What is the image of a linear transformation?

The image (or range) of T is the set of all outputs: Im(T) = {T(v) : v ∈ V}. It is a subspace of the codomain. For matrix transformations, the image is the column space of A. Its dimension equals the rank of A, and a vector w is in the image if and only if Ax = w has a solution.

What is the kernel of a linear transformation?

The kernel of T is the set of all inputs that map to zero: ker(T) = {v ∈ V : T(v) = 0}. It is a subspace of the domain. For matrix transformations, the kernel is the null space of A. Its dimension (the nullity) equals n minus the rank.

How do you determine if a linear transformation is injective?

A linear transformation is injective (one-to-one) if and only if its kernel is trivial: ker(T) = {0}. For matrix transformations, this is equivalent to full column rank, linearly independent columns, no free variables, and (for square matrices) nonzero determinant.

What is the rank-nullity theorem?

The rank-nullity theorem states that dim(Im(T)) + dim(ker(T)) = dim(V) for any linear transformation T: V → W. The domain dimensions split between what the map preserves (the image) and what it destroys (the kernel). For matrices, rank + nullity = n.

When is a linear transformation an isomorphism?

A linear transformation is an isomorphism when it is both injective and surjective (bijective). For maps between spaces of equal dimension, any one of the three conditions implies the other two. For matrix transformations, this is equivalent to the matrix being square and invertible.

Image and Kernel of Linear Transformation

Bijectivity and Isomorphisms

The Rank-Nullity Theorem for Maps

Dimension Constraints

Computing the Image and Kernel

The Fundamental Decomposition

Summary: Image and Kernel Side by Side

What a Transformation Hits and What It Kills

Every linear transformation partitions its domain into two complementary pieces: the kernel, consisting of everything that maps to zero, and a complement that maps bijectively onto the image. The dimensions of the kernel and image are locked together by the rank-nullity theorem, and their relationship determines whether the transformation is injective, surjective, or neither.

The Image

The image (or range) of a linear transformation

T: V \to W

is the set of all outputs:

\text{Im}(T) = \{T(\mathbf{v}) : \mathbf{v} \in V\}

The image is a subspace of

W

. It contains

T(\mathbf{0}) = \mathbf{0}

, and if

T(\mathbf{u})

and

T(\mathbf{v})

are in the image, then so is

cT(\mathbf{u}) + dT(\mathbf{v}) = T(c\mathbf{u} + d\mathbf{v})

— closure under both operations follows from linearity.

When

T(\mathbf{x}) = A\mathbf{x}

for a matrix

A

, the image is the column space of

A

: the set of all vectors expressible as linear combinations of the columns. The dimension of the image equals the rank of

A

.

The image answers the reachability question: a vector

\mathbf{w} \in W

is in the image if and only if the equation

T(\mathbf{v}) = \mathbf{w}

— equivalently,

A\mathbf{x} = \mathbf{w}

— has a solution.

The Kernel

The kernel (or null space) of

T: V \to W

is the set of all inputs that map to zero:

\ker(T) = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}\}

The kernel is a subspace of

V

. It contains

\mathbf{0}

(since

T(\mathbf{0}) = \mathbf{0}

), and if

T(\mathbf{u}) = \mathbf{0}

and

T(\mathbf{v}) = \mathbf{0}

, then

T(c\mathbf{u} + d\mathbf{v}) = cT(\mathbf{u}) + dT(\mathbf{v}) = \mathbf{0}

, so

c\mathbf{u} + d\mathbf{v} \in \ker(T)

.

When

T(\mathbf{x}) = A\mathbf{x}

, the kernel is the null space of

A

: all solutions to the homogeneous system

A\mathbf{x} = \mathbf{0}

. Its dimension is the nullity, equal to

n - \text{rank}(A)

.

The kernel measures the information lost by

T

. Vectors in the kernel are collapsed to

\mathbf{0}

— they represent directions that the transformation annihilates. A larger kernel means more information is destroyed.

Injectivity

A linear transformation

T

is injective (one-to-one) if different inputs always produce different outputs:

T(\mathbf{u}) = T(\mathbf{v})

implies

\mathbf{u} = \mathbf{v}

.

For linear maps, injectivity has an elegant equivalent:

T

is injective if and only if

\ker(T) = \{\mathbf{0}\}

. The proof is short. If

T(\mathbf{u}) = T(\mathbf{v})

, then

T(\mathbf{u} - \mathbf{v}) = T(\mathbf{u}) - T(\mathbf{v}) = \mathbf{0}

, so

\mathbf{u} - \mathbf{v} \in \ker(T)

. If the kernel is trivial,

\mathbf{u} - \mathbf{v} = \mathbf{0}

and

\mathbf{u} = \mathbf{v}

.

For matrix transformations, injectivity is equivalent to full column rank:

\text{rank}(A) = n

. This means every column is a pivot column, no free variables exist in

A\mathbf{x} = \mathbf{0}

, the columns are linearly independent, and the determinant is nonzero (in the square case).

Injectivity means the transformation preserves distinctness — no two different inputs are confused with each other.

Surjectivity

A linear transformation

T: V \to W

is surjective (onto) if

\text{Im}(T) = W

— every vector in the codomain is the image of some vector in the domain.

For matrix transformations, surjectivity is equivalent to full row rank:

\text{rank}(A) = m

. This means every row contains a pivot, the column space is all of

\mathbb{R}^m

, and the system

A\mathbf{x} = \mathbf{b}

has a solution for every right-hand side

\mathbf{b}

.

Surjectivity means the transformation has no blind spots — every output is reachable from some input. Failure of surjectivity means the image is a proper subspace of the codomain: certain vectors in

W

are inherently unreachable, no matter what input is chosen.

Equivalent condition	Injectivity (one-to-one)	Surjectivity (onto)
Map-level definition	T(u) = T(v) ⇒ u = v	every w ∈ W is the image of some v
Kernel / image	ker(T) = {0}	Im(T) = W
Rank condition	rank(A) = n (full column rank)	rank(A) = m (full row rank)
Columns / rows of A	columns are linearly independent	columns span ℝᵐ (column space = ℝᵐ)
System Ax = b behavior	at most one solution for every b	at least one solution for every b
Echelon-form trait	no free variables in Ax = 0	every row contains a pivot
Determinant (square A only)	det(A) ≠ 0	det(A) ≠ 0 (collapses to the same test)

Bijectivity and Isomorphisms

A linear transformation that is both injective and surjective is bijective. A bijective linear transformation is called an isomorphism — it establishes that the domain and codomain are structurally identical as vector spaces.

For a map

T: V \to W

between spaces of equal dimension (

\dim(V) = \dim(W) = n

), the three conditions collapse: injective

\iff

surjective

\iff

bijective. Checking any one of the three establishes the other two. This is because the rank-nullity theorem forces

\dim(\text{Im}(T)) + \dim(\ker(T)) = n

, and

\dim(\text{Im}(T)) \leq n = \dim(W)

. If the kernel is trivial (injective), the image has dimension

n

and must equal all of

W

(surjective). If the image is all of

W

(surjective), the kernel must have dimension

0

(injective).

For matrix transformations between spaces of the same dimension, bijectivity is equivalent to the matrix being square and invertible.

The Rank-Nullity Theorem for Maps

For a linear transformation

T: V \to W

with

V

finite-dimensional:

\dim(\text{Im}(T)) + \dim(\ker(T)) = \dim(V)

The domain dimensions split between what the map preserves and what it destroys. The image captures the dimensions that survive; the kernel captures the dimensions that are annihilated.

For matrix transformations

T(\mathbf{x}) = A\mathbf{x}

, this becomes

\text{rank}(A) + \text{nullity}(A) = n

— the familiar rank-nullity theorem in concrete language.

The theorem constrains the interplay between injectivity and surjectivity. If

\dim(V) > \dim(W)

, the image can have at most

\dim(W)

dimensions, forcing the kernel to have at least

\dim(V) - \dim(W)

dimensions — the map cannot be injective. If

\dim(V) < \dim(W)

, the image cannot fill all of

W

— the map cannot be surjective.

Dimension Constraints

The rank-nullity theorem imposes hard limits on what a linear transformation can achieve.

T: V \to W

can be injective only if

\dim(V) \leq \dim(W)

. A map from a larger space to a smaller one must collapse some directions — the kernel is forced to be nontrivial.

T: V \to W

can be surjective only if

\dim(V) \geq \dim(W)

. A map from a smaller space to a larger one cannot cover all directions — the image is a proper subspace.

T

can be bijective only if

\dim(V) = \dim(W)

. This is necessary but not sufficient — even with equal dimensions, the map must still have full rank.

These constraints apply to all linear maps, not just matrix transformations. They are consequences of the rank-nullity theorem and the dimension theory of vector spaces.

Dimension relation	Injective possible?	Surjective possible?	Bijective possible?	Forced consequence
dim(V) < dim(W)	✓	✗	✗	image is a proper subspace of W
dim(V) = dim(W)	✓	✓	✓	injective ⇔ surjective ⇔ bijective (one check establishes all three)
dim(V) > dim(W)	✗	✓	✗	kernel forced to have dim ≥ dim(V) − dim(W)

Computing the Image and Kernel

For a matrix transformation

T(\mathbf{x}) = A\mathbf{x}

, the image and kernel are computed by row reduction.

The kernel is the null space of

A

: solve

A\mathbf{x} = \mathbf{0}

, reduce to echelon form, and express the solution in parametric vector form. Each free variable contributes one basis vector for

\ker(T)

.

The image is the column space of

A

: row reduce

A

, identify the pivot columns, and take the corresponding columns of the original matrix

A

as a basis for

\text{Im}(T)

Worked Example

For

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 1 & 3 & 4 \end{pmatrix}

, row reduction gives

\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}

. Pivots in columns

1

and

2

. The image has basis

\{(1, 0, 1), (2, 1, 3)\}

— two-dimensional. The kernel has one free variable (

x_3 = t

), giving

\ker(T) = \text{Span}\{(-1, -1, 1)\}

— one-dimensional. Check:

2 + 1 = 3 = n

The Fundamental Decomposition

The rank-nullity theorem has a structural interpretation that goes beyond dimension counting. The domain

V

decomposes as a direct sum:

V = \ker(T) \oplus (\text{a complement of } \ker(T))

The transformation

T

kills everything in the kernel and maps the complement bijectively onto the image. Every vector

\mathbf{v} \in V

splits as

\mathbf{v} = \mathbf{v}_k + \mathbf{v}_c

where

\mathbf{v}_k \in \ker(T)

and

\mathbf{v}_c

is in the complement. Then

T(\mathbf{v}) = T(\mathbf{v}_c)

, and the restriction of

T

to the complement is a bijection onto

\text{Im}(T)

.

For matrix transformations, the four fundamental subspaces provide the natural complement: the row space of

A

is the orthogonal complement of the null space in

\mathbb{R}^n

, and

A

maps the row space bijectively onto the column space. The null-space component is destroyed; the row-space component survives intact.

Summary: Image and Kernel Side by Side

The whole page revolves around a single duality: the image lives in the codomain and records what T reaches, while the kernel lives in the domain and records what T destroys. Every concept introduced — injectivity, surjectivity, bijectivity, rank-nullity, dimension constraints, the fundamental decomposition — is a relation between these two subspaces. The table below sets them side by side across every attribute the page has touched, so the symmetry and the link supplied by rank-nullity are visible at a glance.

Attribute	Image of T	Kernel of T
Set-theoretic definition	{ T(v) : v ∈ V }	{ v ∈ V : T(v) = 0 }
Where it lives	subspace of the codomain W	subspace of the domain V
Matrix counterpart (T(x) = Ax)	column space of A	null space of A
Dimension	rank(A)	nullity(A) = n − rank(A)
What it measures	reachable outputs — directions T can produce	information destroyed — directions T collapses to 0
Computed by	pivot columns of A (original columns, not RREF)	parametric solution of Ax = 0 from RREF
Trivial case	{0} ⇔ T is the zero map	{0} ⇔ T is injective
Full case	= W ⇔ T is surjective	= V ⇔ T is the zero map
Linked by rank-nullity	dim(Im(T)) + dim(ker(T)) = dim(V) — what survives plus what is destroyed equals the input dimension