What is a change-of-basis matrix?

The change-of-basis matrix P converts coordinates from one basis to another. Column j of P is the coordinate vector of the j-th basis vector of the source basis expressed in the target basis. The inverse P⁻¹ converts in the reverse direction.

What does it mean for two matrices to be similar?

Two matrices A and A' are similar if A' = P⁻¹AP for some invertible matrix P. Similar matrices represent the same linear transformation in different bases. They share the same determinant, trace, eigenvalues, characteristic polynomial, and rank.

What properties are preserved by similarity?

Similar matrices share every property intrinsic to the transformation: determinant, trace, eigenvalues with multiplicities, characteristic polynomial, and rank. Individual entries, symmetry, and sparsity are generally not preserved unless the change-of-basis matrix has special structure.

How is diagonalization a change of basis?

Diagonalization uses eigenvectors as the new basis. In this basis, the transformation acts by scaling each basis vector by its eigenvalue, so the matrix becomes diagonal: D = P⁻¹AP where P has eigenvectors as columns. This reduces powers to A^k = PD^kP⁻¹.

What is orthogonal similarity?

Orthogonal similarity uses an orthogonal change-of-basis matrix P (where P⁻¹ = Pᵀ), giving A' = PᵀAP. It preserves symmetry and is the basis of the spectral theorem: every real symmetric matrix is orthogonally similar to a diagonal matrix of its eigenvalues.

Change of Basis & Similarity | Learn Math Class

The Problem

A linear transformation

T: V \to V

is a fixed geometric object — it sends each vector to a definite image regardless of how coordinates are assigned. But the matrix that represents

T

depends on the choice of basis. Different bases assign different coordinates to the same vectors, and the matrix that converts input coordinates to output coordinates changes accordingly.

This raises a natural question: if

T

has matrix

A

in one basis and matrix

A'

in another, how are

A

and

A'

related? The answer is the similarity relation

A' = P^{-1}AP

, where

P

is the change-of-basis matrix. Understanding this relation is the key to choosing bases strategically — picking the basis that makes the matrix as simple as possible.

The Change-of-Basis Matrix

If

\mathcal{B}

and

\mathcal{C}

are two bases for

V

, the change-of-basis matrix

P_{\mathcal{C} \leftarrow \mathcal{B}}

converts

\mathcal{B}

-coordinates to

\mathcal{C}

-coordinates:

[\mathbf{v}]_\mathcal{C} = P_{\mathcal{C} \leftarrow \mathcal{B}} \, [\mathbf{v}]_\mathcal{B}

Column

j

of

P

is the

\mathcal{C}

-coordinate vector of the

j

-th basis vector of

\mathcal{B}

. The reverse conversion uses the inverse:

P_{\mathcal{B} \leftarrow \mathcal{C}} = P^{-1}

.

Worked Example

In

\mathbb{R}^2

, let

\mathcal{B} = \{(1, 1), (1, -1)\}

and let

\mathcal{C}

be the standard basis. The

\mathcal{C}

-coordinates of the

\mathcal{B}

-basis vectors are just their components:

(1, 1)

and

(1, -1)

. So

P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

To find the

\mathcal{B}

-coordinates of

\mathbf{v} = (5, 1)

: solve

P\mathbf{c} = (5, 1)

. Using

P^{-1} = \frac{1}{-2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix}

, we get

\mathbf{c} = (3, 2)

. So

\mathbf{v} = 3(1, 1) + 2(1, -1)

.

The Similarity Relation

If

T: V \to V

has matrix

A

in basis

\mathcal{B}

and matrix

A'

in basis

\mathcal{C}

, then

A' = P^{-1}AP

where

P = P_{\mathcal{C} \leftarrow \mathcal{B}}

is the change-of-basis matrix from

\mathcal{B}

to

\mathcal{C}

.

The derivation is direct. For any vector

\mathbf{v}

, the transformation in

\mathcal{B}

-coordinates reads

[T(\mathbf{v})]_\mathcal{B} = A[\mathbf{v}]_\mathcal{B}

. Converting to

\mathcal{C}

-coordinates:

[T(\mathbf{v})]_\mathcal{C} = P^{-1}[T(\mathbf{v})]_\mathcal{B} = P^{-1}A[\mathbf{v}]_\mathcal{B} = P^{-1}AP[\mathbf{v}]_\mathcal{C}

. Since this holds for every

\mathbf{v}

, the matrix of

T

in basis

\mathcal{C}

is

P^{-1}AP

.

Two matrices related by

A' = P^{-1}AP

for some invertible

P

are called similar. Similarity is an equivalence relation: every matrix is similar to itself (

P = I

), similarity is symmetric (

A' = P^{-1}AP

implies

A = PA'P^{-1}

), and it is transitive.

Properties Preserved by Similarity

Similar matrices represent the same transformation, so they share every property that is intrinsic to the transformation rather than to a particular coordinate system.

The determinant is preserved:

\det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A)

.

The trace is preserved:

\text{tr}(P^{-1}AP) = \text{tr}(APP^{-1}) = \text{tr}(A)

by the cyclic property.

The eigenvalues are preserved:

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

, so the characteristic polynomial — and therefore all eigenvalues with their multiplicities — is the same.

The rank is preserved: multiplying by invertible matrices cannot change the rank.

Individual matrix entries, symmetry, and sparsity are generally not preserved. A symmetric matrix

A

can become non-symmetric under

P^{-1}AP

if

P

is not orthogonal.

Quantity / property of A	Preserved under A ↦ P⁻¹AP?	Reason
Determinant	✓	det(P⁻¹AP) = det(P⁻¹) det(A) det(P) = det(A)
Trace	✓	tr(P⁻¹AP) = tr(APP⁻¹) = tr(A) by cyclic property
Eigenvalues (with algebraic multiplicities)	✓	characteristic polynomial is preserved
Characteristic polynomial	✓	det(P⁻¹AP − λI) = det(P⁻¹(A − λI)P) = det(A − λI)
Rank and nullity	✓	multiplication by invertible matrices cannot change rank
Individual entries	✗	entries are basis-dependent coordinates; they change with P
Symmetry (A = Aᵀ)	✗	preserved only when P is orthogonal (P⁻¹ = Pᵀ)
Sparsity / triangular structure	✗	generally destroyed unless P is itself sparse / triangular

Diagonalization as a Change of Basis

If

T

has

n

linearly independent eigenvectors

\mathbf{v}_1, \dots, \mathbf{v}_n

with eigenvalues

\lambda_1, \dots, \lambda_n

, use them as the basis

\mathcal{B}

. In this eigenvector basis,

T

acts by scaling each basis vector:

T(\mathbf{v}_i) = \lambda_i \mathbf{v}_i

The matrix of

T

in this basis is diagonal:

D = \text{diag}(\lambda_1, \dots, \lambda_n)

.

The change-of-basis matrix

P

has the eigenvectors as columns:

P = [\mathbf{v}_1 \; \cdots \; \mathbf{v}_n]

. The similarity relation gives

A = PDP^{-1}

, or equivalently

D = P^{-1}AP

.

Diagonalization is the most powerful application of basis change. It reduces matrix powers to diagonal powers:

A^k = PD^kP^{-1} = P\,\text{diag}(\lambda_1^k, \dots, \lambda_n^k)\,P^{-1}

. It simplifies differential equations, recurrence relations, and any computation involving repeated application of the same transformation.

When Diagonalization Fails

Not every matrix is diagonalizable. A transformation may not have

n

linearly independent eigenvectors — this happens when the geometric multiplicity of some eigenvalue is strictly less than its algebraic multiplicity.

For example,

A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}

has eigenvalue

\lambda = 2

with algebraic multiplicity

2

, but the eigenspace is one-dimensional (spanned by

(1, 0)

). There is no basis of eigenvectors, so

A

cannot be diagonalized.

In such cases, the best achievable form under similarity is the Jordan normal form: a block-diagonal matrix where each block is an upper triangular matrix with a single eigenvalue on the diagonal and ones on the superdiagonal. The Jordan form is unique up to ordering of blocks and is the canonical representative of the similarity class. Its full development belongs to advanced linear algebra.

Orthogonal Similarity

When the change-of-basis matrix

P

is orthogonal (

P^{-1} = P^T

), the similarity relation becomes

A' = P^TAP

. This is called orthogonal similarity.

Orthogonal similarity preserves more than ordinary similarity. If

A

is symmetric, then

P^TAP

is also symmetric — a property that ordinary similarity does not guarantee.

The Spectral Theorem states that every real symmetric matrix is orthogonally similar to a diagonal matrix. The eigenvectors of a symmetric matrix can be chosen orthonormal, and the columns of

P

form an orthonormal basis. This is a stronger conclusion than ordinary diagonalizability — the diagonalizing basis is not just independent but orthonormal, which simplifies projections, least squares, and numerical computation.

Worked Example: Full Basis Change

Let

A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}

. Find a diagonalization

A = PDP^{-1}

.

The characteristic polynomial is

\det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 2)(\lambda - 5)

. Eigenvalues:

\lambda_1 = 2

,

\lambda_2 = 5

.

For

\lambda_1 = 2

:

(A - 2I)\mathbf{v} = \mathbf{0}

gives

\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}

, so

\mathbf{v}_1 = (1, -2)

.

For

\lambda_2 = 5

:

(A - 5I)\mathbf{v} = \mathbf{0}

gives

\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}

, so

\mathbf{v}_2 = (1, 1)

.

P = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}

Verification:

P^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix}

, and

PDP^{-1} = \begin{pmatrix} 1 & 1 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}\frac{1}{3}\begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A

.

Application:

A^{10} = PD^{10}P^{-1} = P\begin{pmatrix} 2^{10} & 0 \\ 0 & 5^{10} \end{pmatrix}P^{-1} = P\begin{pmatrix} 1024 & 0 \\ 0 & 9765625 \end{pmatrix}P^{-1}

.

Why Basis Choice Matters

The standard basis is the default, but it is rarely the best choice for a given problem.

An eigenvector basis diagonalizes the matrix, reducing powers and exponentials to operations on diagonal entries. A system of differential equations

\mathbf{x}' = A\mathbf{x}

decouples into independent scalar equations when

A

is diagonal.

An orthonormal basis simplifies projections and least-squares computations. Coordinates relative to an orthonormal basis are computed by dot products rather than by solving systems, and numerical errors are minimized because the change-of-basis matrix has condition number

1

.

A Jordan basis achieves the simplest possible form for non-diagonalizable matrices, isolating the defective eigenvalues into small blocks.

Choosing the right basis is often the key insight that converts a hard problem into an easy one. The transformation does not change — only its numerical description does — but the right description can make all the difference between a tractable computation and an intractable one.

New basis	Resulting matrix form	What it simplifies
Eigenvector basis (when n indep. eigenvectors exist)	diagonal D = diag(λ₁, ..., λₙ)	matrix powers Aᵏ = PDᵏP⁻¹; exponentials e^(At); differential systems x' = Ax decouple into scalar ODEs
Orthonormal eigenbasis (symmetric A)	diagonal D, with P orthogonal (P⁻¹ = Pᵀ)	projections, least-squares, numerical stability (condition number κ(P) = 1)
Jordan basis (when diagonalization fails)	block-diagonal Jordan normal form	canonical representative for defective matrices; isolates each defective eigenvalue into a small block
Standard basis (default)	the original matrix A	rarely the best choice — the natural starting point but the right basis can convert a hard problem into an easy one

Summary: Matrix Structure → Canonical Form

The strategic-basis discussion above was organized by which kind of basis you choose. Reading the page in the other direction — starting from what you know about the matrix and asking what form it can be reduced to — gives a recognition guide for similarity transformations. The table below collects the standard canonical forms: when a matrix's structure permits a stronger reduction, similarity can deliver it; when no diagonalization exists, similarity still delivers the Jordan form (over ℝ or ℂ) and Schur form (always available over ℂ). It is the lookup card to consult when a specific A is in front of you and the question is how far similarity can simplify it.

Structure of A	Canonical form via similarity	Choice of P	Source theorem
All n eigenvalues distinct	diagonal D = P⁻¹AP	columns = the n eigenvectors (automatically independent)	distinct eigenvalues ⇒ independent eigenvectors
Geometric mult. = algebraic mult. for every eigenvalue	diagonal D	columns = n linearly independent eigenvectors	diagonalization theorem
Real and symmetric (A = Aᵀ)	diagonal D via orthogonal similarity D = PᵀAP	columns = orthonormal eigenvectors; P orthogonal	Spectral Theorem (always succeeds)
Defective (geom. mult. < alg. mult. for some eigenvalue)	Jordan normal form J = P⁻¹AP (block-diagonal, with 1's on superdiagonal)	columns = eigenvectors + generalized eigenvectors	Jordan decomposition theorem
General complex square matrix	upper triangular Schur form T = U*AU	unitary U (complex orthogonal)	Schur decomposition (always exists over ℂ)

Change of Basis

Same Transformation, Different Matrix

The Problem

The Change-of-Basis Matrix

Worked Example

The Similarity Relation

Properties Preserved by Similarity

Diagonalization as a Change of Basis

When Diagonalization Fails

Orthogonal Similarity

Worked Example: Full Basis Change

Why Basis Choice Matters

Summary: Matrix Structure → Canonical Form