Linear Transformations: Kernel and Image

A linear transformation T: V → W preserves addition and scalar multiplication, T(au + bv) = aT(u) + bT(v); its kernel ker T = {v : T(v) = 0} and image Im T = {T(v)} are subspaces of V and W, and they satisfy dim V = dim ker T + dim Im T. In brief: for T(x, y, z) = (x + y, y − z), the kernel is the line spanned by (−1, 1, 1) and the image is all of ℝ², so 3 = 1 + 2. A linear map is injective exactly when its kernel is {0}.

Linear maps are the structure-preserving functions between vector spaces: rotations, projections, differentiation and matrix multiplication are all examples. This guide defines linear maps, the kernel and the image, shows how to compute them, relates them to injectivity, surjectivity and isomorphisms, and proves the rank–nullity theorem in its general form, with worked examples and a table.

What is a linear transformation?

A map T from a vector space V to a vector space W is linear if T(u + v) = T(u) + T(v) and T(cv) = cT(v) for all vectors u, v and scalars c. In particular T(0) = 0. Once a basis is chosen, T is described by a matrix, as, and for T(x) = Ax the matrix is A. A linear map is determined by its values on a basis.

Example 1: testing linearity

T(x, y) = (x + 2y, 3x) is linear. S(x) = x + 1 is not (S(0) = 1 ≠ 0). U(x, y) = (xy, x) is not linear, since U(2(1, 1)) = U(2, 2) = (4, 2) ≠ 2U(1, 1) = (2, 2). The map f(x) = |x| on ℝ fails additivity: |−1 + 1| = 0 but |−1| + |1| = 2.

What are the kernel and the image?

The kernel (null space) is ker T = {v in V : T(v) = 0}, a subspace of V. The image (range) is Im T = {T(v) : v in V}, a subspace of W. For T(x) = Ax, the kernel is the null space of A and the image is its column space. The dimension of the image is the rank of T, and that of the kernel is the nullity.

Example 2: a map from ℝ³ to ℝ²

T(x, y, z) = (x + y, y − z), with the matrix [[1, 1, 0], [0, 1, −1]]. Kernel: x + y = 0 and y − z = 0, so (x, y, z) = t(−1, 1, 1). Image: the columns (1, 0), (1, 1), (0, −1) span ℝ² (the rank is 2). Thus dim ker = 1, dim Im = 2, and 1 + 2 = 3 = dim ℝ³ ✓.

Example 3: a projection

The projection P(x, y, z) = (x, y, 0) onto the xy-plane has the kernel {(0, 0, z)} (the z-axis, dimension 1) and the image the xy-plane (dimension 2).

Example 4: the matrix of Example 1 from the fundamental subspaces

For T(x) = Ax with A = [[1, 2, 1, 4], [2, 4, 0, 2], [1, 2, 3, 10]], T: ℝ⁴ → ℝ³. The kernel is spanned by (−2, 1, 0, 0) and (−1, 0, −3, 1), and the image is a plane in ℝ³ spanned by (1, 2, 1) and (1, 0, 3). So 2 + 2 = 4 ✓.

When is a map injective, surjective or an isomorphism?

T is injective (one-to-one) if and only if ker T = {0}: then T(u) = T(v) implies T(u − v) = 0 and so u = v. T is surjective (onto) if and only if Im T = W. A bijective linear map is an isomorphism, and its inverse is also linear. Two finite-dimensional spaces are isomorphic exactly when their dimensions are equal, so every n-dimensional real vector space looks like ℝⁿ via coordinates. For a map between spaces of the same finite dimension, injective, surjective and bijective are equivalent.

Example 5: an isomorphism

T(x, y) = (x + y, x − y) has the kernel {0}, since x + y = 0 and x − y = 0 force x = y = 0, so T is injective, hence an isomorphism of ℝ². Its inverse is T⁻¹(u, v) = ((u + v)/2, (u − v)/2).

Example 6: the coordinate map

The map from P₂ to ℝ³ sending a + bx + cx² to (a, b, c) is an isomorphism, which shows that P₂ and ℝ³ have the same linear structure.

What is the rank–nullity theorem for maps?

For a linear map T on a finite-dimensional space V: dim V = dim ker T + dim Im T. Proof sketch: choose a basis u₁, …, u_k of the kernel and extend it by w₁, …, w_m to a basis of V. Then T(w₁), …, T(w_m) is a basis of the image: it spans the image, because the kernel vectors map to 0, and it is independent, because a relation Σc_jT(w_j) = 0 would put Σc_jw_j in the kernel and contradict the independence of the extended basis. So dim V = k + m.

Example 7: a consequence

A linear map from ℝ⁵ to ℝ³ cannot be injective (the kernel has dimension at least 5 − 3 = 2), and a map from ℝ² to ℝ³ cannot be surjective (the image has dimension at most 2).

Which examples are worth knowing?

Example 8: differentiation

D: P₂ → P₁, D(p) = p′. The kernel is the constants (dimension 1), the image is all of P₁ (dimension 2: every a + bx is the derivative of ax + bx²/2). And 1 + 2 = 3 = dim P₂ ✓. On the infinite-dimensional space of all polynomials, D is surjective but not injective.

Example 9: matrix maps

On the 2 × 2 matrices (dimension 4), T(A) = A − A^T has the kernel of the symmetric matrices (dimension 3) and the image of the skew-symmetric matrices [[0, b], [−b, 0]] (dimension 1): 3 + 1 = 4 ✓. The trace map tr: M₂ → ℝ is surjective with the kernel of the trace-zero matrices (dimension 3).

Example 10: a rotation

The rotation by an angle θ has the kernel {0} and the image ℝ²: an isomorphism, with the inverse rotation by −θ.

Table of kernels and images

Where does the concept come from?

Linear substitutions were studied in the eighteenth and nineteenth centuries (Euler, Cayley, Sylvester), and the abstract notion of a linear operator emerged with Peano’s axioms in 1888 and with the functional analysis of Hilbert and Banach in the early twentieth century. The words “kernel” and “image” were adopted with the abstract algebra of Noether and her school, which emphasized homomorphisms and their kernels.

Which mistakes are common?

Calling an affine map linear

A map x ↦ Ax + b with b ≠ 0 does not send 0 to 0, so it is not linear.

Mixing up the kernel and the image

The kernel is a subspace of the domain, the image a subspace of the codomain. For a 3 × 4 matrix, the kernel lives in ℝ⁴ and the image in ℝ³.

Equating injective with surjective in general

The equivalence holds only for maps between spaces of the same finite dimension. A map from ℝ² to ℝ³ can be injective but never surjective.

Forgetting the zero vector in the kernel

The kernel always contains 0, and a trivial kernel does not mean an empty one.