Commuting Operators

This post continues Nilpotent Operators and Eigenvalues and Eigenvectors, and asks which operators commute with a given $T \in \mathcal{L}(V)$. The first part collects what can be said about commuting operators in general, and ends with a reduction: over $\mathbb{C}$, describing the operators that commute with $T$ comes down to the same question for a nilpotent operator. The second part answers that question completely using the Young diagram, and the last section carries the answer back to an arbitrary $T$.

A map of the post:

flowchart TD
    P["Commuting families"] --> Q["Joint generalized eigenspaces"]
    Q --> R["Simultaneous diagonalization"]
    P --> R2["Simultaneous upper triangularization"]
    Q -.->|"one operator"| B
    A["Operator T over ℂ"] --> B["Split: V = ⊕ G(λ,T)<br/>reduction to the nilpotent case"]
    B --> C["On each G(λ,T): T = λI + N_λ<br/>C(λI + N) = C(N)"]
    C --> D["Nilpotent N with a Jordan basis<br/>tops v_i, lengths m_i"]
    D --> E["Description of the commutant<br/>S ↔ (Sv_i), Sv_i ∈ null N^(m_i)"]
    E --> F["dim C(N) = Σ (μ'_j)²<br/>between n and n²"]
    F --> G{"Single chain?"}
    G -->|"Yes: one column"| H["C(N) = F[N]"]
    G -->|No| I["F[N] ⊊ C(N)"]
    E --> K["Assemble the blocks<br/>C(T) ≅ ∏ C(N_λ)"]
    K --> L{"Every N_λ a single chain?<br/>i.e. min poly = char poly"}
    H -.-> L
    L -->|Yes| M["C(T) = F[T]"]
    L -->|No| O["F[T] ⊊ C(T)"]

Notation

Symbol	Meaning
$n = \dim V$	dimension of $V$
$\mathcal{L}(V)$	linear operators $V \to V$
$\mathcal{P}(\mathbf{F})$	polynomials with coefficients in $\mathbf{F}$
$E(\lambda,T) = \operatorname{null}(T - \lambda I)$	eigenspace; $\lambda$ is an eigenvalue $\iff E(\lambda,T) \ne {0}$
$G(\lambda,T) = \operatorname{null}(T - \lambda I)^n$	generalized eigenspace
$\sigma(T)$	spectrum: the set of eigenvalues of $T$; $\lvert \sigma(T) \rvert$ is the number of distinct eigenvalues
$p_T$	minimal polynomial of $T$
$\mu = (m_1 \ge \dots \ge m_k)$	Jordan chain lengths of a nilpotent operator, as a partition of $\dim V$; the Segre characteristic
$\mu’$	conjugate partition of $\mu$; $\mu’_j$ is the size of level $j$; the Weyr characteristic
$p$	index of nilpotency: the smallest exponent with $N^p = 0$, equal to $m_1$
\(\mathcal{C}(T) = \{S \in \mathcal{L}(V) : ST = TS\}\)	commutant (or centralizer) of $T$; the operator-algebra notation ${T}^{\prime}$ is avoided here since $T^{\prime}$ is the dual map
\(\mathbf{F}[T] = \{q(T) : q \in \mathcal{P}(\mathbf{F})\}\)	the algebra of polynomials in $T$: the operators expressible as $q(T)$

Both $\mathcal{C}(T)$ and $\mathbf{F}[T]$ are subspaces of $\mathcal{L}(V)$, and in fact subalgebras: they are closed under composition.

$\mathbf{F}[T] \subseteq \mathcal{C}(T)$ always, since $T$ commutes with $I$ and with itself, hence with every polynomial in itself. That direction is free, and it is not the interesting one. The question this post answers is the converse: when is every operator commuting with $T$ a polynomial in $T$?

Basic Facts

The smaller of the two spaces has a dimension that is easy to read off: $I, T, \dots, T^{\deg p_T - 1}$ is a basis of $\mathbf{F}[T]$, so $\dim\mathbf{F}[T] = \deg p_T$.

Shifting an operator by a scalar changes neither space.

For every $\lambda \in \mathbf{F}$, $\mathcal{C}(T + \lambda I) = \mathcal{C}(T)$ and $\mathbf{F}[T + \lambda I] = \mathbf{F}[T]$.

$S(T + \lambda I) - (T + \lambda I)S = ST - TS$, so $S$ commutes with $T + \lambda I$ iff it commutes with $T$. For the second, $q(T + \lambda I) = r(T)$ for the polynomial $r(z) := q(z + \lambda)$, so $\mathbf{F}[T + \lambda I] \subseteq \mathbf{F}[T]$; the same with $-\lambda$ gives the reverse inclusion. $\blacksquare$

Commuting Families

Call a subset $\mathcal{E} \subseteq \mathcal{L}(V)$ a commuting family if $ST = TS$ for all $S, T \in \mathcal{E}$. The family may be infinite. The first lemma needs only two of its members.

Invariant Subspaces

Suppose $ S,T \in \mathcal{L}(V) $ are such that $ ST = TS $. Suppose $ q \in \mathcal{P}(\mathbf{F}) $. Then

$ \operatorname{null} q(S) $ and $ \operatorname{range} q(S) $ are invariant under $ T $.

$T$ commutes with every power of $S$, hence with $q(S)$. If $q(S)v = 0$, then $q(S)(Tv) = T\,q(S)v = 0$; and $T\,q(S)v = q(S)(Tv) \in \operatorname{range}q(S)$. $\blacksquare$

Special cases:

• $ q(z) = z - \lambda $, then $ E(\lambda, S) $ is invariant under $ T $.
• $ q(z) = (z - \lambda)^{\dim V} $, then $ G(\lambda, S) $ is invariant under $ T $.

Commuting operators share the subspaces $\operatorname{null}q(S)$ and $\operatorname{range}q(S)$, but not invariant subspaces in general. On $\mathbf{F}^4$, let $N$ send $e_2 \mapsto e_1 \mapsto 0$ and $e_4 \mapsto e_3 \mapsto 0$, and let $S$ swap $e_1 \leftrightarrow e_3$ and $e_2 \leftrightarrow e_4$. Then $SN = NS$: both send $e_2 \mapsto e_3$, $e_4 \mapsto e_1$, and $e_1, e_3 \mapsto 0$. But

• $\operatorname{span}(e_1)$ is invariant under $N$, since $Ne_1 = 0$, but not under $S$, since $Se_1 = e_3$;
• $\operatorname{span}(e_2 + e_4)$ is invariant under $S$, since $S(e_2 + e_4) = e_2 + e_4$, but not under $N$, since $N(e_2 + e_4) = e_1 + e_3$.

Restrictions and Quotients

Suppose $U$ is invariant under every element of a commuting family $\mathcal{E}$. On the operators leaving $U$ invariant, restriction $T \mapsto \left.T\right\rvert_U$ and passage to the quotient operator $T \mapsto T/U$ are unital algebra homomorphisms. So the restrictions, and likewise the quotient operators, again form a commuting family, and the minimal polynomials of $\left.T\right\rvert_U$ and $T/U$ divide $p_T$. In particular, restricting or passing to the quotient keeps split minimal polynomials split.

Joint Generalized Eigenspaces

The primary decomposition extends from one operator to a commuting family. Simultaneous diagonalization will follow from it at once.

Suppose $\mathcal{E} \subseteq \mathcal{L}(V)$ is a commuting family whose minimal polynomials all split. For each function $\alpha : \mathcal{E} \to \mathbf{F}$, define the joint generalized eigenspace

\[V_\alpha = \bigcap_{T \in \mathcal{E}} G(\alpha(T), T).\]

Then only finitely many $V_\alpha$ are nonzero, and

\[V = \bigoplus_\alpha V_\alpha.\]

Each $V_\alpha$ is invariant under every element of $\mathcal{E}$, and $\left.T\right\rvert_{V_\alpha} - \alpha(T)I$ is nilpotent for each $T \in \mathcal{E}$.

Two claims follow directly from the definition. Each $G(\mu, T)$ is invariant under every element of $\mathcal{E}$ by Invariant Subspaces, hence so is the intersection $V_\alpha$. And $V_\alpha \subseteq G(\alpha(T), T)$, so $\left.T\right\rvert_{V_\alpha} - \alpha(T)I$ is nilpotent (vacuously when $V_\alpha = \{0\}$, which happens whenever some $\alpha(T)$ is not an eigenvalue of $T$).

It remains to show $V = \bigoplus_\alpha V_\alpha$ with finitely many nonzero terms. Induct on $\dim V$.

Case 1: every $T \in \mathcal{E}$ has a single eigenvalue $\beta(T)$. Its minimal polynomial splits with the single zero $\beta(T)$, so $G(\beta(T), T) = V$ for every $T$, and $V_\beta = V$. For $\alpha \ne \beta$, some $T$ has $\alpha(T) \ne \beta(T)$. Then $\alpha(T)$ is not an eigenvalue of $T$, so $G(\alpha(T), T) = \{0\}$ and $V_\alpha = \{0\}$.

Case 2: some $S \in \mathcal{E}$ has distinct eigenvalues $\lambda_1, \dots, \lambda_m$ with $m \ge 2$. Then $V = \bigoplus_{j=1}^m G_j$ with $G_j = G(\lambda_j, S)$, each a proper subspace invariant under every element of $\mathcal{E}$, and by Restrictions and Quotients the restrictions $\left.T\right\rvert_{G_j}$, for $T \in \mathcal{E}$, form a commuting family with split minimal polynomials. Since $\dim G_j < \dim V$, the induction hypothesis applies to this family on $G_j$ and gives

\[G_j = \bigoplus_\alpha \bigcap_{T \in \mathcal{E}} G\big(\alpha(T), \left.T\right\rvert_{G_j}\big).\]

For each $T \in \mathcal{E}$ and scalar $\mu$, $G(\mu, \left.T\right\rvert_{G_j}) = G_j \cap G(\mu, T)$, since both consist of the vectors of $G_j$ killed by a power of $T - \mu I$. Intersecting over all $T \in \mathcal{E}$ turns each summand into $G_j \cap V_\alpha$, so

\[G_j = \bigoplus_\alpha \big(G_j \cap V_\alpha\big).\]

Now $V_\alpha \subseteq G(\alpha(S), S)$, which is $G_j$ if $\alpha(S) = \lambda_j$ and $\{0\}$ if $\alpha(S)$ is not an eigenvalue of $S$. Since distinct $G_j$ meet only in $0$, $G_j \cap V_\alpha$ is $V_\alpha$ when $\alpha(S) = \lambda_j$ and $\{0\}$ otherwise, and

\[V = \bigoplus_{j=1}^m G_j = \bigoplus_{j=1}^m \bigoplus_{\alpha \,:\, \alpha(S) = \lambda_j} V_\alpha = \bigoplus_\alpha V_\alpha,\]

with finitely many nonzero terms because each $G_j$ has finitely many. $\blacksquare$

For a single operator this is the primary decomposition $V = \bigoplus_{\lambda \in \sigma(T)} G(\lambda, T)$.

Simultaneous Diagonalization

A subset $\mathcal{E} \subseteq \mathcal{L}(V)$ is simultaneously diagonalizable if there is a basis of $V$ with respect to which every element of $\mathcal{E}$ has a diagonal matrix.

Suppose every element of $\mathcal{E} \subseteq \mathcal{L}(V)$ is diagonalizable. Then $\mathcal{E}$ is a commuting family $\iff$ $\mathcal{E}$ is simultaneously diagonalizable.

($\Rightarrow$) Suppose $\mathcal{E}$ is a commuting family. A diagonalizable operator’s minimal polynomial is a product of distinct linear factors, so it splits, and Joint Generalized Eigenspaces gives $V = \bigoplus_\alpha V_\alpha$. Each $T \in \mathcal{E}$ is diagonalizable, so none of its eigenvalues is defective: $G(\lambda, T) = E(\lambda, T)$ for every $\lambda$. Hence $V_\alpha \subseteq G(\alpha(T), T) = E(\alpha(T), T)$, that is, $\left.T\right\rvert_{V_\alpha} = \alpha(T)I$.

So every nonzero vector of $V_\alpha$ is an eigenvector of every element of $\mathcal{E}$. Concatenating bases of the nonzero $V_\alpha$ gives a basis of $V$ made of common eigenvectors, with respect to which every element of $\mathcal{E}$ has a diagonal matrix.

($\Leftarrow$) For $R, T \in \mathcal{E}$, the matrices $\mathcal{M}(R)$ and $\mathcal{M}(T)$ with respect to the common basis are diagonal, and diagonal matrices commute, so $\mathcal{M}(RT) = \mathcal{M}(R)\mathcal{M}(T) = \mathcal{M}(T)\mathcal{M}(R) = \mathcal{M}(TR)$ and $RT = TR$. $\blacksquare$

Simultaneous Upper Triangularization

Without diagonalizability, commuting operators can still be triangularized together, as long as each one can be triangularized on its own. A subset $\mathcal{E} \subseteq \mathcal{L}(V)$ is simultaneously upper-triangularizable if there is a basis of $V$ with respect to which every element of $\mathcal{E}$ has an upper-triangular matrix.

Suppose every element of $\mathcal{E} \subseteq \mathcal{L}(V)$ is upper-triangularizable (automatic when $\mathbf{F} = \mathbb{C}$). Then $\mathcal{E}$ is a commuting family $\implies$ $\mathcal{E}$ is simultaneously upper-triangularizable.

An operator is upper-triangularizable iff its minimal polynomial splits, so every minimal polynomial in $\mathcal{E}$ splits. Suppose $\mathcal{E}$ is a commuting family, and induct on $n = \dim V$; the cases $n \le 1$ are trivial. By Restrictions and Quotients, the induction hypothesis applies to the restrictions to, and quotients by, any subspace invariant under all of $\mathcal{E}$.

A common eigenvector. If every element of $\mathcal{E}$ is a scalar multiple of $I$, any nonzero $v_1$ is an eigenvector of all of them. Otherwise pick $S \in \mathcal{E}$ that is not. Its minimal polynomial splits and has positive degree, so $S$ has an eigenvalue $\lambda$, and $W = E(\lambda, S)$ is nonzero, not all of $V$ since $S \ne \lambda I$, and invariant under every element of $\mathcal{E}$ by Invariant Subspaces, since each element commutes with $S$. By the induction hypothesis, $W$ has a basis making every restriction to $W$ upper triangular. Its first vector $v_1$ is an eigenvector of every restriction, since the first column of an upper-triangular matrix is zero below the diagonal, hence of every element of $\mathcal{E}$.

The remaining vectors. Let $U = \operatorname{span}(v_1)$, which is invariant under every $T \in \mathcal{E}$. By the induction hypothesis, $V/U$ has a basis $v_2 + U, \dots, v_n + U$ making every $T/U$ upper triangular. Each $v_j + U$ is nonzero, so $v_j \notin U = \operatorname{span}(v_1)$ for $j \ge 2$; by Block Matrix, the whole list $v_1, v_2, \dots, v_n$ is a basis of $V$, and with respect to it every $T \in \mathcal{E}$ has matrix

\[\mathcal{M}(T) = \begin{pmatrix} c_T & \ast \\ 0 & \mathcal{M}(T/U) \end{pmatrix},\]

where $\mathcal{M}(\left.T\right\rvert_U) = (c_T)$ is $1 \times 1$ because $Tv_1 = c_Tv_1$. Since $\mathcal{M}(T/U)$ is upper triangular, so is $\mathcal{M}(T)$. $\blacksquare$

In particular, a commuting family of upper-triangularizable operators on a nonzero $V$ has a common eigenvector: the first vector of such a basis, since $Tv_1 \in \operatorname{span}(v_1)$ for every $T \in \mathcal{E}$.

Only diagonalization comes with a converse. Diagonal matrices commute, so a common diagonal basis forces commuting; upper-triangular matrices generally do not commute, so a common triangular basis forces nothing.

Jordan form does not pass to commuting families. Let $N$ be a single chain on $\mathbf{F}^3$, which commutes with $N^2$. The only Jordan form of $N$ is $J_3(0)$, so any Jordan basis $b_1, b_2, b_3$ for $N$ has $Nb_3 = b_2$, $Nb_2 = b_1$, $Nb_1 = 0$. In it $N^2b_3 = b_1$, a $1$ in the top-right corner, which is not a Jordan matrix. So $N$ and $N^2$ have no common Jordan basis.

Reduction to the Nilpotent Case

The results above concern commuting families. To describe the commutant of a single operator, the first move is to break $V$ into generalized eigenspaces, on each of which the operator is a scalar plus a nilpotent.

Suppose $\mathbf{F} = \mathbb{C}$. For each $\lambda \in \sigma(T)$ write $G_\lambda = G(\lambda, T)$, so that $V = \bigoplus_{\lambda \in \sigma(T)} G_\lambda$, and let $N_\lambda = \left.(T - \lambda I)\right\rvert_{G_\lambda}$, which is nilpotent. For $S$ leaving every $G_\lambda$ invariant, let

\[\Psi(S) = \big(\left.S\right\rvert_{G_\lambda}\big)_{\lambda \in \sigma(T)}.\]

(a) $ST = TS \iff$ each $G_\lambda$ is invariant under $S$ and $\left. S \right\rvert_{G_\lambda}$ commutes with $N_\lambda$.

(b) $\Psi$ is an isomorphism

\[\mathcal{C}(T) \;\cong\; \prod_{\lambda \in \sigma(T)} \mathcal{C}(N_\lambda), \qquad \text{so} \qquad \dim\mathcal{C}(T) = \sum_{\lambda \in \sigma(T)} \dim\mathcal{C}(N_\lambda).\]

(c) $\Psi$ restricts to an isomorphism

\[\mathbf{F}[T] \;\cong\; \prod_{\lambda \in \sigma(T)} \mathbf{F}[N_\lambda].\]

(a) ($\Rightarrow$) $G_\lambda = \operatorname{null}(T - \lambda I)^n$ is invariant under $S$ by Invariant Subspaces, with the roles of $S$ and $T$ swapped. Restricting $ST = TS$ to $G_\lambda$ and subtracting $\lambda \left.S\right\rvert_{G_\lambda}$ from both sides gives $\left.S\right\rvert_{G_\lambda}N_\lambda = N_\lambda\left.S\right\rvert_{G_\lambda}$.

($\Leftarrow$) $\left.S\right\rvert_{G_\lambda}$ commutes with $N_\lambda + \lambda I = \left.T\right\rvert_{G_\lambda}$, so $STv = TSv$ for $v$ in each $G_\lambda$, hence for every $v \in V$.

(b) By (a), $\Psi$ maps $\mathcal{C}(T)$ into the product. It is linear. It is injective: if $\Psi(S) = 0$, then $S$ is $0$ on every $G_\lambda$, and every $v \in V$ is a sum of vectors from the $G_\lambda$, so $S = 0$. For surjectivity, given $R = (R_\lambda)$ with $R_\lambda \in \mathcal{C}(N_\lambda)$, define $S_R$ componentwise: write each $v \in V$ uniquely as $v = \sum_\lambda v_\lambda$ with $v_\lambda \in G_\lambda$, and set $S_Rv = \sum_\lambda R_\lambda v_\lambda$. For $v \in G_\lambda$ this gives $S_Rv = R_\lambda v \in G_\lambda$, so each $G_\lambda$ is $S_R$-invariant with $\left.S_R\right\rvert_{G_\lambda} = R_\lambda$, and $S_R \in \mathcal{C}(T)$ by (a), and $\Psi(S_R) = R$.

(c) For every polynomial $q$, $\left.q(T)\right\rvert_{G_\lambda} = q(\lambda I + N_\lambda)$, which lies in $\mathbf{F}[\lambda I + N_\lambda] = \mathbf{F}[N_\lambda]$ by shift invariance. So $\Psi$ maps $\mathbf{F}[T]$ into $\prod_\lambda \mathbf{F}[N_\lambda]$, injectively by (b). Write $p_T = \prod_\lambda (z - \lambda)^{e_\lambda}$. The exponent $e_\lambda$ is the index of nilpotency of $N_\lambda$, so the minimal polynomial of $N_\lambda$ is $z^{e_\lambda}$. Since $\dim\mathbf{F}[T] = \deg p_T$, both sides have dimension $\sum_\lambda e_\lambda$, and an injective linear map between spaces of the same finite dimension is an isomorphism. $\blacksquare$

By shift invariance, $\mathcal{C}(\left.T\right\rvert_{G_\lambda}) = \mathcal{C}(\lambda I + N_\lambda) = \mathcal{C}(N_\lambda)$. Everything below therefore takes $N$ nilpotent, with no loss; Back to Arbitrary Operators assembles the blocks.

(c) is a Chinese remainder theorem for operators: choose a polynomial in $N_\lambda$ independently for each eigenvalue, and a single polynomial in $T$ realizes all the choices at once. Choosing $I$ on one $G_\lambda$ and $0$ on the others recovers the fact that the projections onto the generalized eigenspaces are polynomials in $T$.

The Commutant of a Nilpotent Operator

From here on $N \in \mathcal{L}(V)$ is nilpotent, with a Jordan basis of tops $v_1, \dots, v_k$ and lengths $m_1 \ge \dots \ge m_k$, so the vectors $N^jv_i$ with $0 \le j \le m_i - 1$ form a basis of $V$.

Description of the Commutant

Unlike the generalized eigenspaces in the reduction, the Jordan chains are not preserved by commuting operators (see the warning below), so the commutant cannot be split chain by chain. Instead, a commuting operator is pinned down by where it sends the $k$ tops, and each top may go anywhere killed by the right power of $N$.

The map

\[\Phi : \mathcal{C}(N) \to \operatorname{null}N^{m_1} \times \dots \times \operatorname{null}N^{m_k}, \qquad \Phi(S) = (Sv_1,\dots,Sv_k)\]

is an isomorphism. Its inverse sends $w = (w_1, \dots, w_k)$ to the operator $S_w$ defined on the Jordan basis by

\[S_w(N^jv_i) := N^j w_i, \qquad 0 \le j \le m_i - 1.\]

Into the product. If $SN = NS$, then $SN^j = N^jS$ for all $j$. Since $N^{m_i}v_i = 0$,

\[0 = S(N^{m_i}v_i) = N^{m_i}(Sv_i),\]

so $Sv_i \in \operatorname{null}N^{m_i}$. $\Phi$ is clearly linear.

Injective. Suppose $\Phi(S) = 0$, i.e. $Sv_1 = \dots = Sv_k = 0$. Then on every basis vector $S(N^jv_i) = N^j(Sv_i) = 0$, so $S = 0$.

Surjective. Given $w = (w_1, \dots, w_k)$ with $w_i \in \operatorname{null}N^{m_i}$, define $S_w$ as in the statement. Taking $j = 0$ gives $S_wv_i = w_i$, so $\Phi(S_w) = w$ once $S_w \in \mathcal{C}(N)$. Check $S_wN = NS_w$ on a basis vector $N^jv_i$:

• If $j < m_i - 1$: $\;S_wN(N^jv_i) = S_w(N^{j+1}v_i) = N^{j+1}w_i$, while $NS_w(N^jv_i) = N(N^jw_i) = N^{j+1}w_i$. Equal.
• If $j = m_i-1$ (bottom of the chain): $\;S_wN(N^{m_i-1}v_i) = S_w(0) = 0$, while $NS_w(N^{m_i-1}v_i) = N^{m_i}w_i = 0$ by the constraint. $\blacksquare$

In practice: find a Jordan basis, choose each $w_i$ in $\operatorname{null}N^{m_i}$, and extend by $S_w(N^jv_i) := N^jw_i$. Every commuting operator arises this way, exactly once.

The invertible elements of $\mathcal{C}(N)$ stay invertible when $N$ is added, with an explicit inverse; see Invertible Plus Nilpotent.

Nothing in the description of the commutant says $Sv_i$ has to lie in chain $i$. The constraint space $\operatorname{null}N^{m_i}$ is not a subspace of chain $i$: it is spanned by the bottom $\min(m_i, m_j)$ vectors of every chain $j$, so it reaches across the whole array. When all lengths equal $p$, $\operatorname{null}N^{m_i} = \operatorname{null}N^p = V$ and there is no constraint whatsoever; the swap $S$ in Invariant Subspaces is such an operator, carrying one chain of $N$ onto the other.

With unequal lengths the constraint bites. Take $\mathbf{F}^3$ with chains

\[e_2 \mapsto e_1 \mapsto 0, \qquad f \mapsto 0,\]

so $m_1 = 2$ and $m_2 = 1$. The top of the long chain is unconstrained, since $\operatorname{null}N^2 = \mathbf{F}^3$, so $Se_2$ may be $f$. But the top of the short chain satisfies $Sf \in \operatorname{null}N = \operatorname{span}(e_1, f)$, so $Sf$ cannot be $e_2$: a short chain can only be sent into the bottom portion of a longer one, never onto its top. The freedom is asymmetric, and $\min(m_i, m_j)$ is exactly the bookkeeping for it.

Dimension

Recall from Nilpotent Operators that $d_m = \dim\operatorname{null}N^m = \sum_j \min(m, m_j)$: $\operatorname{null}N^m$ is spanned by the bottom $\min(m, m_j)$ vectors of every chain $j$.

\[\dim\mathcal{C}(N) = \sum_{i,j}\min(m_i,m_j) = \sum_l \left(\mu'_l\right)^2.\]

By the description, $\dim\mathcal{C}(N) = \sum_i d_{m_i} = \sum_{i,j}\min(m_i, m_j)$. Writing $\min(m_i,m_j) = \#\{l : l \le m_i \text{ and } l \le m_j\}$ turns the double sum into a count of triples $(i,j,l)$:

\[\sum_{i,j}\min(m_i,m_j) \;=\; \sum_{l \ge 1}\#\{i : m_i \ge l\}\cdot\#\{j : m_j \ge l\} \;=\; \sum_{l\ge1}\left(\mu'_l\right)^2. \qquad \blacksquare\]

Here are the windows $\operatorname{null}N^{m_i}$ for $\mu = (3,2,1)$:

So $\dim\mathcal{C}(N)$ is the sum of squares of the level sizes: the commutant is large when the diagram is short and wide, with many chains of similar length, and small when it is tall and thin. The extremes are as far apart as they could be.

For every nilpotent $N$ on $V$ with $\dim V = n$,

\[n \;\le\; \dim\mathcal{C}(N) \;\le\; n^2,\]

with equality on the left exactly when the diagram is a single column ($k = 1$), and on the right exactly when it is a single level ($N = 0$).

Since each nonzero $\mu’_l$ is at least $1$,

\[\sum_l (\mu'_l)^2 \;\ge\; \sum_l \mu'_l \;=\; n,\]

with equality iff every level has size $1$, i.e. $k = \mu’_1 = 1$. And since all terms are non-negative,

\[\sum_l (\mu'_l)^2 \;\le\; \left(\sum_l \mu'_l\right)^2 \;=\; n^2,\]

with equality iff only one term is nonzero, i.e. $p = 1$ and $N = 0$. $\blacksquare$

Worked example

Take $N = \partial/\partial x$ on $V = \operatorname{span}(1,x,y,x^2,xy)$, with the Jordan basis found in Nilpotent Operators: chain $x^2 \mapsto 2x \mapsto 2$ and chain $xy \mapsto y$. So $v_1 = x^2$ with $m_1 = 3$, and $v_2 = xy$ with $m_2 = 2$.

Constraints: $w_1 \in \operatorname{null}N^3 = V$, unrestricted; $w_2 \in \operatorname{null}N^2 = \operatorname{span}(1,y,x,xy)$. The asymmetry from the warning is visible: $w_1$ may be $xy$, but $w_2$ may not be $x^2$, since $N^2x^2 = 2 \neq 0$.

Choose $w_1 = xy$ and $w_2 = 2x$. Then

\(S(x^2) = xy, \quad S(2x) = N(xy) = y, \quad S(2) = N^2(xy) = 0,\) \(S(xy) = 2x, \quad S(y) = N(2x) = 2.\)

Spot-check: $SN(x^2) = S(2x) = y$ and $NS(x^2) = N(xy) = y$; also $SN(xy) = S(y) = 2$ and $NS(xy) = N(2x) = 2$.

Dimension check: $\dim\operatorname{null}N^3 + \dim\operatorname{null}N^2 = 5 + 4 = 9$, agreeing with $\sum_{i,j}\min(m_i,m_j) = 3+2+2+2$ and with $\sum_l(\mu’_l)^2 = 4 + 4 + 1$, out of $25$ for all of $\mathcal{L}(V)$.

The isomorphism also gives a basis of the commutant for free: run over a basis of $\operatorname{null}N^{m_i}$ for one index with the other $w$’s set to $0$. Here that gives $5 + 4 = 9$ explicit commuting operators spanning the commutant.

The two extremes are worked out below.

Freedom in a Jordan Basis

Nilpotent Operators measures how far a Jordan basis is from unique by a number $r$: listed in a fixed order, a Jordan basis is a point of $V^n$, and $r$ is the dimension of a linear subspace of $V^n$ in which the Jordan bases form a nonempty open subset. It finds $r = n$ for one column and $r = n^2$ for one level by direct arguments. The description of the commutant gives $r$ for every shape.

Fix a Jordan basis of $N$ with tops $v_1, \dots, v_k$. Then $S \mapsto \big(S(N^jv_i)\big)$ is a one-to-one correspondence between the invertible elements of $\mathcal{C}(N)$ and the Jordan bases of the same shape. Consequently

\[r = \dim\mathcal{C}(N) = \sum_j (\mu'_j)^2.\]

A Jordan basis of the same shape is a list $(N^jw_i)$, with $w = (w_1, \dots, w_k) \in \prod_i \operatorname{null}N^{m_i}$, that is a basis of $V$. For such $w$, the description of the commutant gives $S_w \in \mathcal{C}(N)$ with $S_w(N^jv_i) = N^jw_i$, so the list is the image of the fixed basis under $S_w$, and it is a basis iff $S_w$ is invertible. Since $\Phi$ is an isomorphism, $S \leftrightarrow \Phi(S)$ matches the invertible elements of $\mathcal{C}(N)$ one-to-one with these Jordan bases.

For $r$: the linear map $w \mapsto (N^jw_i)$ from $\prod_i \operatorname{null}N^{m_i}$ into $V^n$ is injective (take $j = 0$), so its image is a linear subspace of dimension $\dim\mathcal{C}(N)$. Inside it, the Jordan bases are the points where $\det S_w \ne 0$. This determinant is a polynomial in the coordinates of $w$, and it is nonzero at $w = (v_1, \dots, v_k)$, where $S_w = I$. So the Jordan bases form a nonempty open subset of this subspace, and $r = \dim\mathcal{C}(N)$. $\blacksquare$

For one column this gives $r = n$, and for one level $r = n^2$, the two values found directly in Nilpotent Operators. In group language, the invertible elements of $\mathcal{C}(N)$ form a group under composition, and the theorem says it acts simply transitively on the Jordan bases of a given shape.

The similarity class. The operators similar to $N$, called the nilpotent orbit of $N$ in Lie theory, form a set of dimension

\[\dim\mathcal{L}(V) - \dim\mathcal{C}(N) \;=\; n^2 - \sum_j (\mu'_j)^2 .\]

This is a dimension count, in the spirit of rank–nullity. The map $C \mapsto CNC^{-1}$, defined on the invertible $C \in \mathcal{L}(V)$ (the changes of basis, a nonempty open subset of $\mathcal{L}(V)$, so of dimension $n^2$), sends them onto the similarity class of $N$, and $CNC^{-1} = DND^{-1}$ exactly when $D^{-1}C \in \mathcal{C}(N)$. So the operators giving any one output form a copy of the invertible elements of $\mathcal{C}(N)$, which by the same determinant argument are a nonempty open subset of $\mathcal{C}(N)$, of dimension $\dim\mathcal{C}(N)$. Subtracting this redundancy leaves $n^2 - \dim\mathcal{C}(N)$. Here the dimension of an open set means its dimension as a manifold over $\mathbb{R}$ or $\mathbb{C}$, or as an algebraic variety over a general field; making the count rigorous takes the orbit–stabilizer theorem for Lie groups or its algebraic-geometry analogue, since the map is not linear.

Freedom in the basis and size of the similarity class always sum to $n^2$. For $N = 0$ the class is a single point, since nothing is similar to the zero operator but itself; for a single chain it is $n^2 - n$, as large as a nilpotent operator’s class can be.

Two Extremes

Nilpotent Operators singles out two shapes. One column, $\mu = (n)$ and $\mu’ = (1^n)$, is a single chain. One level, $\mu = (1^n)$ and $\mu’ = (n)$, is the zero operator. They are conjugate partitions, and they are the two equality cases of $n \le \dim\mathcal{C}(N) \le n^2$: one commutant is as small as it can be, the other as large.

One Column: a Single Chain

If $k = 1$, the single chain $v, Nv, \dots, N^{n-1}v$ is a basis of $V$, and $\operatorname{null}N^n = V$, so $w = Sv$ is arbitrary. Expand it in the chain:

\[w = c_0v + c_1Nv + \dots + c_{n-1}N^{n-1}v = q(N)v, \qquad q(z) = c_0 + c_1z + \dots + c_{n-1}z^{n-1}.\]

Then on each basis vector, using $SN^t = N^tS$ and the fact that $q(N)$ commutes with $N^t$,

\[S(N^tv) \;=\; N^t(Sv) \;=\; N^tq(N)v \;=\; q(N)(N^tv),\]

so $S = q(N) = c_0I + c_1N + \dots + c_{n-1}N^{n-1}$.

In the basis $(N^{n-1}v,\dots,Nv,v)$, $N$ is $J_n(0)$ and $N^t$ is the band of $1$s on the $t$-th superdiagonal (Nilpotent Block), so $S$ has $c_t$ on the $t$-th superdiagonal: upper-triangular Toeplitz, the same shape as the powers of $J_n(\lambda)$ in Non-nilpotent Block. For $n = 3$, with $e_3 \mapsto e_2 \mapsto e_1 \mapsto 0$ and $Se_3 = w = ae_1 + be_2 + ce_3$,

\[N = \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}, \qquad S = \begin{pmatrix} c&b&a\\ 0&c&b\\ 0&0&c\end{pmatrix} = cI + bN + aN^2.\]

For $N$ nilpotent with $k$ chains,

\[\mathcal{C}(N) = \mathbf{F}[N] \iff k = 1,\]

that is, iff $N$ has a single Jordan block, equivalently $\dim\operatorname{null}N = 1$.

($\Leftarrow$) is the computation above.

($\Rightarrow$) The minimal polynomial of $N$ is $z^p$, so $\dim\mathbf{F}[N] = p$. If $k \ge 2$, then $n = \sum_i m_i > m_1 = p$, since the omitted lengths are positive, so by the lower bound

\[\dim\mathcal{C}(N) \ \ge\ n \ >\ p \ =\ \dim\mathbf{F}[N]\]

and the inclusion $\mathbf{F}[N] \subseteq \mathcal{C}(N)$ is strict. $\blacksquare$

One Level: the Zero Operator

The opposite shape is $N = 0$, where every chain has length $1$. Now the constraint $w_i \in \operatorname{null}N^{m_i}$ reads $w_i \in \operatorname{null}N = V$ for every $i$, so there is no constraint at all and every operator commutes:

\[\mathcal{C}(0) = \mathcal{L}(V), \qquad \dim\mathcal{C}(0) = n^2.\]

Meanwhile $N^1 = 0$ already, so $p = 1$ and $\mathbf{F}[N] = \{c I : c \in \mathbf{F}\}$ is just the scalars, of dimension $1$. This is the failure of $\mathcal{C}(N) = \mathbf{F}[N]$ at its most extreme: $n^2$ against $1$.

Comparing the Ends

	one column	one level
$\dim\mathcal{C}(N)$	$n$	$n^2$
$\dim\mathbf{F}[N] = p$	$n$	$1$
$\mathcal{C}(N) = \mathbf{F}[N]$?	yes	no, unless $n = 1$

Back to Arbitrary Operators

Over $\mathbb{C}$, the commutant of an arbitrary $T$ is found in three steps.

1. Split. Write $V = \bigoplus_{\lambda \in \sigma(T)} G_\lambda$. By the reduction, a commuting operator preserves each $G_\lambda$, and on $G_\lambda$ commuting with $\left.T\right\rvert_{G_\lambda} = \lambda I + N_\lambda$ is the same as commuting with $N_\lambda$.
2. Solve each block. Choose $R_\lambda \in \mathcal{C}(N_\lambda)$ by the description of the commutant: pick where the tops of a Jordan basis of $N_\lambda$ go.
3. Assemble. The operator $S_R$ acting as $R_\lambda$ on each $G_\lambda$ is the general element of $\mathcal{C}(T)$. In a basis made of bases of the $G_\lambda$, its matrix is block diagonal, with one block per eigenvalue.

Example

Take $T = \operatorname{diag}(2, 2, 5)$ on $\mathbf{F}^3$. Then $G_2 = \operatorname{span}(e_1, e_2)$ with $N_2 = 0$, so $R_2$ is any operator on $G_2$; and $G_5 = \operatorname{span}(e_3)$, so $R_5$ is any scalar. Hence

\[\mathcal{C}(T) = \left\{\begin{pmatrix} a&b&0\\ c&d&0\\ 0&0&e \end{pmatrix} : a,b,c,d,e \in \mathbf{F}\right\}, \qquad \dim\mathcal{C}(T) = 5,\]

while $p_T = (z-2)(z-5)$ gives $\dim\mathbf{F}[T] = 2$. So $\mathbf{F}[T] \subsetneq \mathcal{C}(T)$: here $N_2 = 0$ has two chains rather than one.

In general, with $\mu^{(\lambda)}$ the diagram of $N_\lambda$,

\[\dim\mathcal{C}(T) = \sum_{\lambda \in \sigma(T)} \dim\mathcal{C}(N_\lambda) = \sum_{\lambda \in \sigma(T)} \sum_l \left(\mu^{(\lambda)\prime}_l\right)^2.\]

The question from the start of the post now has a complete answer.

Suppose $\mathbf{F} = \mathbb{C}$. Then $\mathcal{C}(T) = \mathbf{F}[T]$ iff the minimal and characteristic polynomials of $T$ coincide.

By (b) and (c) of the reduction, $\Psi$ is injective on $\mathcal{C}(T)$ and carries $\mathcal{C}(T)$ onto $\prod_\lambda \mathcal{C}(N_\lambda)$ and $\mathbf{F}[T]$ onto $\prod_\lambda \mathbf{F}[N_\lambda]$. So $\mathcal{C}(T) = \mathbf{F}[T]$ iff $\mathcal{C}(N_\lambda) = \mathbf{F}[N_\lambda]$ for every $\lambda$, iff every $N_\lambda$ is a single chain, by the One Column theorem.

Finally, the exponents of $z - \lambda$ in the characteristic and minimal polynomials are the number of boxes and the height of the tallest column of the diagram of $N_\lambda$, which are equal iff that diagram is a single column. So the two polynomials coincide iff every $N_\lambda$ is a single chain. $\blacksquare$

The minimal and characteristic polynomials coincide iff $T$ is cyclic. For cyclic $T$, the One Column computation also gives $\mathcal{C}(T) = \mathbf{F}[T]$ directly, with $T$ in place of $N$: write $Sv = q(T)v$, then $S(T^jv) = T^jq(T)v = q(T)(T^jv)$, so $S = q(T)$.