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:

Change of Basis Formula

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

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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

The Change-of-Basis Matrix

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

Pitfall: Converting in the wrong direction

Multiplying

P

by the standard coordinates of

(3, 1)

gives

(4, 2)

, but

4(1, 1) + 2(1, -1) = (6, 2)

, not

(3, 1)

. The

\mathcal{B}

-coordinates come from

P^{-1}

:

P^{-1}(3, 1) = (2, 1)

.

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 same point, two sets of coordinates

The grid has been redrawn along the new basis vectors and the point read off against it instead of against the axes. The point has not moved; only the description has. The change-of-basis matrix is simply the dictionary between the two readings. Swap between bases and watch the numbers change on the change of basis explorer.

Its inverse is the dictionary read the other way, which is why change of basis always comes as a conjugating pair.

Basis Change Notation

Notation

Basis Change Notation

The coordinate machinery is borrowed from elsewhere. What belongs to this page is a prime that is not a derivative, an arrow whose two ends are deliberately the same, and a relation that is never given a symbol.

The coordinate brackets

[\mathbf{v}]_\mathcal{B}

, the script letters for bases, and the backward-arrow subscript on

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

with its domino cancellation are all set out at basis. The same device around a map,

[T]_{\mathcal{C} \leftarrow \mathcal{B}}

, belongs to matrix representation. The letter contract of

A = PDP^{-1}

is at diagonalization, and the transpose standing in for an inverse at the spectral decomposition.

A' = P^{-1}AP

A prime equals P inverse A P

The prime says same transformation, different basis. It is not a derivative, not a transpose, and not a new matrix built from

A

by some operation on its entries.

A

and

A'

are two descriptions of one object, and the prime is the only thing recording that they describe the same thing.

CasesWhich side

P^{-1}

takes is decided by the direction

P

converts, not by preference. With

P

carrying

\mathcal{C}

-coordinates into

\mathcal{B}

-coordinates,

P

on the right converts the input into

\mathcal{B}

-coordinates,

A

acts, and

P^{-1}

on the left converts the result back, which puts the inverse on the left. Reverse the direction of

P

and the same relation is written

A = PA'P^{-1}

, with the letters unchanged and the sides swapped.

Also writtenBoth forms appear on this page:

A' = P^{-1}AP

when a basis is being changed, and

A = PDP^{-1}

when a matrix is being built out of eigen-data. They are one relation read in opposite directions.

Do not confuseTaking the visible symmetry of

P^{-1}AP

— the same letter on both sides — for a promise that structure survives. Symmetry of

A

does not:

P^{-1}AP

is generally not symmetric unless

P

is orthogonal, and entries, zeros and sparsity are lost as a matter of course.

Same glyph elsewhereA prime marks differentiation elsewhere on this site, including inside the Wronskian at determinant applications. Nothing is being differentiated here.

T: V \to V

T maps V to V

Both ends of the arrow name the same space, and that repetition is what makes this page possible. Because input and output live in one space, a single basis choice governs both, so a single

P

appears twice in the relation — once undoing the choice and once redoing it.

CasesA map between different spaces,

T: V \to W

, needs two independent basis choices and the relation becomes

A' = Q^{-1}AP

with two unrelated matrices. Similarity is the special case in which the two collapse into one.

Also writtenSuch a map is called an operator on

V

rather than a transformation between spaces, and the word carries exactly the information the repeated

V

does.

Do not confuseReading the two

V

symbols as decoration. If they differ,

A

and

A'

need not even be the same size, the invariants below stop being invariants, and the notation

P^{-1}AP

does not typecheck.

A \sim B

A is similar to B

This page names the relation in words and never writes it with a symbol, which is a deliberate omission. The tilde is already spoken for: at Gaussian elimination

A \sim B

means row equivalence, a different relation entirely.

CasesThe two relations disagree on the things this page cares about. Row-equivalent matrices generally have different eigenvalues and different determinants; similar matrices always share both. A tilde read with the wrong meaning turns a true statement into a false one without changing a symbol.

Also writtenWhere a symbol is wanted,

A \simeq B

or

A \approx B

appear for similarity, and some texts write

A \sim_{s} B

with a subscript naming the relation. None is standard enough to rely on.

Do not confuseReading similarity as a loose word for resemblance. It is an equivalence relation with a precise definition — reflexive with

P = I

, symmetric by inverting

P

, transitive by multiplying the two conversions — and matrices that merely look alike are not similar.

Same glyph elsewhereThe tilde also marks approximation, asymptotic equivalence, and distribution membership in probability. Its meaning here would come entirely from context, which is the reason for avoiding it.

All linear algebra symbols →↑ Linear transformations

The Similarity Relation

If

T: V \to V

has matrix

A

in basis

\mathcal{B}

and matrix

A'

in basis

\mathcal{C}

, then

Similarity Relation

A' = P^{-1}AP

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where

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

is the change-of-basis matrix from

\mathcal{C}

to

\mathcal{B}

:

[\mathbf{v}]_\mathcal{B} = P[\mathbf{v}]_\mathcal{C}

, and its columns are the

\mathcal{C}

basis vectors written in

\mathcal{B}

-coordinates.

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 with

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

:

[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

.

Fact: Every matrix is similar to its transpose

For every square

A

there is an invertible

P

with

A^T = P^{-1}AP

, even though rows and columns trade places. Transposing does not change the rank of

A - \lambda I

or of any of its powers, and those ranks are exactly what decides similarity.

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.

Similarity can join matrices whose entries look nothing alike. Because

P

may be any invertible matrix, a triangular matrix can be similar to one with no zero entries at all. Deciding whether two given matrices are similar means finding such a

P

, and the entries alone give little hint of where to look.

P^{-1}AP

translates new coordinates into old ones, applies

A

, and translates the answer back.

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. Formally:

Similarity Invariants

A' = P^{-1}AP \Rightarrow \begin{cases} \det(A') = \det(A) \\ \text{tr}(A') = \text{tr}(A) \\ \text{rank}(A') = \text{rank}(A) \\ \text{eigenvalues}(A') = \text{eigenvalues}(A) \end{cases}

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Pitfall: Treating shared invariants as proof of similarity

Calling

I

and

\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}

similar because both have trace

2

, determinant

1

, rank

2

and eigenvalues

1, 1

reads the rule backwards.

P^{-1}IP = I

for every

P

, so

I

is similar only to itself.

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:

Diagonalization as a Change of Basis

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. The similarity relation then gives the diagonalization:

Diagonalization Formula

A = PDP^{-1}, \quad D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n), \quad P = [\mathbf{v}_1 \;\cdots\; \mathbf{v}_n]

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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.

A basis in which the map acts along the axes

In this basis the transformation stretches along each grid direction independently, with no mixing between them — which is exactly what a diagonal matrix does. Diagonalisation is therefore not a computation performed on a matrix but a search for the basis in which the matrix was always going to look simple. Hunt for it yourself on the change of basis explorer.

When no such basis exists the matrix is defective, and the next section is about what can be done instead.

When Diagonalization Fails

Not every matrix is diagonalizable.

THEOREM

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.

Pitfall: Blaming the repeated eigenvalue

Ruling out diagonalization because

\lambda = 2

is repeated would also rule out

2I

, which has the same repeated eigenvalue and is already diagonal. What fails for

A

is the eigenspace: dimension

1

, below the multiplicity

2

.

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.

THEOREMSpectral Theorem

The Spectral Theorem states that every real symmetric matrix is orthogonally similar to a diagonal matrix.

Pitfall: Transposing before normalizing

For

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

, using

P^T

as

P^{-1}

with the eigenvectors

(1, 1)

and

(1, -1)

, orthogonal but not unit, gives

P^TAP = \begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix}

instead of

D

. Dividing each column by

\sqrt{2}

makes

P

orthogonal, and then

P^TAP = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}

.

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)

.

Worked Example: Full Basis Change

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

Pitfall: Swapping P and its inverse

Recombining as

P^{-1}DP

instead of

PDP^{-1}

gives

\begin{pmatrix} 4 & -1 \\ -2 & 3 \end{pmatrix}

, not

A

. With the eigenvectors as its columns,

P

stands on the left of

D

:

A = PDP^{-1}

.

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 ℂ, or over ℝ when every eigenvalue is real) 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.

Every row below is the same operation — conjugate

A

by some

P

and see what comes out — and they are ordered by how much structure

A

has to have for the result to be simple. The split is between the cases that reach a diagonal matrix and the cases that reach the closest thing available, which is where most matrices actually land.

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Transformations · canonical forms

How simple the matrix can be made

Choosing a basis means choosing $P$ , and the question is how simple $P^{-1}AP$ can be made. The answer depends entirely on what $A$ satisfies — and the list is ordered from the strongest structure to no structure at all.

5cases

Diagonal — the best case3

1

Distinct eigenvalues§ 5

n

distinct

\lambda_i

D = P^{-1}AP

Eigenvectors for distinct eigenvalues are automatically independent, so

P

is invertible without further checking. Sufficient but not necessary — the row below covers the repeated case, and

I

diagonalizes trivially with one eigenvalue.

2

Multiplicities agree§ 5

m_g(\lambda_i) = m_a(\lambda_i)

for every

i

D = P^{-1}AP

The necessary and sufficient condition. Columns of

P

are

n

independent eigenvectors; the question is only whether that many exist, and the multiplicity comparison answers it eigenvalue by eigenvalue.

3

Real symmetric§ 7

A = A^{\mathsf{T}}

— always succeeds

D = P^{\mathsf{T}}AP

The spectral theorem guarantees not just diagonalizability but an orthonormal eigenbasis, so

P^{-1} = P^{\mathsf{T}}

and nothing needs inverting. The only row where the hypothesis is checkable at a glance.

As close as the matrix allows2

4

Defective§ 6

m_g(\lambda) < m_a(\lambda)

for some

\lambda

J = P^{-1}AP

Block diagonal with ones on the superdiagonal. The columns of

P

are eigenvectors plus generalized eigenvectors, and the number of ones counts exactly how many eigenvectors are missing — the Jordan form measures the failure rather than working around it.

5

Any complex square matrix§ 3

no hypotheses;

U

unitary

T = U^{*}AU

, upper triangular

Schur decomposition. Triangular rather than diagonal, but it exists for every square matrix over

\mathbb{C}

and

U

is unitary, which makes it numerically stable where the Jordan form is not — the reason libraries compute Schur and not Jordan.

Every row is the same operation with a different guarantee: conjugate by

P

and read the result. Similarity preserves eigenvalues, trace, determinant, rank and characteristic polynomial, so nothing about the transformation is lost in any of these — only the coordinates change, and the entries below say how much simplicity that buys.

How simple the matrix can be made·/linear-algebra/transformations/basis-changeLearn Math Class

Read downward and the hypotheses weaken while the form degrades in step. Distinct eigenvalues give a diagonal matrix and are easy to check; symmetry gives a diagonal matrix and an orthogonal

P

, so the inverse is a transpose; matching multiplicities is the exact condition and is the one that actually has to be tested. Below that line no diagonal form exists, and the question changes from how do I diagonalize this to what is the best I can do.

The last two rows answer that differently and it is worth knowing which is used. The Jordan form is the more informative — the ones on its superdiagonal count precisely how many eigenvectors each eigenvalue is missing — but it is numerically unusable, since an arbitrarily small perturbation can change the block structure entirely. The Schur form gives up on diagonal blocks and settles for triangular, in exchange for a unitary

P

and stability. Libraries compute Schur; textbooks prove Jordan.

Common Pitfalls

Five mistakes account for many wrong answers on this topic. Each is marked with a box where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Converting in the wrong direction

The matrix

P

is built from the vectors of the new basis

\mathcal{B}

, so it takes

\mathcal{B}

-coordinates to standard coordinates. To find the

\mathcal{B}

-coordinates of a vector given in standard coordinates,

P

is applied to them anyway.

Wrong working: with

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

,

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

and

\mathbf{v} = (3, 1)

,

P\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}

taken as the

\mathcal{B}

-coordinates of

\mathbf{v}

.

Right working: solve

P\mathbf{c} = (3, 1)

, that is

c_1 + c_2 = 3

and

c_1 - c_2 = 1

. This gives

c_1 = 2

,

c_2 = 1

, the same as

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

Why it happens:

P

is made of the

\mathcal{B}

vectors, so it feels like the matrix that produces

\mathcal{B}

-coordinates. It does the opposite: it takes

\mathcal{B}

-coordinates as its input.

How to check: rebuild the vector from the coordinates.

2(1, 1) + 1(1, -1) = (3, 1)

, as it should be, while

4(1, 1) + 2(1, -1) = (6, 2)

is a different vector.

Treating shared invariants as proof of similarity

Two matrices turn out to have the same trace, determinant, rank and eigenvalues, and they are declared similar. These invariants are necessary for similarity, not sufficient.

Wrong working:

I

and

N = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}

both have trace

2

, determinant

1

, rank

2

and characteristic polynomial

(\lambda - 1)^2

, so they are similar.

Right working: for every invertible

P

,

P^{-1}IP = P^{-1}P = I

so the only matrix similar to

I

is

I

itself, and

N

is not similar to

I

.

N

is a defective matrix of the kind described in the section on when diagonalization fails.

Why it happens: the list of invariants is read in both directions, as something similar matrices share and as a test that proves similarity.

How to check: a differing invariant proves two matrices are not similar; matching invariants prove nothing. Here the eigenspaces differ:

N - I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}

has rank

1

, so the eigenspace of

N

for

\lambda = 1

is one-dimensional, while for

I

it is all of

\mathbb{R}^2

.

Blaming the repeated eigenvalue

A repeated root of the characteristic polynomial is taken as the reason a matrix cannot be diagonalized.

Wrong working:

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

and

2I

both have characteristic polynomial

(\lambda - 2)^2

, so neither is diagonalizable.

Right working: compare the dimension of each eigenspace with the multiplicity. For

A

,

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

has rank

1

, so the eigenspace has dimension

2 - 1 = 1 < 2

, and

A

is not diagonalizable. For

2I

,

2I - 2I

is the zero matrix, the eigenspace has dimension

2

, and

2I

is diagonalizable; it is already diagonal, with

P = I

.

Why it happens: distinct eigenvalues guarantee a diagonalization, and the converse is assumed without checking, as if a repeated eigenvalue prevented it.

How to check: for each repeated eigenvalue

\lambda

, compute

n - \text{rank}(A - \lambda I)

and compare it with the multiplicity of

\lambda

.

Transposing before normalizing

For a symmetric matrix the eigenvectors are orthogonal, and

P^T

is used in place of

P^{-1}

. But the eigenvectors were not scaled to length

1

, so

P

is not an orthogonal matrix.

Wrong working:

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

has eigenvalues

3

and

1

, with eigenvectors

(1, 1)

and

(1, -1)

. With

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

,

P^TAP = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix}

Right working: divide each eigenvector by its length

\sqrt{2}

.

Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad Q^TAQ = \frac{1}{2}\begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix} = D

Why it happens: perpendicular columns sound like enough for an orthogonal matrix, but an orthogonal matrix needs orthonormal columns, each of length

1

.

How to check: compute

P^TP

. Here

P^TP = 2I

, not

I

, so

P^T = 2P^{-1}

, and

P^TAP

comes out as

2D

.

Swapping P and its inverse

The eigenvalues and eigenvectors are correct, but

P

,

D

and

P^{-1}

are multiplied in the wrong order, as

P^{-1}DP

instead of

PDP^{-1}

.

Wrong working: with

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

,

D = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}

and

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

from the worked example,

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

Right working:

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

, as in the verification step of the worked example.

Why it happens: both orders put

D

between

P

and

P^{-1}

, and the page uses both

A = PDP^{-1}

and

D = P^{-1}AP

, which have the inverse on opposite sides.

How to check: trace and determinant do not catch this, since the wrong matrix also has trace

7

and determinant

10

. Check an eigenvector instead:

A(1, -2) = (2, -4) = 2(1, -2)

, while the wrong matrix sends

(1, -2)

to

(6, -8)

.

Change of Basis FAQ

Does similarity preserve symmetry, zeros or sparsity?

+

None of them in general. The expression

P^{-1}AP

looks symmetric because the same letter appears on both sides, but that visual balance promises nothing about the matrix. A symmetric

A

usually comes out non-symmetric unless

P

is orthogonal, and entries, zero patterns and sparsity are lost as a matter of course.Read more →

Is "similar" the same as "row-equivalent"?

+

No, and the two disagree on exactly the properties this topic cares about. Row-equivalent matrices generally have different eigenvalues and different determinants, while similar matrices always share both. Since a tilde is sometimes used for either relation, reading it with the wrong meaning turns a true statement false without a symbol changing.Read more →

Does similarity apply to maps between two different spaces?

+

No. A map

T : V \to W

needs an independent basis choice at each end, giving

A' = Q^{-1}AP

with two unrelated matrices. Similarity is the special case where domain and codomain coincide so the two collapse into one. If the spaces differ,

A

and

A'

need not even be the same size.Read more →

Change of Basis

Same Transformation, Different Matrix

The Problem

The Change-of-Basis Matrix

Worked Example

Basis Change Notation

Basis Change Notation

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

How simple the matrix can be made

Common Pitfalls

Change of Basis FAQ

Does similarity preserve symmetry, zeros or sparsity?

Is "similar" the same as "row-equivalent"?

Does similarity apply to maps between two different spaces?