Commuting Operators

This post continues Nilpotent Operators. It has two halves. The first collects what can be said about commuting operators in general, and ends with a reduction: for a complex vector space, the whole question collapses to the case of a nilpotent operator. The second half answers that case completely, using the Young diagram from Nilpotent Operators.

Notation

For $T \in \mathcal{L}(V)$, write

\[\mathcal{C}(T) = \{S \in \mathcal{L}(V) : ST = TS\}, \qquad \mathcal{P}(T) = \{q(T) : q \in \mathcal{P}(\mathbf{F})\}\]

for the commutant of $T$ and the operators expressible as polynomials in $T$. Both are subspaces of $\mathcal{L}(V)$, and in fact subalgebras: they are closed under composition.

$\mathcal{P}(T) \subseteq \mathcal{C}(T)$ 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 for nilpotent operators is the converse: when is every operator commuting with $T$ a polynomial in $T$?

Notation recall. A nilpotent $N \in \mathcal{L}(V)$ splits $V$ into Jordan chains with tops $v_1,\dots,v_k$ and lengths $m_1 \ge \dots \ge m_k$, forming a partition $\mu$ of $\dim V$ drawn as a bottom-aligned array of boxes with the chains as columns. Its conjugate $\mu’$ has $\mu’_j = \#\{i : m_i \ge j\}$, the size of level $j$, and $d_j = \dim\operatorname{null}N^j = \mu’_1 + \dots + \mu’_j$. Throughout, $p = m_1$ is the index of nilpotency and $n = \dim V$.

Commuting Operators in General

Nothing in this section assumes nilpotency.

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 $.

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 $.

simultaneous diagonalizability $\iff$ commutativity

Suppose $ \mathcal{E} $ is a subset of $ \mathcal{L}(V) $ and every element of $ \mathcal{E} $ is diagonalizable.

There exists a basis of $ V $ with respect to which every element of $ \mathcal{E} $ has a diagonal matrix $\iff$ every pair of elements of $ \mathcal{E} $ commutes.

($\Leftarrow$) Suppose every pair in $\mathcal{E}$ commutes. Induct on $n = \dim V$.

Case 1: every $T \in \mathcal{E}$ is a scalar multiple of $I$. Then every basis of $V$ works. (This covers $n = 1$, so the base case is free.)

Case 2: some $S \in \mathcal{E}$ is not a scalar multiple of $I$. Since $S$ is diagonalizable with eigenvalues $\lambda_1,\dots,\lambda_m$,

\[V = E(\lambda_1,S) \oplus \dots \oplus E(\lambda_m,S),\]

and $m \geq 2$ (otherwise $S = \lambda_1 I$). So each $E(\lambda_j, S)$ is a subspace of dimension strictly less than $n$.

Fix $j$ and write $W = E(\lambda_j, S)$. $W$ is invariant under every $T \in \mathcal{E}$ and each $\left. T \right\rvert_W$ is diagonalizable, so the restricted family commutes: for $T, R \in \mathcal{E}$ and $w \in W$, invariance gives $(\left. T \right\rvert_W)(\left. R \right\rvert_W)w = T(Rw) = (TR)w = (RT)w = (\left. R \right\rvert_W)(\left. T \right\rvert_W)w$.

So $\mathcal{E}_j = {\left. T\right\rvert_W : T \in \mathcal{E}}$ is a commuting family of diagonalizable operators on a space of dimension $< n$. By the induction hypothesis there is a basis $\mathcal{B}_j$ of $W$ making every element of $\mathcal{E}_j$ diagonal — that is, every vector of $\mathcal{B}_j$ is an eigenvector of $T$ for every $T \in \mathcal{E}$ simultaneously.

Now let $\mathcal{B} = \mathcal{B}_1 \cup \dots \cup \mathcal{B}_m$. Because $V$ is the direct sum of the $E(\lambda_j,S)$, this is a basis of $V$, and each of its vectors is an eigenvector of every $T \in \mathcal{E}$. So every element of $\mathcal{E}$ has a diagonal matrix with respect to $\mathcal{B}$. $\blacksquare$

Suppose $V$ is a finite-dimensional nonzero complex vector space. Suppose that $ \mathcal{E} \subset \mathcal{L}(V) $ is such that $S$ and $T$ commute for all $S,T \in \mathcal{E}$.

(a) There is a vector in $V$ that is an eigenvector for every element of $\mathcal{E}$.

(b) There is a basis of $V$ with respect to which every element of $\mathcal{E}$ has an upper-triangular matrix.

Reduction to the Nilpotent Case

The results above describe 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 $\mathbf{F} = \mathbb{C}$ and $V = \bigoplus_{\lambda_k} G(\lambda_k, T)$.

$ST = TS \iff G(\lambda_k,T) $ is invariant under $S$ and $\left. S \right\rvert_{G(\lambda_k,T)}$ commutes with $\left. (T-\lambda_k I) \right\rvert_{G(\lambda_k,T)}$ for each $ k = 1, \dots, m $.

This is the reduction that governs the rest of the post. On $G(\lambda_k, T)$ the operator $\left. T \right\rvert_{G(\lambda_k,T)}$ is $\lambda_k I + N_k$ with $N_k$ nilpotent, and adding a scalar multiple of $I$ changes nothing about what commutes with it:

\[\mathcal{C}(\lambda I + N) = \mathcal{C}(N).\]

So computing $\mathcal{C}(T)$ means computing $\mathcal{C}(N_k)$ for each block and assembling. Everything below therefore takes $N$ nilpotent, with no loss.

Parametrization

Suppose $SN = NS$. Then $SN^j = N^jS$ for all $j$, so for every basis vector

\[S(N^j v_i) = N^j (S v_i).\]

The left side is $S$ on an arbitrary basis vector; the right side only involves the $k$ vectors $Sv_1,\dots,Sv_k$. So a commuting $S$ is pinned down by its values on the $k$ tops.

Write $w_i = Sv_i$. Since $N^{m_i}v_i = 0$,

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

so the constraint is

\[\boxed{\,w_i \in \operatorname{null} N^{m_i}\,}\]

Each top may be sent anywhere killed by $N^{m_i}$.

Conversely, pick any $w_1,\dots,w_k$ with $w_i \in \operatorname{null}N^{m_i}$ and define $S$ on the basis by

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

$S$ commutes with $N$. Check on a basis vector $N^jv_i$:

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

So the constraint is precisely what’s needed at the bottom of each chain, and nowhere else. Conclusion:

\[\mathcal{C}(N) \;\cong\; \operatorname{null}N^{m_1} \times \dots \times \operatorname{null}N^{m_k}, \qquad S \mapsto (Sv_1,\dots,Sv_k),\] \[\dim\mathcal{C}(N) = \sum_{i,j}\min(m_i,m_j) = \sum_j \left(\mu'_j\right)^2.\]

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

The same box counting evaluates the double sum. Writing $\min(m_i,m_j) = \#\{l : l \le m_i \text{ and } l \le m_j\}$ turns it into a count of triples $(i,j,l)$:

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

the sum of squares of the level sizes. So the commutant is large exactly when the diagram is short and wide — many chains of similar length — and smallest when the diagram is a single column. The two extremes are worked out at the end of this post.

1. Find a Jordan basis for $N$, with tops $v_1,\dots,v_k$ and lengths $m_1,\dots,m_k$.
2. For each $i$, compute $\operatorname{null}N^{m_i}$ and choose any $w_i$ in it.
3. Define $S(N^tv_i) := N^tw_i$ for $0 \le t \le m_i - 1$. Done: $S$ commutes with $N$, and every commuting operator is obtained this way.

Nothing above says $Sv_i$ has to lie in chain $i$. The reason is visible in the constraint itself: $\operatorname{null}N^{m_i}$ 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. In the extreme case where all lengths equal $p$, we get $\operatorname{null}N^{m_i} = \operatorname{null}N^p = V$ and there is no constraint whatsoever.

Take $\mathbf{F}^4$ with two chains of length $2$:

\[e_2 \to e_1 \to 0, \qquad e_4 \to e_3 \to 0.\]

Here $N^2 = 0$, so both constraints read $w_i \in \mathbf{F}^4$ and any pair of tops is legal. Swapping the chains is the choice $w_1 = e_4$, $w_2 = e_2$, which forces

\[Se_2 = e_4, \quad Se_1 = Ne_4 = e_3, \qquad Se_4 = e_2, \quad Se_3 = Ne_2 = e_1.\]

No verification is needed, since the construction guarantees $SN = NS$, but as a spot-check: $SNe_2 = Se_1 = e_3$ and $NSe_2 = Ne_4 = e_3$. So $S$ commutes with $N$ while carrying chain $1$ onto chain $2$.

Shearing works the same way. Take $w_1 = e_2 + e_4$ and $w_2 = e_4$, giving

\[Se_2 = e_2 + e_4, \quad Se_1 = e_1 + e_3, \qquad Se_4 = e_4, \quad Se_3 = e_3.\]

This blends chain $1$ into chain $2$ without being a permutation of the basis.

The most extreme case is the one-level shape: $N = 0$ on $\mathbf{F}^2$, two chains of length $1$. Every operator commutes, so the commutant is all of $\mathcal{L}(\mathbf{F}^2)$, of dimension $4$; the ones preserving $\operatorname{span}(e_1)$ are the upper triangular matrices, a proper subspace of dimension $3$. Preserving a chosen block is a genuine restriction that most commuting operators fail.

Unequal lengths are where the constraint bites. Take $\mathbf{F}^3$ with chains

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

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.

Worked example

Take $N = \partial/\partial x$ on $V = \operatorname{span}(1,x,y,x^2,xy)$ again, with the Jordan basis found earlier: chain $x^2 \to 2x \to 2$ and chain $xy \to y$. 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 is visible: $w_1$ may be $xy$, but $w_2$ may not be $x^2$, since $N^2x^2 = 2 \neq 0$. The short chain cannot be sent to the top of the long one.

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 handles the 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.

Exercise

Give an example of two commuting operators $S,T$ on $\mathbf{F}^4$ such that there is a subspace of $\mathbf{F}^4$ that is invariant under $S$ but not under $T$ and there is a subspace of $\mathbf{F}^4$ that is invariant under $T$ but not under $S$.

Take $\mathbf{F}^4$ with $N e_2 = e_1$, $Ne_4 = e_3$, $Ne_1 = Ne_3 = 0$: two chains of length $2$, tops $e_2, e_4$. Both nulls are all of $\mathbf{F}^4$, so choose $w_1 = e_4$, $w_2 = e_2$, giving the chain swap

\[S: e_2 \mapsto e_4,\ e_1 \mapsto e_3,\ e_4 \mapsto e_2,\ e_3 \mapsto e_1.\]

Then $SN = NS$ by construction, and

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

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 the commutant sees them as opposite too — one as small as it can be, the other as large. Each is worth working out on its own before comparing them.

One Column: a Single Chain

If $k = 1$ with length $m$, then $\operatorname{null}N^m = V$, so there is no constraint at all and $S$ is determined by an arbitrary $w = Sv$. Since $k = 1$, the chain

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

is a basis of $V$, so expand $w$ in it:

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

Now compare $S$ with $q(N)$ on each basis vector $N^tv$, for $0 \le t \le m-1$:

\[S(N^tv) \;=\; N^t(Sv) \;=\; N^t\big(q(N)v\big) \;=\; q(N)(N^tv).\]

The first equality is $SN^t = N^tS$, from $SN = NS$. The second substitutes $Sv = w = q(N)v$. The third is $N^tq(N) = q(N)N^t$, since polynomials in $N$ commute with powers of $N$.

So $S$ and $q(N)$ agree on a basis of $V$, hence are equal as operators:

\[S = c_0I + c_1N + \dots + c_{m-1}N^{m-1}.\]

In the basis $(N^{m-1}v,\dots,Nv,v)$ this is the upper-triangular Toeplitz matrix with $c_0$ down the diagonal, $c_1$ on the next diagonal, and so on.

The inclusion $\mathcal{P}(N) \subseteq \mathcal{C}(N)$ holds for every nilpotent $N$. What the one-chain case gives is equality, and that is what fails when $k \ge 2$: the inclusion becomes strict, so “commutes with $N$” stops being the same condition as “is a polynomial in $N$”.

So for one column the commutant is exactly $\mathcal{P}(N)$, and $\dim\mathcal{C}(N) = n$. That turns out to characterize this shape.

For $N$ nilpotent with $k$ chains,

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

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

Recall $p = m_1$ is the index of nilpotency, so $N^p = 0$, and $\dim V = \sum_i m_i$. The inclusion $\mathcal{P}(N) \subseteq \mathcal{C}(N)$ always holds, so only equality is at issue.

$\dim\mathcal{P}(N) = p$. Spanning: any term of degree $\ge p$ vanishes on substituting $N$, since writing $q(z) = z^pa(z) + r(z)$ with $\deg r < p$ gives $q(N) = r(N)$. So $I, N, \dots, N^{p-1}$ spans $\mathcal{P}(N)$. Independence: choose $i_0$ with $m_{i_0} = p$, so the chain of $v_{i_0}$ has exactly $p$ entries. If $\sum_{t<p} c_tN^t = 0$, applying it to $v_{i_0}$ gives $\sum_{t<p} c_tN^tv_{i_0} = 0$, a relation among $p$ distinct members of the Jordan basis, so every $c_t = 0$.

$\dim\mathcal{C}(N) \ge \dim V$. By the parametrization above, $\dim\mathcal{C}(N) = \sum_{i,j}\min(m_i,m_j)$. View that as a $k \times k$ table with entry $(i,j)$ equal to $\min(m_i,m_j)$. Its diagonal entries are $\min(m_i,m_i) = m_i$, summing to $\dim V$, and every remaining entry is positive. So the bound holds, with equality exactly when the table has no off-diagonal entries, i.e. when $k = 1$.

Conclusion. If $k \ge 2$, then $\dim V = \sum_i m_i > \max_i m_i = p$, since the omitted lengths are positive, so

\[\dim\mathcal{C}(N) \ \ge\ \dim V \ >\ p \ =\ \dim\mathcal{P}(N)\]

and the inclusion is strict. If $k = 1$, the double sum has the single term $\min(m_1,m_1) = m_1 = \dim V = p$, so both spaces have dimension $p$, and an inclusion of subspaces of equal finite dimension is an equality. $\blacksquare$

The general version. For an arbitrary $T \in \mathcal{L}(V)$, the same statement reads: $\mathcal{C}(T) = \mathcal{P}(T)$ iff the minimal and characteristic polynomials of $T$ coincide — iff $T$ is cyclic, meaning some $v$ has $v, Tv, \dots, T^{n-1}v$ a basis of $V$. For nilpotent $N$ the minimal polynomial is $z^p$ and the characteristic polynomial is $z^n$, so the condition is $p = n$: a single chain filling all of $V$.

($\Leftarrow$) If $\deg(\text{min poly}) = n$, then $V$ is cyclic: there is $v$ with $v, Tv, \dots, T^{n-1}v$ a basis. Given $S \in \mathcal{C}(T)$, write $Sv = q(T)v$ for some polynomial $q$ (possible since that list spans $V$). Then for each $j$,

\[S(T^j v) = T^j(Sv) = T^j q(T) v = q(T)(T^j v),\]

so $S$ and $q(T)$ agree on a basis, hence $S = q(T)$.

($\Rightarrow$) Use $\dim \mathcal{P}(T) = \deg(\text{min poly})$ together with $\dim \mathcal{C}(T) \geq n$, with equality exactly when $T$ is cyclic — proved for nilpotent $T$ in One Column below. If $\mathcal{C}(T) = \mathcal{P}(T)$ then $\deg(\text{min poly}) = \dim\mathcal{C}(T) \geq n$, and since the minimal polynomial always divides the characteristic one, degree $n$ forces them equal. $\blacksquare$

Worked example

Take the one-column case, $n = 3$: $e_3 \to e_2 \to e_1 \to 0$; that is, $Ne_3 = e_2$, $Ne_2 = e_1$, $Ne_1 = 0$, so

\[N = \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}.\]

Suppose $SN = NS$. Whatever $S$ does to the top vector, say $Se_3 = w$, everything else is forced:

\[Se_2 = S(Ne_3) = N(Se_3) = Nw, \qquad Se_1 = S(Ne_2) = N(Se_2) = N^2w.\]

So $S$ is completely determined by the single vector $w$. Conversely, any $w$ works: the only condition left to check is $S(Ne_1) = N(Se_1)$, i.e. $0 = N^3w$, which is automatic.

Write $w = a e_1 + b e_2 + c e_3$. Then $Se_3 = ae_1+be_2+ce_3$, $Se_2 = be_1 + ce_2$, $Se_1 = ce_1$, so

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

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 $\mathcal{P}(N) = \{c I : c \in \mathbf{F}\}$ is just the scalars, of dimension $1$. This is the failure of $\mathcal{C}(N) = \mathcal{P}(N)$ at its most extreme: $n^2$ against $1$.

The $\mathbf{F}^2$ case appears in the warning box above: with $N = 0$ every operator commutes, yet only the upper-triangular ones preserve $\operatorname{span}(e_1)$. Commuting with $N$ constrains an operator not at all here, which is exactly why it cannot force any chain structure to be preserved.

Comparing the Ends

Side by side, the two shapes bracket every invariant in play.

	one column	one level
$\mu$	$(n)$	$(1^n)$
$\mu’$	$(1^n)$	$(n)$
$\dim\mathcal{C}(N)$	$n$	$n^2$
$\dim\mathcal{P}(N) = p$	$n$	$1$
$\mathcal{C}(N) = \mathcal{P}(N)$?	yes	no, unless $n = 1$

The two dimensions computed above, $n$ and $n^2$, are not merely far apart: they are the smallest and largest values $\dim\mathcal{C}(N)$ can take.

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 for one column and on the right exactly for one level.

Both bounds fall out of $\dim\mathcal{C}(N) = \sum_j(\mu’_j)^2$. Since each $\mu’_j \ge 1$,

\[\sum_j (\mu'_j)^2 \;\ge\; \sum_j \mu'_j \;=\; n,\]

with equality iff every level has size $1$ — a single column. And since all terms are non-negative,

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

with equality iff only one term is nonzero — a single level. $\blacksquare$

Why the commutant measures how non-unique a Jordan basis is.

The invertible elements of $\mathcal{C}(N)$ act simply transitively on the Jordan bases of a fixed shape: given two of them, exactly one invertible operator commuting with $N$ carries the first to the second. So $\dim\mathcal{C}(N)$ is the dimension of the choice available, which is where the counts in Nilpotent Operators came from — every basis of $V$ when $N = 0$ ($n^2$ worth of freedom), and only the choice of a top when $k = 1$ ($n$ worth).

The complementary count is just as clean. The operators similar to $N$ form a set of dimension

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

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.

What the polynomials miss. The characteristic polynomial of a nilpotent $N$ is $z^n$ and its minimal polynomial is $z^p$, so between them they see the total number of boxes and the height of the tallest column — and nothing else. That is not enough to pin down the shape. The smallest example is $n = 4$:

\[\mu = (2,2) \qquad\text{and}\qquad \mu = (2,1,1)\]

both have characteristic polynomial $z^4$ and minimal polynomial $z^2$, yet $\mu’ = (2,2)$ against $(3,1)$, so $\dim\operatorname{null}N$ is $2$ against $3$ and the operators are not similar. Their commutants differ too, of dimension $8$ against $10$.

The diagram is the complete invariant; the two polynomials are its two most easily computed shadows.