the inverse being one is a special case rather than a coincidence.

Consistency with the block formula.

The Non-nilpotent Block section gives $A^r = \sum_{t}\binom{r}{t}\lambda^{r-t}N^t$ for $r \ge 0$. Read that at $r = -1$, where $\binom{-1}{t} = (-1)^t$:

\[A^{-1} = \sum_{t=0}^{m-1}(-1)^t\lambda^{-1-t}N^t,\]

which is the formula above. So the upper-triangular Toeplitz picture extends to negative exponents, and the geometric series is what proves the case $r = -1$.

The same proof gives more. If $S$ is invertible, $N$ is nilpotent, and $SN = NS$, then $S + N$ is invertible with

\[(S+N)^{-1} = \sum_{t=0}^{p-1}(-1)^t S^{-(t+1)}N^t.\]

Take $A = -S^{-1}N$; commutativity is what lets the powers separate, giving $A^p = (-1)^pS^{-p}N^p = 0$, and the telescoping runs unchanged. The theorem is the case $S = \lambda I$.

Commutativity is not decoration. Without it $A^p$ does not collapse, and the conclusion genuinely fails: an invertible operator plus a nilpotent one need not be invertible.

Worked example

Take $N = J_3(0)$ on $\mathbf{F}^3$, so $p = 3$ and $\lambda I + N = J_3(\lambda)$:

\[J_3(\lambda)^{-1} = \frac{1}{\lambda}I - \frac{1}{\lambda^2}N + \frac{1}{\lambda^3}N^2 = \begin{pmatrix} 1/\lambda & -1/\lambda^2 & 1/\lambda^3 \\ 0 & 1/\lambda & -1/\lambda^2 \\ 0 & 0 & 1/\lambda\end{pmatrix}.\]

Multiplying out confirms it, with everything past $N^2$ killed:

\[(\lambda I+N)\left(\tfrac{1}{\lambda}I - \tfrac{1}{\lambda^2}N + \tfrac{1}{\lambda^3}N^2\right) = I - \tfrac{1}{\lambda}N + \tfrac{1}{\lambda^2}N^2 + \tfrac{1}{\lambda}N - \tfrac{1}{\lambda^2}N^2 + \tfrac{1}{\lambda^3}N^3 = I.\]

The inverse is again upper-triangular Toeplitz, with alternating signs and increasing powers of $1/\lambda$ along successive diagonals. That shape is not an accident either: Commuting Operators shows that for a single chain the upper-triangular Toeplitz matrices are exactly the operators commuting with $N$.

What Next

The diagram determines $N$ up to similarity, and this post read it in one direction: from $N$ to its shape. Commuting Operators reads it in the other, showing that the same diagram controls which other operators commute with $N$ — with the two extremes above turning out to be the two extremes there as well.