Diagonalization: Simplifying Matrices by Choice of Basis

A matrix is diagonalizable if and only if the sum of its geometric multiplicities equals the matrix size. We prove this criterion, give a step-by-step diagonalization procedure with applications to computing A^n, and prove Schur's theorem that every complex square matrix is unitarily triangularizable.

Folio Official

March 1, 2026

1. Definition of Diagonalizability

Definition 1 (Diagonalizable matrix).

n \times n

matrix

A

is called diagonalizable if there exists an invertible matrix

P

such that

P^{- 1} A P = λ_{1} O ⋱ O λ_{n}

is a diagonal matrix.

Remark 2.

Diagonalizability is equivalent to the statement that, in a suitably chosen basis,

A

is represented by a diagonal matrix. The columns of

P

are eigenvectors, and the diagonal entries are the corresponding eigenvalues.

2. Conditions for Diagonalizability

Theorem 3 (Necessary and sufficient conditions for diagonalizability).

The following conditions on an

n \times n

matrix

A

are equivalent:

$A$ is diagonalizable.
$A$ possesses $n$ linearly independent eigenvectors.
For every eigenvalue $λ_{i}$ , the geometric multiplicity equals the algebraic multiplicity.
The characteristic polynomial $p_{A} (λ)$ splits completely over $K$ , and for each eigenvalue the geometric multiplicity equals the algebraic multiplicity.

Proof.

(1 \Leftrightarrow 2)

: The equation

P^{- 1} A P = D

is equivalent to

A P = P D

. Writing

P = (v_{1} \dots v_{n})

and

D = diag (λ_{1}, \dots, λ_{n})

, we see that

A v_{i} = λ_{i} v_{i}

for each

i

. The matrix

P

is invertible precisely when

v_{1}, \dots, v_{n}

are linearly independent.

(2 \Leftrightarrow 3)

: The eigenspaces for distinct eigenvalues are always in direct sum, so the total count of linearly independent eigenvectors is the sum of the geometric multiplicities. This sum equals

n

if and only if each geometric multiplicity equals the corresponding algebraic multiplicity. □

Example 4 (A diagonalizable matrix).

Let

A = (41 - 2 1)

. Its characteristic polynomial is

p_{A} (λ) = λ^{2} - 5 λ + 6 = (λ - 2) (λ - 3)

. Since

A

has two distinct eigenvalues, it is diagonalizable.

For

λ_{1} = 2

v_{1} = (1, 1)^{T}

. For

λ_{2} = 3

v_{2} = (2, 1)^{T}

. Setting

P = (1121)

, we obtain

P^{- 1} A P = (2003) .

Example 5 (A non-diagonalizable matrix).

Let

A = (2012)

. The characteristic polynomial is

p_{A} (λ) = (λ - 2)^{2}

, so

λ = 2

has algebraic multiplicity

2

. However,

dim V_{2} = 1

, so the geometric multiplicity is strictly less than the algebraic multiplicity, and

A

is not diagonalizable.

3. The Diagonalization Procedure

To diagonalize an $n \times n$ matrix $A$ , proceed as follows:

Compute the characteristic polynomial $p_{A} (λ) = det (A - λ I)$ .
Solve $p_{A} (λ) = 0$ to find the eigenvalues $λ_{1}, \dots, λ_{k}$ and their algebraic multiplicities.
For each eigenvalue $λ_{i}$ , compute a basis for the eigenspace $V_{λ_{i}} = ker (A - λ_{i} I)$ .
Check that the geometric multiplicity equals the algebraic multiplicity for every eigenvalue. (If not, $A$ is not diagonalizable.)
Form $P$ by arranging the eigenvectors as columns; then $D = P^{- 1} A P$ is diagonal.

4. Computing Powers of a Matrix

When $A$ is diagonalizable, writing $A = P D P^{- 1}$ gives

A^{n} = P D^{n} P^{- 1} = P λ_{1}^{n} O ⋱ O λ_{k}^{n} P^{- 1} .

Example 6.

For

A = (41 - 2 1)

with

P = (1121)

and

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

, we obtain

A^{n} = (1121) (2^{n} 0 0 3^{n}) (- 1 1 2 - 1) = (- 2^{n} + 2 \cdot 3^{n} - 2^{n} + 3^{n} 2^{n + 1} - 2 \cdot 3^{n} 2^{n + 1} - 3^{n}) .

5. Triangularization

Theorem 7 (Schur's triangularization theorem).

Over

K = C

, every

n \times n

matrix

A

is unitarily triangularizable: there exists a unitary matrix

U

such that

U^{*} A U

is upper triangular.

Remark 8.

Even matrices that cannot be diagonalized can always be triangularized (provided the base field is algebraically closed). The diagonal entries of the resulting upper triangular matrix are the eigenvalues of

A

, which confirms the identities

tr A = \sum λ_{i}

and

det A = \prod λ_{i}

Theorem 9.

A

is triangularized as

T = P^{- 1} A P

with

T

upper triangular, then

T^{n}

is again upper triangular with diagonal entries

λ_{i}^{n}

. The factorization

A^{n} = P T^{n} P^{- 1}

therefore provides a method for computing powers even for non-diagonalizable matrices.

Linear Algebra Algebra Textbook Diagonalization Triangularization Matrix Powers

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Diagonalization: Simplifying Matrices by Choice of Basis

Folio Official

March 1, 2026

1. Definition of Diagonalizability

Definition 1 (Diagonalizable matrix).

n \times n

matrix

A

is called diagonalizable if there exists an invertible matrix

P

such that

P^{- 1} A P = λ_{1} O ⋱ O λ_{n}

is a diagonal matrix.

Remark 2.

Diagonalizability is equivalent to the statement that, in a suitably chosen basis,

A

is represented by a diagonal matrix. The columns of

P

are eigenvectors, and the diagonal entries are the corresponding eigenvalues.

2. Conditions for Diagonalizability

Theorem 3 (Necessary and sufficient conditions for diagonalizability).

The following conditions on an

n \times n

matrix

A

are equivalent:

$A$ is diagonalizable.
$A$ possesses $n$ linearly independent eigenvectors.
For every eigenvalue $λ_{i}$ , the geometric multiplicity equals the algebraic multiplicity.
The characteristic polynomial $p_{A} (λ)$ splits completely over $K$ , and for each eigenvalue the geometric multiplicity equals the algebraic multiplicity.

Proof.

(1 \Leftrightarrow 2)

: The equation

P^{- 1} A P = D

is equivalent to

A P = P D

. Writing

P = (v_{1} \dots v_{n})

and

D = diag (λ_{1}, \dots, λ_{n})

, we see that

A v_{i} = λ_{i} v_{i}

for each

i

. The matrix

P

is invertible precisely when

v_{1}, \dots, v_{n}

are linearly independent.

(2 \Leftrightarrow 3)

: The eigenspaces for distinct eigenvalues are always in direct sum, so the total count of linearly independent eigenvectors is the sum of the geometric multiplicities. This sum equals

n

if and only if each geometric multiplicity equals the corresponding algebraic multiplicity. □

Example 4 (A diagonalizable matrix).

Let

A = (41 - 2 1)

. Its characteristic polynomial is

p_{A} (λ) = λ^{2} - 5 λ + 6 = (λ - 2) (λ - 3)

. Since

A

has two distinct eigenvalues, it is diagonalizable.

For

λ_{1} = 2

v_{1} = (1, 1)^{T}

. For

λ_{2} = 3

v_{2} = (2, 1)^{T}

. Setting

P = (1121)

, we obtain

P^{- 1} A P = (2003) .

Example 5 (A non-diagonalizable matrix).

Let

A = (2012)

. The characteristic polynomial is

p_{A} (λ) = (λ - 2)^{2}

, so

λ = 2

has algebraic multiplicity

2

. However,

dim V_{2} = 1

, so the geometric multiplicity is strictly less than the algebraic multiplicity, and

A

is not diagonalizable.

3. The Diagonalization Procedure

To diagonalize an $n \times n$ matrix $A$ , proceed as follows:

Compute the characteristic polynomial $p_{A} (λ) = det (A - λ I)$ .
Solve $p_{A} (λ) = 0$ to find the eigenvalues $λ_{1}, \dots, λ_{k}$ and their algebraic multiplicities.
For each eigenvalue $λ_{i}$ , compute a basis for the eigenspace $V_{λ_{i}} = ker (A - λ_{i} I)$ .
Check that the geometric multiplicity equals the algebraic multiplicity for every eigenvalue. (If not, $A$ is not diagonalizable.)
Form $P$ by arranging the eigenvectors as columns; then $D = P^{- 1} A P$ is diagonal.

4. Computing Powers of a Matrix

When $A$ is diagonalizable, writing $A = P D P^{- 1}$ gives

A^{n} = P D^{n} P^{- 1} = P λ_{1}^{n} O ⋱ O λ_{k}^{n} P^{- 1} .

Example 6.

For

A = (41 - 2 1)

with

P = (1121)

and

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

, we obtain

A^{n} = (1121) (2^{n} 0 0 3^{n}) (- 1 1 2 - 1) = (- 2^{n} + 2 \cdot 3^{n} - 2^{n} + 3^{n} 2^{n + 1} - 2 \cdot 3^{n} 2^{n + 1} - 3^{n}) .

5. Triangularization

Theorem 7 (Schur's triangularization theorem).

Over

K = C

, every

n \times n

matrix

A

is unitarily triangularizable: there exists a unitary matrix

U

such that

U^{*} A U

is upper triangular.

Remark 8.

A

, which confirms the identities

tr A = \sum λ_{i}

and

det A = \prod λ_{i}

Theorem 9.

A

is triangularized as

T = P^{- 1} A P

with

T

upper triangular, then

T^{n}

is again upper triangular with diagonal entries

λ_{i}^{n}

. The factorization

A^{n} = P T^{n} P^{- 1}

therefore provides a method for computing powers even for non-diagonalizable matrices.

Linear Algebra Algebra Textbook Diagonalization Triangularization Matrix Powers

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Diagonalization: Simplifying Matrices by Choice of Basis

1. Definition of Diagonalizability

2. Conditions for Diagonalizability

3. The Diagonalization Procedure

4. Computing Powers of a Matrix

5. Triangularization

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Diagonalization: Simplifying Matrices by Choice of Basis

1. Definition of Diagonalizability

2. Conditions for Diagonalizability

3. The Diagonalization Procedure

4. Computing Powers of a Matrix

5. Triangularization

Share your expertise with the world

More from Folio Official

Matrices and Representation of Linear Maps

The Determinant: A Scalar Invariant of Square Matrices

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The Singular Value Decomposition: Structure of Arbitrary Matrices