Matrix Representation of Linear Operators and Coordinate Transformations

Once a basis is chosen, a linear operator T on a vector space is described by a matrix whose j-th column is the coordinate vector of T applied to the j-th basis vector; if the basis changes with a transition matrix P, the matrix of the operator changes to P⁻¹AP. In brief: the operator T(x, y) = (2x + y, x + 3y) has the matrix [[2, 1], [1, 3]] in the standard basis, and in the basis (1, 1), (1, −1) it has the matrix [[7/2, −1/2], [−1/2, 3/2]]; both matrices have trace 5 and determinant 5.

A linear operator is a basis-free object, and its matrix is a representation that depends on the chosen basis. This guide explains how to build the matrix of an operator, how coordinates of vectors change, how the matrix changes under a change of basis, and what stays the same, with worked examples and a table.

How do you find the matrix of an operator?

Let T be a linear operator on an n-dimensional space V with the basis B = (b₁, …, b_n). Every vector v has the coordinates [v]_B (the numbers in v = c₁b₁ + … + c_nb_n). The matrix [T]_B is the n × n matrix whose j-th column is [T(b_j)]_B. Then [T(v)]_B = [T]_B [v]_B. In the standard basis of ℝⁿ, the columns are simply the images of the unit vectors.

Example 1: an operator on the plane

T(x, y) = (2x + y, x + 3y). T(1, 0) = (2, 1) and T(0, 1) = (1, 3), so the standard matrix is [[2, 1], [1, 3]].

Example 2: differentiation of polynomials

Let D = d/dx act on the polynomials of degree at most 2, with the basis (1, x, x²). D(1) = 0 = (0, 0, 0), D(x) = 1 = (1, 0, 0), D(x²) = 2x = (0, 2, 0). The columns give [D] = [[0, 1, 0], [0, 0, 2], [0, 0, 0]]. Applying it to p = 3 + 5x + 4x² = (3, 5, 4): [[0, 1, 0], [0, 0, 2], [0, 0, 0]] · (3, 5, 4) = (5, 8, 0), that is 5 + 8x, the derivative ✓. The matrix is nilpotent, since the third derivative of a quadratic is zero.

Example 3: a rotation and a reflection

The rotation by 90° has the matrix [[0, −1], [1, 0]]. The reflection in the line y = x has [[0, 1], [1, 0]]. In the basis (1, 1), (1, −1) the reflection acts as J(1, 1) = (1, 1) and J(1, −1) = (−1, 1) = −(1, −1), so its matrix is the diagonal matrix [[1, 0], [0, −1]], a much simpler description.

How do coordinates change with the basis?

Let B and B′ be two bases of ℝⁿ, and let P be the matrix whose columns are the vectors of B′ written in B (the transition matrix from B′ to B). Then [v]_B = P [v]_B′, that is [v]_B′ = P⁻¹ [v]_B. The matrix P is invertible because its columns are independent.

Example 4: new coordinates

With B the standard basis and B′ = ((1, 1), (1, −1)), P = [[1, 1], [1, −1]] and P⁻¹ = [[1/2, 1/2], [1/2, −1/2]]. The vector v = (3, 1) has [v]_B′ = P⁻¹(3, 1) = (2, 1). Check: 2(1, 1) + 1(1, −1) = (3, 1) ✓.

How does the matrix change with the basis?

If A = [T]_B and A′ = [T]_B′, then A′ = P⁻¹AP. Matrices related in this way are called similar. The reason: to compute T on B′-coordinates, convert to B (multiply by P), apply A, and convert back (multiply by P⁻¹).

Example 5: changing the basis of Example 1

A = [[2, 1], [1, 3]], P = [[1, 1], [1, −1]]. AP = [[2 + 1, 2 − 1], [1 + 3, 1 − 3]] = [[3, 1], [4, −2]]. Then P⁻¹AP = [[1/2, 1/2], [1/2, −1/2]] · [[3, 1], [4, −2]] = [[7/2, −1/2], [−1/2, 3/2]]. Direct check: T(1, 1) = (3, 4) = (7/2)(1, 1) − (1/2)(1, −1), the first column ✓, and T(1, −1) = (1, −2) = −(1/2)(1, 1) + (3/2)(1, −1), the second column ✓.

What does not depend on the basis?

Quantities defined by the operator itself are the same for all similar matrices: the determinant, the trace, the rank, the eigenvalues and the characteristic polynomial. In Example 5, both matrices have the trace 5 and the determinant 5 (7/2 · 3/2 − 1/4 = 21/4 − 1/4 = 5), while the entries themselves differ. Choosing a basis in which the matrix is as simple as possible (diagonal or Jordan form) is a central goal.

How do operations on operators correspond to matrices?

In a fixed basis, the sum T + S has the matrix [T] + [S], a scalar multiple has k[T], the composition S ∘ T has the product [S][T] (first T, then S), and the inverse operator has the matrix inverse [T]⁻¹. This is why matrix multiplication is defined as it is.

Example 6: composition

Let S be the reflection in y = x, [S] = [[0, 1], [1, 0]], and T from Example 1. S ∘ T(x, y) = S(2x + y, x + 3y) = (x + 3y, 2x + y), and [S][T] = [[0, 1], [1, 0]] · [[2, 1], [1, 3]] = [[1, 3], [2, 1]] ✓, which matches. The other order T ∘ S has the matrix [[2, 1], [1, 3]] · [[0, 1], [1, 0]] = [[1, 2], [3, 1]].

Example 7: a map between different spaces

For a linear map T: ℝ³ → ℝ², T(x, y, z) = (x + y, y − z), the matrix is 2 × 3: [[1, 1, 0], [0, 1, −1]]. With different bases in domain and codomain, A′ = Q⁻¹AP, where P and Q are the two transition matrices.

Table of examples

Where does the idea come from?

Cayley’s 1858 memoir introduced matrices as representations of linear substitutions and defined their product to correspond to composition. The abstract notion of a linear operator on a vector space, independent of coordinates, was developed in the late nineteenth and early twentieth centuries by Peano, Weyl and the school of abstract algebra.

Which mistakes are common?

Writing the images as rows

With the convention [T(v)] = A[v], the images T(b_j) form the columns of A, not the rows.

Using P instead of P⁻¹

Check the direction of the transition matrix. If P has the new basis vectors as columns, then new coordinates are P⁻¹ times old coordinates, and A′ = P⁻¹AP.

Expressing the images in the wrong basis

When computing [T]_B′ directly, express T(b′_j) in the basis B′, not in the standard basis.

Reversing the order of composition

The matrix of “first T, then S” is [S][T], with the later operator on the left.