Matrices and Representation of Linear Maps

Every linear map between finite-dimensional spaces is uniquely represented by a matrix once bases are chosen, and composition of maps corresponds to matrix multiplication. We derive the change-of-basis formula, characterize invertible matrices, and show that matrix rank equals the rank of the corresponding linear map.

Folio Official

March 1, 2026

1. Representation Matrices

Definition 1 (Representation matrix).

Let

V

and

W

be finite-dimensional vector spaces with bases

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

and

C = {w_{1}, \dots, w_{m}}

, respectively. For a linear map

T : V \to W

, write

T (v_{j}) = i = 1 \sum m a_{ij} w_{i} (j = 1, \dots, n) .

The

m \times n

matrix

A = (a_{ij})

is called the representation matrix (or matrix representation) of

T

with respect to the bases

B

and

C

, and is denoted

[T]_{B}^{C}

Remark 2.

The

j

-th column of the representation matrix is the coordinate vector of

T (v_{j})

with respect to

C

. In short,

[T]_{B}^{C}

is the matrix whose columns are the images of the basis vectors, expressed in coordinates.

Example 3.

Let

T : R^{2} \to R^{3}

be defined by

T (x, y) = (x + y, 2 x, - y)

. With respect to the standard bases

{e_{1}, e_{2}}

and

{f_{1}, f_{2}, f_{3}}

T (e_{1}) = (1, 2, 0), T (e_{2}) = (1, 0, - 1) ⟹ [T] = 120 10 - 1 .

Theorem 4 (Composition corresponds to matrix multiplication).

Let

T : V \to W

and

S : W \to U

be linear maps, and let

A = [T]_{B}^{C}

and

B^{'} = [S]_{C}^{D}

be their representation matrices with respect to bases

B, C, D

. Then the representation matrix of the composition

S \circ T : V \to U

[S \circ T]_{B}^{D} = B^{'} A .

In other words, composition of linear maps corresponds to multiplication of their matrices.

2. Change of Basis

Definition 5 (Transition matrix).

Let

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

and

B^{'} = {v_{1}^{'}, \dots, v_{n}^{'}}

be two bases for

V

. The matrix

P = (p_{ij})

defined by

v_{j}^{'} = i = 1 \sum n p_{ij} v_{i} (j = 1, \dots, n)

is called the transition matrix (or change-of-basis matrix) from

B

B^{'}

Theorem 6 (Change-of-basis formula).

Let

T : V \to V

be a linear map with representation matrix

A

with respect to a basis

B

, and representation matrix

A^{'}

with respect to a basis

B^{'}

. If

P

is the transition matrix from

B

B^{'}

, then

A^{'} = P^{- 1} A P .

Proof.

Let

x

be the coordinate vector of

v \in V

with respect to

B

, and let

x^{'}

be its coordinate vector with respect to

B^{'}

. Then

x = P x^{'}

. The image of

v

under

T

has

B

-coordinates

A x

, and the

B^{'}

-coordinates of this image are

P^{- 1} A x = P^{- 1} A P x^{'}

. Therefore

A^{'} = P^{- 1} A P

. □

Definition 7 (Similar matrices).

Two square matrices

A

and

A^{'}

are said to be similar if there exists an invertible matrix

P

such that

A^{'} = P^{- 1} A P

Remark 8.

Similar matrices represent the same linear map with respect to different bases. Consequently, similar matrices share the same eigenvalues, determinant, rank, and trace.

3. The Rank of a Matrix

Definition 9 (Column rank and row rank).

The column rank of an

m \times n

matrix

A

is the dimension of the subspace of

R^{m}

spanned by the columns of

A

. The row rank is the dimension of the subspace of

R^{n}

spanned by the rows of

A

Theorem 10 (Column rank equals row rank).

For any matrix, the column rank and the row rank are equal. This common value is denoted

rank A

Proof.

View

A

as the representation matrix of the linear map

T_{A} : K^{n} \to K^{m}

defined by

T_{A} (x) = A x

. The column rank is

dim Im T_{A}

. Elementary row operations do not change the dimension of the image, and the number of nonzero rows in the reduced row echelon form equals the row rank. Hence the two ranks coincide. □

4. Invertible Matrices

Definition 11 (Invertible matrix).

n \times n

matrix

A

is called invertible (or nonsingular) if there exists a matrix

B

such that

A B = B A = I_{n}

. The matrix

B

is called the inverse of

A

and is denoted

A^{- 1}

Theorem 12 (Equivalent conditions for invertibility).

For an

n \times n

matrix

A

, the following are equivalent:

$A$ is invertible.
$rank A = n$ .
$det A \neq = 0$ .
The only solution to $A x = 0$ is $x = 0$ .
The columns of $A$ are linearly independent.
The reduced row echelon form of $A$ is $I_{n}$ .
The map $T_{A} : K^{n} \to K^{n}$ is an isomorphism.

Theorem 13 (Uniqueness of the inverse).

The inverse of an invertible matrix is unique.

Proof.

A B = B A = I

and

A C = C A = I

, then

B = B I = B (A C) = (B A) C = I C = C

. □

Theorem 14 (Properties of the inverse).

A

and

B

are invertible matrices, then:

$(A^{- 1})^{- 1} = A$ .
$(A B)^{- 1} = B^{- 1} A^{- 1}$ .
$(A^{T})^{- 1} = (A^{- 1})^{T}$ .

Proof.

(1) Since

A^{- 1} \cdot A = I

and

A \cdot A^{- 1} = I

, the matrix

A

serves as the inverse of

A^{- 1}

. By uniqueness,

(A^{- 1})^{- 1} = A

(2) We verify directly:

(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I

. Similarly,

(B^{- 1} A^{- 1}) (A B) = I

. By uniqueness,

(A B)^{- 1} = B^{- 1} A^{- 1}

(3) We compute

(A^{T}) (A^{- 1})^{T} = (A^{- 1} A)^{T} = I^{T} = I

. Likewise,

(A^{- 1})^{T} A^{T} = (A A^{- 1})^{T} = I

. By uniqueness,

(A^{T})^{- 1} = (A^{- 1})^{T}

. □

Linear Algebra Algebra Textbook Representation Matrix Change of Basis Invertible Matrices

Folio Official

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

1 followers·107 articles

Matrices and Representation of Linear Maps

Folio Official

March 1, 2026

1. Representation Matrices

Definition 1 (Representation matrix).

Let

V

and

W

be finite-dimensional vector spaces with bases

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

and

C = {w_{1}, \dots, w_{m}}

, respectively. For a linear map

T : V \to W

, write

T (v_{j}) = i = 1 \sum m a_{ij} w_{i} (j = 1, \dots, n) .

The

m \times n

matrix

A = (a_{ij})

is called the representation matrix (or matrix representation) of

T

with respect to the bases

B

and

C

, and is denoted

[T]_{B}^{C}

Remark 2.

The

j

-th column of the representation matrix is the coordinate vector of

T (v_{j})

with respect to

C

. In short,

[T]_{B}^{C}

is the matrix whose columns are the images of the basis vectors, expressed in coordinates.

Example 3.

Let

T : R^{2} \to R^{3}

be defined by

T (x, y) = (x + y, 2 x, - y)

. With respect to the standard bases

{e_{1}, e_{2}}

and

{f_{1}, f_{2}, f_{3}}

T (e_{1}) = (1, 2, 0), T (e_{2}) = (1, 0, - 1) ⟹ [T] = 120 10 - 1 .

Theorem 4 (Composition corresponds to matrix multiplication).

Let

T : V \to W

and

S : W \to U

be linear maps, and let

A = [T]_{B}^{C}

and

B^{'} = [S]_{C}^{D}

be their representation matrices with respect to bases

B, C, D

. Then the representation matrix of the composition

S \circ T : V \to U

[S \circ T]_{B}^{D} = B^{'} A .

In other words, composition of linear maps corresponds to multiplication of their matrices.

2. Change of Basis

Definition 5 (Transition matrix).

Let

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

and

B^{'} = {v_{1}^{'}, \dots, v_{n}^{'}}

be two bases for

V

. The matrix

P = (p_{ij})

defined by

v_{j}^{'} = i = 1 \sum n p_{ij} v_{i} (j = 1, \dots, n)

is called the transition matrix (or change-of-basis matrix) from

B

B^{'}

Theorem 6 (Change-of-basis formula).

Let

T : V \to V

be a linear map with representation matrix

A

with respect to a basis

B

, and representation matrix

A^{'}

with respect to a basis

B^{'}

. If

P

is the transition matrix from

B

B^{'}

, then

A^{'} = P^{- 1} A P .

Proof.

Let

x

be the coordinate vector of

v \in V

with respect to

B

, and let

x^{'}

be its coordinate vector with respect to

B^{'}

. Then

x = P x^{'}

. The image of

v

under

T

has

B

-coordinates

A x

, and the

B^{'}

-coordinates of this image are

P^{- 1} A x = P^{- 1} A P x^{'}

. Therefore

A^{'} = P^{- 1} A P

. □

Definition 7 (Similar matrices).

Two square matrices

A

and

A^{'}

are said to be similar if there exists an invertible matrix

P

such that

A^{'} = P^{- 1} A P

Remark 8.

Similar matrices represent the same linear map with respect to different bases. Consequently, similar matrices share the same eigenvalues, determinant, rank, and trace.

3. The Rank of a Matrix

Definition 9 (Column rank and row rank).

The column rank of an

m \times n

matrix

A

is the dimension of the subspace of

R^{m}

spanned by the columns of

A

. The row rank is the dimension of the subspace of

R^{n}

spanned by the rows of

A

Theorem 10 (Column rank equals row rank).

For any matrix, the column rank and the row rank are equal. This common value is denoted

rank A

Proof.

View

A

as the representation matrix of the linear map

T_{A} : K^{n} \to K^{m}

defined by

T_{A} (x) = A x

. The column rank is

dim Im T_{A}

. Elementary row operations do not change the dimension of the image, and the number of nonzero rows in the reduced row echelon form equals the row rank. Hence the two ranks coincide. □

4. Invertible Matrices

Definition 11 (Invertible matrix).

n \times n

matrix

A

is called invertible (or nonsingular) if there exists a matrix

B

such that

A B = B A = I_{n}

. The matrix

B

is called the inverse of

A

and is denoted

A^{- 1}

Theorem 12 (Equivalent conditions for invertibility).

For an

n \times n

matrix

A

, the following are equivalent:

$A$ is invertible.
$rank A = n$ .
$det A \neq = 0$ .
The only solution to $A x = 0$ is $x = 0$ .
The columns of $A$ are linearly independent.
The reduced row echelon form of $A$ is $I_{n}$ .
The map $T_{A} : K^{n} \to K^{n}$ is an isomorphism.

Theorem 13 (Uniqueness of the inverse).

The inverse of an invertible matrix is unique.

Proof.

A B = B A = I

and

A C = C A = I

, then

B = B I = B (A C) = (B A) C = I C = C

. □

Theorem 14 (Properties of the inverse).

A

and

B

are invertible matrices, then:

$(A^{- 1})^{- 1} = A$ .
$(A B)^{- 1} = B^{- 1} A^{- 1}$ .
$(A^{T})^{- 1} = (A^{- 1})^{T}$ .

Proof.

(1) Since

A^{- 1} \cdot A = I

and

A \cdot A^{- 1} = I

, the matrix

A

serves as the inverse of

A^{- 1}

. By uniqueness,

(A^{- 1})^{- 1} = A

(2) We verify directly:

(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I

. Similarly,

(B^{- 1} A^{- 1}) (A B) = I

. By uniqueness,

(A B)^{- 1} = B^{- 1} A^{- 1}

(3) We compute

(A^{T}) (A^{- 1})^{T} = (A^{- 1} A)^{T} = I^{T} = I

. Likewise,

(A^{- 1})^{T} A^{T} = (A A^{- 1})^{T} = I

. By uniqueness,

(A^{T})^{- 1} = (A^{- 1})^{T}

. □

Linear Algebra Algebra Textbook Representation Matrix Change of Basis Invertible Matrices

Folio Official

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

1 followers·107 articles

Matrices and Representation of Linear Maps

1. Representation Matrices

2. Change of Basis

3. The Rank of a Matrix

4. Invertible Matrices

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Matrices and Representation of Linear Maps

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Share your expertise with the world

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