Similarity

Two operators are similar when they are the same operator seen in different coordinates. This post makes that precise through change of basis, then asks the natural question: given $S$ and $T$, how can you tell whether they are similar?

The earlier posts collected the pieces of the answer. Here they are put together: first the quantities similarity preserves, then the theorem that some of them suffice when the minimal polynomial splits, and finally what changes over $\mathbb{R}$ and over other fields.

Notation

Symbol	Meaning
$\mathbf{F}$	the field $\mathbb{R}$ or $\mathbb{C}$
$n = \dim V$	dimension of $V$
$\mathcal{L}(V)$	linear operators $V \to V$
$\mathcal{M}(T)$	matrix of $T$ with respect to a chosen basis
$E(\lambda,T) = \operatorname{null}(T - \lambda I)$	eigenspace; $\lambda$ is an eigenvalue $\iff E(\lambda,T) \ne {0}$
$G(\lambda,T) = \operatorname{null}(T - \lambda I)^n$	generalized eigenspace
$g_\lambda = \dim E(\lambda,T)$	geometric multiplicity
$d_\lambda = \dim G(\lambda,T)$	algebraic multiplicity
$e_\lambda$	exponent of $(z-\lambda)$ in the minimal polynomial
$J_s(\lambda)$	Jordan block of size $s$ for $\lambda$
$\sigma(T)$	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
\(\mathcal{C}(T) = \{S \in \mathcal{L}(V) : ST = TS\}\)	commutant (or centralizer) of $T$; the operator-algebra notation ${T}^{\prime}$ is avoided here since $T^{\prime}$ is the dual map
$T_{\mathbb{C}}$	complexification of a real operator $T$

Throughout, $S, T \in \mathcal{L}(V)$ and $n = \dim V$.

Change of Basis and Similarity

Write $\mathcal{M}(T, (u_1, \dots, u_n), (v_1, \dots, v_n))$ for the matrix of $T$ with the $u$’s as the basis of the domain and the $v$’s as the basis of the target, and $\mathcal{M}(T, (u_1, \dots, u_n))$ when both are the $u$’s.

Change of basis. Let $u_1, \dots, u_n$ and $v_1, \dots, v_n$ be bases of $V$, and let $C = \mathcal{M}(I, (u_1, \dots, u_n), (v_1, \dots, v_n))$, the change-of-basis matrix (or transition matrix) whose $k$-th column expresses $u_k$ in the $v$’s. Then $C$ is invertible with $C^{-1} = \mathcal{M}(I, (v_1, \dots, v_n), (u_1, \dots, u_n))$, and for every $T \in \mathcal{L}(V)$

\[\mathcal{M}(T, (u_1, \dots, u_n)) = C^{-1}\,\mathcal{M}(T, (v_1, \dots, v_n))\,C.\]

Conversely, given the $v$’s, every invertible $C \in \mathbf{F}^{n,n}$ is a change-of-basis matrix: set $u_k = \sum_j C_{j,k}v_j$.

Matrices multiply like the maps they represent: the matrix of a composition, with matching bases in the middle, is the product of the matrices. So $\mathcal{M}(I, (v), (u))\,\mathcal{M}(I, (u), (v)) = \mathcal{M}(I, (u)) = I$, and similarly in the other order, which proves the claim about $C^{-1}$. Writing $T = ITI$,

\[\mathcal{M}(T, (u)) = \mathcal{M}(I, (v), (u))\;\mathcal{M}(T, (v))\;\mathcal{M}(I, (u), (v)) = C^{-1}\,\mathcal{M}(T, (v))\,C.\]

For the converse, the $u_k$ are the images of the $v_k$ under the operator whose matrix in the $v$’s is $C$; that operator is invertible, so the $u_k$ form a basis, and $\mathcal{M}(I, (u), (v)) = C$ by construction. $\blacksquare$

Similarity. $A, B \in \mathbf{F}^{n,n}$ are similar, written $A \sim B$, if $A = C^{-1}BC$ for some invertible $C \in \mathbf{F}^{n,n}$. Likewise $S, T \in \mathcal{L}(V)$ are similar if $S = C^{-1}TC$ for some invertible $C \in \mathcal{L}(V)$.

The two readings agree: for any basis $v_1, \dots, v_n$, $S \sim T$ iff $\mathcal{M}(S, (v)) \sim \mathcal{M}(T, (v))$, iff

\[\mathcal{M}(S, (v_1, \dots, v_n)) = \mathcal{M}(T, (u_1, \dots, u_n))\]

for some basis $u_1, \dots, u_n$. The conjugating matrix is then the change-of-basis matrix $\mathcal{M}(I, (u), (v))$.

$\mathcal{M}(\cdot, (v))$ is a bijection $\mathcal{L}(V) \to \mathbf{F}^{n,n}$ that turns composition into product, so $S = C^{-1}TC$ iff $\mathcal{M}(S, (v)) = \mathcal{M}(C, (v))^{-1}\,\mathcal{M}(T, (v))\,\mathcal{M}(C, (v))$. This gives the first equivalence. For the second, apply the change-of-basis formula and its converse: $\mathcal{M}(S, (v)) = C^{-1}\mathcal{M}(T, (v))C$ for an invertible $C$ iff $\mathcal{M}(S, (v)) = \mathcal{M}(T, (u))$ for the basis $u$ with $\mathcal{M}(I, (u), (v)) = C$. $\blacksquare$

So similar matrices are the matrices of one operator in two bases, and similar operators are those with the same matrix in suitable bases; everything below can be read for matrices or for operators. The map $T \mapsto C^{-1}TC$ is called conjugation by $C$, or a similarity transformation, and the similarity class of $T$ is the set $\{C^{-1}TC : C \text{ invertible}\}$ of operators similar to $T$.

Similarity Invariants

A quantity attached to operators is a similarity invariant if similar operators always share it.

Suppose $S = C^{-1}TC$ with $C$ invertible. Then:

(a) $q(S) = C^{-1}q(T)C$ for every polynomial $q$, so $S$ and $T$ have the same minimal polynomial.

(b) For every $\lambda \in \mathbf{F}$ and $j \ge 0$, $C^{-1}$ maps $\operatorname{null}(T - \lambda I)^j$ isomorphically onto $\operatorname{null}(S - \lambda I)^j$, so

\[\dim\operatorname{null}(S - \lambda I)^j = \dim\operatorname{null}(T - \lambda I)^j.\]

(a) is the conjugation row of the table in Polynomials: $T \mapsto C^{-1}TC$ is an algebra isomorphism, so it commutes with substituting into $q$, and $q(S) = 0$ iff $q(T) = 0$.

(b) By (a) with $q(z) = (z - \lambda)^j$, $(S - \lambda I)^j = C^{-1}(T - \lambda I)^jC$. So $(S - \lambda I)^jx = 0$ iff $(T - \lambda I)^j(Cx) = 0$, that is, $x \in \operatorname{null}(S - \lambda I)^j$ iff $Cx \in \operatorname{null}(T - \lambda I)^j$. Hence $C^{-1}$ maps the second null space onto the first, injectively since $C^{-1}$ is invertible. $\blacksquare$

Every invariant met so far in the series is read off from (a) and (b):

invariant	read off from	defined in
eigenvalues $\sigma(T)$	$\lambda$ with $\dim\operatorname{null}(T - \lambda I) > 0$	Eigenvalues and Eigenvectors
geometric multiplicity $g_\lambda$	$j = 1$	Generalized Eigenspaces
algebraic multiplicity $d_\lambda$	$j = n$	Generalized Eigenspaces
minimal polynomial $p_T$ and its exponents $e_\lambda$	(a)	Polynomials
characteristic polynomial $\prod_\lambda (z - \lambda)^{d_\lambda}$ (split case)	the $d_\lambda$	Generalized Eigenspaces
diagram $\mu^{(\lambda)}$ of the nilpotent part at each $\lambda$	all $j$: $\mu^{(\lambda)\prime}_j$ is the jump from $j-1$ to $j$	The Nilpotent Part
$\dim\mathbf{F}[T] = \deg p_T$	(a)	Polynomials
$\dim\mathcal{C}(T)$	$\mathcal{C}(C^{-1}TC) = C^{-1}\mathcal{C}(T)C$	Commuting Operators

The diagram row is the finest: every other row in the split case is computed from the eigenvalues and their diagrams. The diagram of the nilpotent part $N_\lambda$ is determined by $\dim\operatorname{null}N_\lambda^j$, and $\operatorname{null}N_\lambda^j = \operatorname{null}(T - \lambda I)^j$ because these null spaces already lie in $G(\lambda, T)$.

The trace and the determinant are also similarity invariants: in the split case they are the sum and the product of the eigenvalues, each repeated $d_\lambda$ times, so they are read off from the characteristic polynomial. Like the characteristic polynomial, they are far from complete: $I$ and $J_2(1)$ on $\mathbf{F}^2$ share both, yet $I$ is similar only to itself.

Classification When the Minimal Polynomial Splits

The diagrams are not only the finest invariant on the list. Together with the eigenvalues, they are a complete invariant: they decide similarity.

Suppose the minimal polynomials of $S$ and $T$ split over $\mathbf{F}$ (automatic when $\mathbf{F} = \mathbb{C}$). The following are equivalent:

(a) $S \sim T$.

(b) $\dim\operatorname{null}(S - \lambda I)^j = \dim\operatorname{null}(T - \lambda I)^j$ for every $\lambda \in \mathbf{F}$ and every $j \ge 1$.

(c) $\sigma(S) = \sigma(T)$, and for each $\lambda$ the nilpotent parts of $S$ and $T$ at $\lambda$ have the same diagram.

(a) $\Rightarrow$ (b) is part (b) of the previous theorem.

(b) $\Rightarrow$ (c): taking $j = 1$, the same $\lambda$ have nonzero eigenspaces, so $\sigma(S) = \sigma(T)$. For each $\lambda$, the null spaces of the powers of the nilpotent part at $\lambda$ are the $\operatorname{null}(\,\cdot\, - \lambda I)^j$, as noted after the table, so their dimensions agree for every $j$. The diagram is determined by these dimensions, since $\mu’_j$ is the jump from $j-1$ to $j$.

(c) $\Rightarrow$ (a): since the minimal polynomials split, $V = \bigoplus_{\lambda \in \sigma(T)} G(\lambda, S) = \bigoplus_{\lambda \in \sigma(T)} G(\lambda, T)$. Fix $\lambda$ with diagram $\mu = (m_1, \dots, m_k)$. The nilpotent parts of $S$ on $G(\lambda, S)$ and of $T$ on $G(\lambda, T)$ have Jordan bases of the same shape, and with respect to them both have the matrix $J_{m_1}(0) \oplus \dots \oplus J_{m_k}(0)$. Adding $\lambda I$, the restrictions of $S$ and $T$ both have the matrix $J_{m_1}(\lambda) \oplus \dots \oplus J_{m_k}(\lambda)$. Concatenating these bases over all $\lambda$ gives a basis of $V$ for $S$ and one for $T$ with respect to which the two operators have the same block diagonal matrix, so $S \sim T$ by the characterization of similarity. $\blacksquare$

Condition (b) is a finite computation: $\lambda$ only needs to range over the eigenvalues, and $j$ only up to $n$, after which the null spaces stop growing. In matrix terms, it compares the ranks of $(A - \lambda I)^j$ and $(B - \lambda I)^j$.

The proof of (c) $\Rightarrow$ (a) also produces a canonical representative of each similarity class.

Jordan form. If the minimal polynomial of $T$ splits, then $T$ has a matrix of the form

\[\bigoplus_{\lambda \in \sigma(T)} \Big( J_{m^{(\lambda)}_1}(\lambda) \oplus \dots \oplus J_{m^{(\lambda)}_{k_\lambda}}(\lambda) \Big),\]

where \(\mu^{(\lambda)} = \big(m^{(\lambda)}_1 \ge \dots \ge m^{(\lambda)}_{k_\lambda}\big)\) is the diagram of the nilpotent part at $\lambda$. Two such operators are similar iff their Jordan forms agree up to the order of the eigenvalues.

Similarity Classes

By the classification, the similarity classes of operators whose minimal polynomial splits are indexed by the data in (c): a set of eigenvalues with a diagram for each, or equivalently a Jordan form up to the order of the eigenvalues.

The operator conjugating $T$ to a given $S$ is far from unique. If $S = C^{-1}TC$ and $C’ = ZC$, then $C’^{-1}TC’ = C^{-1}(Z^{-1}TZ)C$, so $C’$ also conjugates $T$ to $S$ iff $Z \in \mathcal{C}(T)$. The conjugators are therefore the $ZC$ with $Z \in \mathcal{C}(T)$ invertible. Commuting Operators turns this into a dimension count: the invertible $C$ have $n^2$ degrees of freedom, and each member of the class absorbs $\dim\mathcal{C}(T)$ of them, so the class has dimension

\[n^2 - \dim\mathcal{C}(T).\]

Since $\dim\mathcal{C}(T)$ is computed from the diagrams, so is the size of each class. At one extreme, every invertible $C$ fixes a scalar operator $\lambda I$, so its class is the single point $\{\lambda I\}$. At the other, an operator whose minimal and characteristic polynomials coincide has $\dim\mathcal{C}(T) = n$ and the largest class, of dimension $n^2 - n$.

When the Polynomials Suffice

The classification explains when the cheap invariants, the two polynomials, already decide similarity. The characteristic polynomial records the eigenvalues and each $d_\lambda$, the number of boxes in the diagram at $\lambda$; the minimal polynomial records each $e_\lambda$, the height of its tallest column. So the two polynomials are a complete invariant exactly when no two diagrams with the same number of boxes have the same tallest column.

Suppose the minimal polynomials of $S$ and $T$ split.

(a) If $S$ and $T$ are diagonalizable, then $S \sim T$ iff they have the same characteristic polynomial.

(b) If $n \le 3$, then $S \sim T$ iff they have the same characteristic and minimal polynomials.

(a) A diagonalizable operator has every diagram a single level, $\mu^{(\lambda)} = (1, \dots, 1)$ with $d_\lambda$ boxes. So the diagrams are determined by the $d_\lambda$, which the characteristic polynomial records.

(b) Each diagram has $d_\lambda \le 3$ boxes. The partitions of $1$, $2$, $3$ are $(1)$; $(2), (1,1)$; $(3), (2,1), (1,1,1)$, and within each size the largest part tells them apart. So each diagram is determined by $d_\lambda$ and $e_\lambda$. $\blacksquare$

At $n = 4$ this fails for the first time: the nilpotent diagrams $(2,2)$ and $(2,1,1)$ have the same number of boxes and the same tallest column, so they share both polynomials, but they are not similar.

A Matrix Is Similar to Its Transpose

If the minimal polynomial of $A \in \mathbf{F}^{n,n}$ splits, then $A \sim A^{t}$.

By the Jordan form, $A = C^{-1}JC$ with $J = \bigoplus J_m(\lambda)$, so $A^{t} = C^{t}J^{t}(C^{t})^{-1}$ is similar to $J^{t} = \bigoplus J_m(\lambda)^{t}$. It is enough to show $J_m(\lambda)^{t} \sim J_m(\lambda)$ for each block. Let $P$ be the $m \times m$ matrix with $1$s on the antidiagonal, so $P^{-1} = P$ and $PXP$ has entries $(PXP){i,j} = X{m+1-i,\,m+1-j}$: it reverses the order of the basis. The superdiagonal entries $(i, i+1)$ of $J_m(\lambda)$ move to positions $(m+1-i, m-i)$, which is the subdiagonal, and the diagonal stays put. So $PJ_m(\lambda)P = J_m(\lambda)^{t}$. $\blacksquare$

For operators this says $T$ and its dual map $T’$ have similar matrices, since $\mathcal{M}(T’) = \mathcal{M}(T)^t$ for dual bases. The statement holds over every field; that needs the rational canonical form below.

Beyond the Split Case

Without the splitting hypothesis there may be too few eigenvalues for the classification above.

Real Operators

Over $\mathbb{R}$ the minimal polynomial need not split, and rotations have no Jordan form. Complexification resolves this: it does not change whether two real operators are similar.

Suppose $A, B \in \mathbb{R}^{n,n}$. If $A$ and $B$ are similar over $\mathbb{C}$, they are similar over $\mathbb{R}$.

Let $P \in \mathbb{C}^{n,n}$ be invertible with $PA = BP$, and write $P = X + iY$ with $X, Y$ real. Since $A$ and $B$ are real, comparing real and imaginary parts gives $XA = BX$ and $YA = BY$, so $(X + tY)A = B(X + tY)$ for every real $t$.

It remains to make $X + tY$ invertible for some real $t$. The function $f(t) = \det(X + tY)$ is a polynomial in $t$, and $f(i) = \det P \ne 0$, so $f$ is not the zero polynomial and has finitely many zeros. Choose a real $t_0$ that is not one of them. Then $Q = X + t_0Y$ is real and invertible with $QA = BQ$, so $A = Q^{-1}BQ$. $\blacksquare$

With respect to a basis of $V$, a real operator $T$ and its complexification $T_{\mathbb{C}}$ have the same matrix. So two real operators are similar iff their complexifications are, and the classification above, applied over $\mathbb{C}$, decides it: compare the eigenvalues of $S_{\mathbb{C}}$ and $T_{\mathbb{C}}$, now possibly non-real, and their diagrams.

For a real $T$, the complex Jordan form is symmetric under conjugation. Write $K$ for conjugation on $V_{\mathbb{C}}$ (called $C$ in Complexification, but $C$ is taken here). It satisfies $K(T_{\mathbb{C}} - \lambda I)^j = (T_{\mathbb{C}} - \bar\lambda I)^jK$, as in Complexification, so it carries each $\operatorname{null}(T_{\mathbb{C}} - \lambda I)^j$ onto $\operatorname{null}(T_{\mathbb{C}} - \bar\lambda I)^j$ and the diagrams at $\lambda$ and $\bar\lambda$ agree. Pairing each block $J_m(\lambda)$ with its conjugate $J_m(\bar\lambda)$ and choosing real bases of the paired spaces gives the real Jordan form, where each pair becomes a $2m \times 2m$ block built from rotation-scaling $2 \times 2$ blocks.

Other Fields

Over a general field, a Jordan form exists only after passing to a larger field in which the minimal polynomial splits. A canonical form within $\mathbf{F}$ itself replaces the eigenvalues by the monic irreducible factors of the minimal polynomial, and the classification keeps the same shape.

Rational canonical form. For arbitrary $\mathbf{F}$, $S \sim T$ iff

\[\dim\operatorname{null}q(S)^j = \dim\operatorname{null}q(T)^j \quad\text{for every monic irreducible } q \in \mathcal{P}(\mathbf{F}) \text{ and every } j \ge 1.\]

For $q = z - \lambda$ these are the conditions in (b) above. The canonical representative replaces Jordan blocks by companion matrices of the powers $q^m$, using the primary decomposition $V = \bigoplus_q \operatorname{null}q(T)^{r_q}$ in place of the generalized eigenspaces. The proof goes through the structure theorem for finitely generated modules over the polynomial ring $\mathbf{F}[z]$, which is beyond this series. Two consequences: $A \sim A^t$ over every field, and two matrices over $\mathbf{F}$ that are similar over a larger field are already similar over $\mathbf{F}$, generalizing the real case above.