Change of Basis and Similarity Transformations

A linear operator is a geometric object — a map that sends a vector space into itself — and it exists before anyone chooses coordinates. To compute with it we pick a basis and copy its action into a grid of numbers, a matrix. But that grid is a human convenience, glued to a particular choice of axes; it is not part of the physics. The organizing question of this article is the one every working physicist eventually meets: when we swap one basis for another, how do the numbers rearrange, and which combinations of numbers refuse to move — because those are the coordinate-free quantities that can carry physical meaning. The answer is a single operation, the similarity transformation. If $P$ is the matrix whose columns list the new basis vectors written in the old coordinates, then the component column of a fixed vector transforms as $[v]' = P^{-1}[v]$ while the operator matrix transforms by conjugation, $A' = P^{-1}AP$ . The two factors are inverses of one another for a reason that is almost forced: they must conspire so that the abstract statement $w = \mathcal{A}v$ comes out true in every basis at once. The operator never changes; only its bookkeeping does. Everything that survives conjugation is a candidate observable. The determinant and the trace are the two most elementary survivors; the full list is packaged in the characteristic polynomial, whose roots are the eigenvalues. This is precisely why eigenvalues — and not raw matrix entries — carry physics: they are the squared frequencies of a vibrating lattice, the energy levels of a Hamiltonian, the principal moments of an inertia tensor. Choosing a basis well is therefore a strategy. Diagonalization, the spectral theorem, and the Jordan form are all one idea worn three ways: pick $P$ to make $P^{-1}AP$ as simple as the operator will allow.

Key takeaways

A vector exists independently of any basis; only its component column changes when you re-express it, and the change is governed by an invertible change-of-basis matrix $P$ .
If $P$ holds the new basis vectors as columns in the old coordinates, then basis vectors carry $P$ while components carry $P^{-1}$ : $[v]' = P^{-1}[v]$ . This opposite behaviour is the contravariant–covariant duality.
An operator's matrix transforms by a similarity transformation $A' = P^{-1}AP$ . Two matrices related this way are the same operator seen in two bases; similarity is an equivalence relation.
Conjugation preserves every intrinsic property: determinant, trace, characteristic polynomial, eigenvalues, rank, and Jordan structure. These are the labels of the equivalence class.
Between two orthonormal bases $P$ is unitary (orthogonal in the real case), and similarity becomes $A' = U^{\dagger}AU$ — the transformation that preserves inner products, norms, and Hermiticity, hence the one quantum mechanics uses.
Key idea: a change of basis acts on operator matrices by conjugation, $A' = P^{-1}AP$ . The operator itself is untouched — only its coordinate description moves — so the intrinsic content of an operator is exactly what conjugation leaves fixed: det, trace, characteristic polynomial, eigenvalues, rank.

The core insight

A change of basis acts on operator matrices by conjugation: $A' = P^{-1}AP$ . This similarity transformation leaves the operator itself untouched; only its coordinate description changes. Hence the intrinsic content of an operator is precisely the set of quantities invariant under conjugation — determinant, trace, characteristic polynomial, eigenvalues, rank.

Where this lives today

The similarity transformation is the computational heartbeat of modern physics. Every time a quantum chemist or condensed-matter physicist "diagonalizes the Hamiltonian" to find energy levels, they are searching for the unitary $U$ that makes $U^{\dagger}HU$ diagonal, and for huge matrices iterative eigensolvers such as Lanczos build that basis change one Krylov vector at a time. Structural and mechanical engineers run the identical mathematics under the name modal analysis: the normal modes of a bridge, a turbine blade, or a spacecraft are the eigenvectors of $M^{-1}K$ and the resonant frequencies are the eigenvalues, exactly as in Example 2 but with thousands of degrees of freedom. In data science, principal component analysis is an orthogonal change of basis that diagonalizes a covariance matrix. In every case the lesson of this article is the operative one: choose the basis in which the operator is simplest, read off the invariants, and let the physics speak in coordinates that hide nothing.

The physical picture

Think of a vector as an arrow pinned in space, and a basis as a coordinate grid you draw on top of it. Sliding to a different grid does not move the arrow — it only changes the numbers you read off the arrow's tip. An operator is likewise a fixed geometric action: a stretch, a rotation, a shear. Its matrix is that action "read in your grid," and re-drawing the grid reshuffles the matrix entries without altering what the operator does. The first figure freezes the essential fact: one arrow, two grids, two component columns tied together by $P^{-1}$ .

Figure

The same fixed vector v has components (1, 3) in the black standard basis but (2, 1) in the blue skewed basis e′₁ = (1,1), e′₂ = (−1,1). The two component columns are tied by v-new = P⁻¹ v-old, so (2, 1) = ½ times the matrix with rows (1,1) and (−1,1) acting on (1, 3).

The inverse in $[v]' = P^{-1}[v]$ is dictated by consistency. Lengthen the grid lines — scale the basis vectors up by two — and the same physical arrow now needs smaller numbers to describe it, its components halved. Basis vectors and components scale oppositely, "contravariant" against "covariant," so whatever $P$ does to the basis, $P^{-1}$ must do to the components. For an operator we read its action by first stepping into the old coordinates ( $P$ ), applying $A$ , then stepping back out to the new ( $P^{-1}$ ). The sandwich $P^{-1}AP$ is literally "translate, act, translate back," which is exactly why the two flanking factors are forced to be inverses. The second figure draws that sandwich as a commuting square — the single diagram that contains the whole subject.

Figure

The commuting square of a change of basis: applying the operator A then re-coordinatizing by P⁻¹ equals re-coordinatizing first then applying A′, which forces A′ = P⁻¹AP. For A with rows (2,1) and (0,3) the eigenbasis gives A′ = diag(2, 3), and trace 5 and determinant 6 are the same read in either basis.

Invariants are the shadow the operator casts no matter where you stand. The trace (the sum of the eigenvalues) and the determinant (their product) are the two simplest shadows; together with the other coefficients of the characteristic polynomial they pin down the whole spectrum — the operator's basis-free fingerprint.

The core equations

The starting point is that a single abstract vector has two component columns, one per basis, both describing the identical arrow. Everything downstream is a rule enforcing that identity.

v = \sum_{i} c_i\, e_i = \sum_{j} c'_j\, e'_j \qquad (1)

where: $v$ — a fixed abstract vector; $e_i,\ e'_j$ — old and new basis vectors; $c_i,\ c'_j$ — old and new components of $v$ . Reading: the two sums are the same point in the space, expanded on two different sets of axes. The change-of-basis matrix $P$ is defined by expressing each new basis vector in the old basis; its columns are the new basis vectors written in old coordinates. This one convention fixes the direction of every inverse that follows.

e'_j = \sum_i P_{ij}\, e_i \quad\Longleftrightarrow\quad P = \big[\;[e'_1]_{\mathcal B}\;\cdots\;[e'_n]_{\mathcal B}\;\big] \qquad (2)

where: $P$ — matrix whose $j$ -th column lists $e'_j$ in the old basis; $P_{ij}$ — $i$ -th old-basis component of the $j$ -th new basis vector. Reading: to build $P$ , drop the new axes into the old grid and record their coordinates column by column. Because the arrow is fixed, the component column must transform by the inverse of $P$ . This is the single most important — and most error-prone — rule in the subject.

[v]_{\mathcal B} = P\,[v]_{\mathcal B'} \qquad\Longleftrightarrow\qquad [v]_{\mathcal B'} = P^{-1}\,[v]_{\mathcal B} \qquad (3)

where: $[v]_{\mathcal B}$ — column of old components $c_i$ ; $[v]_{\mathcal B'}$ — column of new components $c'_j$ . Reading: basis vectors go one way (by $P$ ), the numbers describing a fixed vector go the other way (by $P^{-1}$ ). An operator matrix carries $P^{-1}$ on the left and $P$ on the right, sandwiching $A$ . This conjugation is the similarity transformation, the definition around which the whole article turns.

A' = P^{-1} A P \qquad (4)

where: $A$ — operator matrix in the old basis; $A'$ — operator matrix in the new basis; $P$ — change-of-basis matrix (new basis in old coordinates). Reading: to see what the operator does in new coordinates, translate in, act, translate back. When both bases are orthonormal, $P$ is no longer a generic invertible matrix but a unitary one (orthogonal, in the real case), and the similarity specializes to a form that preserves inner products.

A' = U^{\dagger} A U,\qquad U^{\dagger}U = UU^{\dagger} = I \qquad (5)

where: $U$ — unitary change-of-basis matrix between orthonormal bases; $U^{\dagger}$ — conjugate transpose of $U$ , equal to $U^{-1}$ here. Reading: rotating between orthonormal frames keeps lengths, angles, and Hermiticity intact — the transformation physics actually performs. The whole reason to care about conjugation is that certain scalars built from $A$ do not move under it. The two most fundamental are the determinant and the trace, and both are symmetric functions of the eigenvalues.

\det A' = \det A = \prod_{i=1}^{n}\lambda_i,\qquad \operatorname{tr} A' = \operatorname{tr} A = \sum_{i=1}^{n}\lambda_i \qquad (6)

where: $\lambda_i$ — the $n$ eigenvalues counted with multiplicity; $n$ — dimension of the space. Reading: the determinant is the product of the eigenvalues, the trace is their sum, and neither notices a change of basis. More is true: the entire characteristic polynomial is invariant, so the full spectrum — not merely its product and sum — belongs to the operator rather than to any grid.

\chi_{A'}(\lambda) = \det(A' - \lambda I) = \det(A - \lambda I) = \chi_A(\lambda) \qquad (7)

where: $\chi_A(\lambda)$ — characteristic polynomial whose roots are the eigenvalues; $\lambda$ — the eigenvalue variable. Reading: similar matrices solve the identical secular equation, so they share every eigenvalue with the same multiplicities.

Derivation AWhy basis vectors and components transform oppositely

The goal is to prove the pivotal asymmetry of equations (2) and (3): the new basis vectors are obtained from the old by $P$ , yet the component column of a fixed vector is obtained by $P^{-1}$ . We derive the rule from a single demand — that the one abstract vector $v$ has consistent expansions in both bases — and never appeal to intuition, so that the notorious "which one gets the inverse" confusion is settled by algebra rather than by memory. Everything hinges on substituting the definition of $P$ into the expansion and matching coefficients.

Step 1. Expand $v$ in the new basis and substitute the definition $e'_j = \sum_i P_{ij} e_i$ from equation (2):

v = \sum_j c'_j\, e'_j = \sum_j c'_j \sum_i P_{ij}\, e_i.

Why: this trades the new basis vectors for old ones, so that both expansions of $v$ are now written over the same set of axes and can be compared term by term.

Step 2. Swap the summation order to collect the coefficient multiplying each old basis vector $e_i$ :

v = \sum_i \Big(\sum_j P_{ij}\, c'_j\Big) e_i.

Why: the object in parentheses is, by construction, the $i$ -th old component of $v$ — we have re-expressed $v$ purely in the old basis with the new components still visible inside.

Step 3. Match against $v = \sum_i c_i\, e_i$ . Because components in a basis are unique, the bracketed sums must equal the old components:

c_i = \sum_j P_{ij}\, c'_j \;\Longrightarrow\; [v]_{\mathcal B} = P\,[v]_{\mathcal B'} \;\Longrightarrow\; [v]_{\mathcal B'} = P^{-1}\,[v]_{\mathcal B}.

Why: uniqueness of the expansion turns the equality of vectors into an equality of column matrices, and inverting delivers the covariant rule — components carry $P^{-1}$ precisely because basis vectors carry $P$ .

Go deeper · Derivation B · 1 minThe conjugation rule and the invariance of the spectrum

Derivation BThe conjugation rule and the invariance of the spectrum

The aim is twofold: first to derive $A' = P^{-1}AP$ as the forced consequence of demanding that the abstract relation $w = \mathcal{A}v$ produce correct components in every basis, and then to show that this conjugation leaves the entire characteristic polynomial — hence every eigenvalue — unchanged. Together these establish that "the operator" is a well-defined object behind its many matrices, and that its spectrum is a genuine coordinate-free label. We start from the matrix statement of the operator in the old basis and push it through the change of coordinates.

Step 1. In the old basis the operator acts as $[w]_{\mathcal B} = A\,[v]_{\mathcal B}$ . Rewrite both component columns in the new basis using equation (3), $[\cdot]_{\mathcal B} = P[\cdot]_{\mathcal B'}$ :

P[w]_{\mathcal B'} = A\,P[v]_{\mathcal B'}.

Why: the same physical relation $w = \mathcal{A}v$ must hold whatever basis we read it in, so we simply translate every column from old coordinates into new ones.

Step 2. Left-multiply by $P^{-1}$ and read off the new-basis matrix:

[w]_{\mathcal B'} = \big(P^{-1}AP\big)\,[v]_{\mathcal B'} \equiv A'\,[v]_{\mathcal B'} \;\Longrightarrow\; A' = P^{-1}AP.

Why: whatever matrix sits between $[v]_{\mathcal B'}$ and $[w]_{\mathcal B'}$ is by definition the operator's matrix in the new basis, so the conjugation is not a convention but the only expression consistent with the fixed operator.

Step 3. Now test invariance of the characteristic polynomial. Factor $P$ out of $A' - \lambda I$ using $\lambda I = P^{-1}(\lambda I)P$ :

A' - \lambda I = P^{-1}AP - \lambda P^{-1}P = P^{-1}(A - \lambda I)P.

Why: writing the shift $\lambda I$ in conjugated form lets the whole matrix $A - \lambda I$ sit inside a single similarity, so its determinant can be handled in one stroke.

Step 4. Take determinants and use multiplicativity, $\det P^{-1}\,\det P = 1$ :

\chi_{A'}(\lambda) = \det P^{-1}\,\det(A - \lambda I)\,\det P = \chi_A(\lambda).

Why: the conjugating determinants are reciprocals and cancel exactly, so the two polynomials are identical — and identical polynomials have identical roots, proving $\operatorname{spec}(A') = \operatorname{spec}(A)$ . Setting $\lambda = 0$ recovers $\det A' = \det A$ , and the coefficient of $\lambda^{n-1}$ recovers $\operatorname{tr}A' = \operatorname{tr}A$ .

Go deeper · Derivation C · 1 minSimilarity is an equivalence relation, and its classes are labelled by invariants

Derivation CSimilarity is an equivalence relation, and its classes are labelled by invariants

The purpose here is to justify the very phrase "the operator behind the matrices." We show that similarity partitions all $n\times n$ matrices into disjoint classes — it is reflexive, symmetric, and transitive — so that a single operator corresponds to exactly one class, and every matrix in that class is one coordinate portrait of it. We then identify what stays constant across a class: precisely the quantities we have been calling invariants, which is why they and not the raw entries deserve to be called intrinsic. Write $A \sim B$ when $B = P^{-1}AP$ for some invertible $P$ .

Step 1. Reflexivity. Take $P = I$ , so $A = I^{-1}AI$ and $A \sim A$ . Why: every matrix is trivially similar to itself — the do-nothing change of basis — which is the minimal sanity check any equivalence relation must pass.

Step 2. Symmetry. If $B = P^{-1}AP$ then, with $Q = P^{-1}$ ,

A = PBP^{-1} = Q^{-1}BQ,

so $B \sim A$ . Why: undoing a change of basis is itself a change of basis, so the relation cannot distinguish "old" from "new" — sameness of operator is a two-way street.

Step 3. Transitivity. If $B = P^{-1}AP$ and $C = Q^{-1}BQ$ , compose the two basis changes:

C = Q^{-1}P^{-1}APQ = (PQ)^{-1}A(PQ),

with $PQ$ invertible, so $A \sim C$ . Why: performing one change of basis and then another is the same as performing their composite in one go, which is exactly what transitivity must express.

Step 4. Consequence. Because det, trace, characteristic and minimal polynomials, rank, and Jordan structure are all fixed under a single conjugation, they are constant on each equivalence class. Why: these invariants are the class labels — two matrices are similar if and only if they share their full Jordan structure — so listing the invariants is the same as naming the operator.

Limiting cases

Specializing the general similarity $A' = P^{-1}AP$ to particular kinds of basis change recovers, one by one, the standard structures of linear algebra.

Case	Result	Meaning
Trivial change $P = I$	$A' = I^{-1}AI = A$	Keeping the same basis leaves the matrix untouched — the identity is the do-nothing similarity, the reflexive case of the relation.
Orthonormal to orthonormal	$A' = U^{\dagger}AU,\ \ U^{\dagger}U = I$	$P$ becomes unitary $U$ (orthogonal $O$ in the real case). Inner products, norms, Hermiticity, and unitarity are all preserved — the right transformation for quantum mechanics and rigid-body geometry.
Eigenbasis (diagonalizable $A$ )	$P^{-1}AP = \operatorname{diag}(\lambda_1,\dots,\lambda_n)$	When the columns of $P$ are a complete set of eigenvectors, similarity diagonalizes $A$ : the eigenvalues line the diagonal and the operator decouples into independent one-dimensional scalings.
Normal operator	$A' = U^{\dagger}AU = \operatorname{diag}(\lambda_i)$	The spectral theorem guarantees an orthonormal eigenbasis (unitary $P$ ) whenever $AA^{\dagger} = A^{\dagger}A$ — the best possible case, diagonal and length-preserving.
Defective operator (no eigenbasis)	$P^{-1}AP = J = \bigoplus_k J_{m_k}(\lambda_k)$	With too few eigenvectors, the simplest reachable form is block-diagonal Jordan blocks: $\lambda$ on the diagonal, $1$ 's on the superdiagonal — the canonical representative of the similarity class.

Common trapMixing up how basis vectors transform with how components transform

The single most common error in the whole subject is putting the inverse in the wrong place. The new basis vectors come from the old by $P$ — the columns of $P$ literally are the new basis — but the components of a fixed vector transform by the inverse, $[v]' = P^{-1}[v]$ . Forget that inverse, or write the operator conjugation as $PAP^{-1}$ instead of $P^{-1}AP$ , and every subsequent number is wrong while looking plausible. The cure is to nail down one convention and never deviate: let $P$ = (new basis vectors written in old coordinates). Then, always, $[v]' = P^{-1}[v]$ and $A' = P^{-1}AP$ . Quick check: if you double the length of every basis vector ( $P = 2I$ ), the components of a fixed arrow must halve — and indeed $P^{-1} = \tfrac12 I$ . If your rule doubles them, the inverse is backwards.

Worked examples

1 · Diagonalizing a 2×2 operator and checking its invariants (confidence builder)

Take the upper-triangular matrix $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ . Find a change of basis that diagonalizes it, and verify explicitly that the trace and determinant survive the transformation.

Step 1. Find the eigenvalues from the characteristic polynomial. For a triangular matrix the diagonal entries are already the roots:

\det(A - \lambda I) = (2-\lambda)(3-\lambda) = 0 \;\Longrightarrow\; \lambda = 2,\ 3.

So the spectrum is $\{2, 3\}$ .

Step 2. Solve $(A - \lambda I)x = 0$ for each eigenvalue. For $\lambda = 2$ , the equations force the second component to zero, giving $x_1 = (1, 0)^{T}$ . For $\lambda = 3$ , $-x_{(1)} + x_{(2)} = 0$ gives $x_2 = (1, 1)^{T}$ . The two eigenvectors are $x_1 = (1,0)^T$ and $x_2 = (1,1)^T$ .

Step 3. Assemble $P$ from the eigenvectors as columns, and invert:

P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \qquad P^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}.

The columns of $P$ are the new basis — the eigenbasis — written in the standard basis.

Step 4. Carry out the similarity transformation. First $AP = \begin{pmatrix} 2 & 3 \\ 0 & 3 \end{pmatrix}$ , then

P^{-1}AP = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.

The operator is now diagonal, $A' = \operatorname{diag}(2, 3)$ : the eigenbasis has decoupled it into two independent scalings.

Step 5. Check the invariants in both bases. The original $A$ has $\operatorname{tr} = 2 + 3 = 5$ and $\det = 2\cdot 3 - 1\cdot 0 = 6$ ; the diagonal $A'$ has $\operatorname{tr} = 2 + 3 = 5$ and $\det = 2\cdot 3 = 6$ . Both give trace 5 and determinant 6, confirming that conjugation left them untouched — exactly as the eigenvalue formulas $\operatorname{tr} = \sum\lambda_i$ and $\det = \prod\lambda_i$ require.

2 · Normal modes of two coupled oscillators (a physical similarity transformation)

Two equal masses $m$ sit in a line, each tied to a fixed wall and to each other by three identical springs of stiffness $k$ . The equations of motion are $\ddot{q} = -M^{-1}K\,q$ with $M = mI$ and $K = k\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ . Diagonalize $M^{-1}K$ to find the normal-mode frequencies.

Step 1. Form the dynamical matrix. With $M = mI$ ,

M^{-1}K = \frac{k}{m}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}.

This matrix is real symmetric, so the spectral theorem promises a real orthogonal eigenbasis — the similarity here will be $O^{T}(M^{-1}K)O$ with $O$ orthogonal.

Step 2. Eigenvalues of the pattern matrix $\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ : from $(2-\mu)^2 - 1 = 0$ we get $2 - \mu = \pm 1$ , so $\mu = 1,\ 3$ . Restoring the prefactor, the squared normal-mode frequencies are

\omega_1^2 = \frac{k}{m}, \qquad \omega_2^2 = \frac{3k}{m}.

Step 3. Eigenvectors give the mode shapes. For $\mu = 1$ the eigenvector is $(1, 1)$ — both masses move together, the in-phase (symmetric) mode, which never stretches the middle spring, hence the lower frequency $\omega_1 = \sqrt{k/m}$ . For $\mu = 3$ the eigenvector is $(1, -1)$ — the masses move oppositely, the out-of-phase (antisymmetric) mode, which compresses the middle spring hardest, hence $\omega_2 = \sqrt{3k/m}$ .

Step 4. Build the orthogonal change of basis from the normalized eigenvectors and diagonalize:

O = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad O^{T}(M^{-1}K)\,O = \frac{k}{m}\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}.

In these normal coordinates the two masses no longer talk to each other — the system is two independent oscillators.

Step 5. Check invariants as a cross-check on the physics. The trace of $M^{-1}K$ is $4k/m$ , and indeed $\omega_1^2 + \omega_2^2 = k/m + 3k/m = 4k/m$ . The determinant is $3(k/m)^2$ , and $\omega_1^2\,\omega_2^2 = (k/m)(3k/m) = 3(k/m)^2$ . The mode frequencies satisfy $\sum\omega_i^2 = \operatorname{tr}$ and $\prod\omega_i^2 = \det$ , the same eigenvalue identities that governed Example 1 — here reading out as measurable frequencies rather than as bare numbers.

Physics of the portrait

The change-of-basis rule looks like bookkeeping, but the recognition that a matrix is a coordinate portrait of an operator — and that the operator, not the portrait, is the real thing — was a genuine conceptual leap, arriving in the mid-nineteenth century from the figure who gave us the algebra of matrices.

Origins · 1 minArthur Cayley and how this was discovered

Arthur Cayley

1821–1895, English mathematician and barrister · founder of the theory of matrices and similarity

Arthur Cayley trained as a lawyer and spent fourteen years at the bar, publishing mathematics in his spare hours, before returning to Cambridge as the first Sadleirian Professor. His 1858 A Memoir on the Theory of Matrices is where the objects of this article are born: he defined how to add and multiply matrices, isolated the identity and the inverse, and — crucially — saw that a linear substitution of variables composes exactly like matrix multiplication, so that changing coordinates and conjugating a matrix are one and the same operation. In the same memoir he stated what we now call the Cayley–Hamilton theorem, that a matrix satisfies its own characteristic equation, and verified it by hand for the $2\times2$ and $3\times3$ cases. The word "matrix" itself had been coined a few years earlier by his close friend James Joseph Sylvester, whose 1852 law of inertia had already identified the first invariants of a quadratic form. What Cayley began as an algebra, others turned into a classification: Weierstrass's elementary divisors (1868) and Camille Jordan's canonical form (1870) answered "when are two matrices similar?" completely, and Frobenius (1878) unified the strands into the modern theory of the minimal polynomial. The result is the viewpoint every physicist now takes for granted — that the eigenvalues, not the entries, are what an operator really is.

History

1829

— Cauchy proves the eigenvalues of a real symmetric matrix are real, the principal-axes theorem

1858

— Cayley's Memoir on the Theory of Matrices founds matrix algebra and the notion of similarity

1868

— Weierstrass classifies matrices up to similarity by their elementary divisors

1870

— Jordan publishes the canonical form of a matrix under similarity

1878

— Frobenius unifies the minimal polynomial and the rational canonical form

Going further

Texts. Stone & Goldbart, Mathematics for Physics, Ch. 1 (vector spaces, operators, change of basis) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 3 (matrices, similarity, diagonalization) · Hassani, Mathematical Physics, Ch. 4–5 (operator algebra and spectral decomposition) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 3–4 · Nakahara, Geometry, Topology and Physics, Ch. 2 (linear maps and their coordinate expressions) · Strang, Linear Algebra and Its Applications, Ch. 5 (eigenvalues and similarity). Key results to keep in mind: the component rule $[v]_{\mathcal B} = P[v]_{\mathcal B'}$ follows from the basis-expansion derivation; determinant and trace invariance follow from $\det(XY) = \det X\det Y$ and the cyclic property of the trace; the worked $2\times2$ example gives $P^{-1}AP = \operatorname{diag}(2,3)$ by direct multiplication, with trace $5$ and determinant $6$ ; the coupled-oscillator spectrum $\omega_1^2 = k/m,\ \omega_2^2 = 3k/m$ are the eigenvalues of $(k/m)\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$ , satisfying $\sum\omega_i^2 = 4k/m$ and $\prod\omega_i^2 = 3(k/m)^2$ . All numerical values are exact.

Prerequisites. Linear Independence, Basis, and Dimension · Linear Operators and Matrix Representations · Determinants and Their Properties · Vector Spaces and Subspaces. Leads to. Eigenvalues, Eigenvectors, and the Characteristic Equation · Diagonalization and the Spectral Decomposition · Jordan Canonical Form and Defective Matrices · Unitary and Orthogonal Operators.

Ask the tutor: Want to see any operator diagonalized step by step, with its trace, determinant, and eigenvalues checked in two different bases?

Why does the operator matrix get P⁻¹ on one side and P on the other, instead of P on both?

Because the operator has to eat a vector and spit out a vector, and both are expressed in the new basis. To act, you first convert the incoming new-basis column into old coordinates ( $P$ ), apply the old matrix $A$ , then convert the result back into new coordinates ( $P^{-1}$ ). That "in-then-out" round trip is what makes the two flanking factors inverses of each other. A quadratic form, by contrast, eats two vectors and returns a scalar; it transforms by congruence, $PᵀAP$ , with a transpose rather than an inverse. The transformation law is dictated by how many vectors an object consumes — operators are one-in, one-out, so they get $P^{-1}\cdots P$ .

Is a similarity transformation the same thing as diagonalization?

No — diagonalization is a special case. A similarity transformation is any conjugation $A \mapsto P^{-1}AP$ , valid for every invertible $P$ , and it always exists. Diagonalization is the lucky situation where you can choose $P$ so that $P^{-1}AP$ comes out diagonal, which requires a complete set of linearly independent eigenvectors. Many operators cannot be diagonalized — the defective ones — yet every operator can still be brought to its Jordan canonical form by an appropriate similarity. So "can I diagonalize $A$ ?" is really the question "does $A$ have enough eigenvectors to span the space?", while "can I apply a similarity to $A$ ?" is always answered yes.

If eigenvalues are basis-independent, why do textbooks bother computing eigenvectors, which are not?

Eigenvalues tell you what the operator does — the scaling factors along its natural directions — but eigenvectors tell you where those directions point, and that is exactly the information you need to build the change of basis $P$ that simplifies the problem. In the coupled-oscillator example the eigenvalues are the frequencies you would measure, but the eigenvectors are the mode shapes: in-phase versus out-of-phase motion. You compute them because the eigenbasis is the coordinate system in which the physics decouples.

When is the change-of-basis matrix guaranteed to be unitary rather than a generic invertible matrix?

Exactly when both the old and the new basis are orthonormal. Unitarity, $U^{\dagger}U = I$ , is precisely the statement that $U$ maps an orthonormal set to an orthonormal set, preserving all inner products and lengths. A generic invertible $P$ between two arbitrary (skewed, unequal-length) bases will distort angles and magnitudes, which is fine for pure linear algebra but ruinous in quantum mechanics, where inner products are probabilities and must be preserved. This is why quantum changes of representation — position to momentum, or into an energy eigenbasis — are always unitary, and why the spectral theorem for normal operators is so prized: it guarantees the diagonalizing basis can be chosen orthonormal.

Trace and determinant are invariant — are they enough to tell whether two matrices are similar?

Not in general. For $2\times2$ matrices trace and determinant fix the characteristic polynomial and hence the eigenvalues, but even there they can fail: $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ and $\lambda I$ share both trace $2\lambda$ and determinant $\lambda^2$ , yet only the second is diagonal, so they are not similar. What is needed is that two matrices share their entire Jordan canonical form — the full list of eigenvalues together with the sizes of their Jordan blocks. Trace and determinant are just two coarse summaries of that structure. They are necessary conditions for similarity and quick to check, but never sufficient on their own.

Does changing basis ever change the physics, or only the description?

Only the description — that is the whole point, and also a powerful correctness check. Any quantity you claim is physical must come out the same in every basis: energies, frequencies, and expectation values are all built from invariants precisely so that they do not depend on your coordinate choice. If an answer changes when you rotate the basis, you have computed something unphysical and made an error. The freedom to change basis is thus both a tool — pick the frame where the work is easiest — and a diagnostic: real observables are the ones that stand still while the coordinates spin.