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Linear Transformations, Rank-Nullity and Quotient Spaces

1 · Prerequisites

2 · Summary

Vector spaces, linear combinations, bases, and finite dimension are the prerequisites for this development. A linear map preserves the vector-space operations, and its kernel and image measure respectively what it collapses and what it reaches.

This page defines linear maps, kernels, images, rank, and nullity. It proves the elementary preservation laws, characterises injectivity by the kernel, extends a kernel basis to obtain a basis of the image, and derives the finite-dimensional rank–nullity theorem.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Linear map between vector spaces over the same field

Definition

Let $V$ and $W$ be vector spaces over the same field $F$ . A function $T : V \to W$ is linear when

$T (a u + b v) = a T (u) + b T (v)$

for all $a, b \in F$ and $u, v \in V$ . Such a function is a linear map, or linear transformation.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

A linear map preserves zero, negatives, and subtraction

Statement

If $T : V \to W$ is linear, then $T (0_{V}) = 0_{W}$ , $T (- v) = - T (v)$ , and $T (u - v) = T (u) - T (v)$ for all $u, v \in V$ .

Facts & Assumptions

Given: Vector spaces $V, W$ over a field $F$ and a linear map $T : V \to W$ .

[L1]

A linear map satisfies $T (a u + b v) = a T (u) + b T (v)$ (Linear map between vector spaces over the same field).

[L2]

The additive structure of a vector space is an abelian group, so it has identities, inverses, cancellation, and $u - v = u + (- v)$ ; also $0_{F} v = 0_{V}$ and $(- 1_{F}) v = - v$ (Vector space over a field, In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ ).

Proof

technique · direct

1.1

Linearity gives $T (0_{V}) = T (0_{F} 0_{V} + 0_{F} 0_{V}) = T (0_{V}) + T (0_{V})$ ; cancellation in $W$ gives $T (0_{V}) = 0_{W}$ .

L1L2given

2.1

Since $0_{W} = T (0_{V}) = T (v + (- v)) = T (v) + T (- v)$ , the inverse law gives $T (- v) = - T (v)$ .

step 1.1L1L2given

3.1

Using $u - v = u + (- v)$ and step 2.1, $T (u - v) = T (u) + T (- v) = T (u) - T (v)$ .

step 2.1L1L2given∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Kernel and image of a linear map

Definition

For a linear map $T : V \to W$ , its kernel and image are respectively

$\ker T : = {v \in V : T (v) = 0_{W}}, im T : = {T (v) : v \in V} .$

That both sets are linear subspaces, and that a trivial kernel characterises injectivity, is proved in The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial ↗.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial

Statement

For a linear map $T : V \to W$ , the sets $\ker T$ and $im T$ are linear subspaces. Moreover, $T$ is injective if and only if $\ker T = {0_{V}}$ .

Facts & Assumptions

Given: A linear map $T : V \to W$ of vector spaces over a field $F$ .

[L1]

The kernel and image have the displayed set descriptions (Kernel and image of a linear map).

[L2]

A nonempty subset is a linear subspace exactly when it is closed under $λ u + v$ (One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $λ u + v \in W$ for all $λ \in F$ and $u, v \in W$ ).

[L3]

A linear map carries $a u + v$ to $a T (u) + T (v)$ , preserves zero, and preserves subtraction (Linear map between vector spaces over the same field, A linear map preserves zero, negatives, and subtraction).

[L4]

A function is injective when equal outputs have equal inputs (Injection, surjection, bijection).

Proof

technique · direct

1.1

The kernel contains $0_{V}$ by [L3]. If $u, v \in \ker T$ and $a \in F$ , then $T (a u + v) = a T (u) + T (v) = 0_{W}$ , so $a u + v \in \ker T$ .

L1L2L3given

1.2

The image contains $0_{W} = T (0_{V})$ . If $T (u), T (v) \in im T$ and $a \in F$ , then $a T (u) + T (v) = T (a u + v)$ lies in the image.

L1L2L3given

1.3

If $T$ is injective and $v \in \ker T$ , then $T (v) = 0_{W} = T (0_{V})$ , so $v = 0_{V}$ and the kernel is trivial.

L1L3L4given

1.4

Conversely, suppose $\ker T = {0_{V}}$ and $T (u) = T (v)$ . Then $T (u - v) = 0_{W}$ by [L3], so $u - v = 0_{V}$ and $u = v$ .

L1L3L4given

2.1

The subspace criterion proves that both $\ker T$ and $im T$ are linear subspaces.

step 1.1step 1.2L2

3.1

Thus $T$ is injective exactly when its kernel is trivial.

step 1.3step 1.4∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Extending a basis of the kernel to a basis of the domain gives a basis of the image

Statement

Let $T : V \to W$ be linear, with $V$ finite-dimensional over $F$ . There are a basis $K$ of $\ker T$ and a basis $B$ of $V$ with $K \subseteq B$ . For this pair, put $C : = B ∖ K$ . Then $C$ is finite, the restriction

$T ∣_{C} : C ⟶ T [C]$

is a bijection, and $T [C]$ is a basis of $im T$ .

More explicitly, if $k : ν \to K$ and $c : r \to C$ are bijections, then the list obtained by placing $k$ before $c$ is an ordered basis of $V$ , while $T \circ c : r \to im T$ is an ordered basis of the image.

Facts & Assumptions

Given: Vector spaces $V, W$ over $F$ , a linear map $T : V \to W$ , and a finite-dimensional domain $V$ .

[L1]

The kernel and image of a linear map are linear subspaces, and $v \in \ker T$ exactly when $T (v) = 0_{W}$ (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial, Kernel and image of a linear map).

[L2]

A linear subspace of a finite-dimensional vector space is finite-dimensional; every linearly independent subset extends to a basis without a choice principle; and every linearly independent subset of a finite-dimensional space is finite (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , claims 1 and 3, If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ ).

[L3]

A basis is a linearly independent spanning subset. An ordered basis is an injective finite list whose image is a basis, equivalently a linearly independent finite list that spans (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L4]

Every vector has exactly one coordinate list with respect to an ordered basis (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[L5]

Finite sums in an abelian group have empty value zero and append one term at a successor; induction is valid on their natural-number length. Natural addition satisfies $p + 0 = p$ and $p + σ (q) = σ (p + q)$ (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, The principle of mathematical induction, Addition of natural numbers).

[L6]

A linear map satisfies $T (a u + b v) = a T (u) + b T (v)$ ; vector spaces have the usual zero, inverse, and distributive laws (Linear map between vector spaces over the same field, In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ ).

[L7]

A subset of a finite set is finite. If two finite sets are disjoint, then their union has cardinality the sum of their cardinalities; finite cardinality is transported by a bijection (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ , The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition, clause 1, The cardinality $∣ A ∣$ of a finite set).

[L8]

A function is bijective exactly when it is injective and surjective (Injection, surjection, bijection).

Proof

technique · direct

1.1

By [L1] and [L2], $\ker T$ is finite-dimensional; choose a basis $K$ of $\ker T$ . It is linearly independent in $V$ , so [L2] extends it to a basis $B$ of $V$ . Put $C : = B ∖ K$ . Since $B$ is finite, so is $C$ , and $B = K \dot{\cup} C$ .

L1L2L7given

1.2

For every finite list $v : n \to V$ and every scalar list $a : n \to F$ , linearity gives $T (\sum_{i < n} a_{i} v_{i}) = \sum_{i < n} a_{i} T (v_{i})$ . This follows by induction on $n$ : the empty case follows by applying [L6] to $0_{V} = 0_{V} + 0_{V}$ and cancelling, and the successor step uses the append recursion and $T (x + a_{n} v_{n}) = T (x) + a_{n} T (v_{n})$ . The same recursion, now by induction on the length of the second list, splits a spliced sum into its first and second blocks.

L5L6discharge-induction

2.1

Write $ν : = ∣ K ∣$ and $r : = ∣ C ∣$ , and choose bijections $k : ν \to K$ and $c : r \to C$ . Splicing the two lists gives $d : ν + r \to B$ , with $d_{i} = k_{i}$ for $i < ν$ and $d_{ν + j} = c_{j}$ for $j < r$ . The disjointness in step 1.1 makes $d$ injective and its image is $B$ , so it is an ordered basis of $V$ .

step 1.1L3L5L7L8construct

3.1

The set $T [C]$ spans $im T$ . Indeed, if $y = T (v)$ lies in the image, [L4] gives coordinates $a : ν + r \to F$ with $v = \sum_{i < ν + r} a_{i} d_{i}$ . Steps 2.1 and 1.2 split this into the $K$ -block and the $C$ -block. Applying $T$ kills the first block because every $k_{i}$ lies in $\ker T$ , leaving $y$ as a finite linear combination of $T (c_{0}), \dots, T (c_{r - 1})$ .

step 2.1step 1.2L1L3L4L6

3.2

The list $T \circ c : r \to W$ is linearly independent. Suppose $\sum_{j < r} b_{j} T (c_{j}) = 0_{W}$ . By step 1.2, $u : = \sum_{j < r} b_{j} c_{j}$ lies in $\ker T$ , so [L4] supplies $a : ν \to F$ with $u = \sum_{i < ν} a_{i} k_{i}$ . Thus the spliced ordered basis $d$ has a vanishing linear combination whose coefficients are $- a_{i}$ on the first block and $b_{j}$ on the second. Its linear independence forces every $b_{j} = 0_{F}$ .

step 2.1step 1.2L1L3L4L6

4.1

Linear independence makes $T \circ c$ injective. It is surjective onto $T [C]$ by that set's definition, so it is a bijection $r \to T [C]$ . Together with steps 3.1 and 3.2, this says that it is an ordered basis of $im T$ and that $T ∣_{C} : C \to T [C]$ is bijective.

step 3.1step 3.2L3L8

5.1

Step 1.1 supplies $K$ and $B$ , and step 4.1 proves all the asserted properties of their complement $C$ .

step 1.1step 4.1∎

Remarks

The construction is valid when $V$ is the zero space: then $K = B = \emptyset$ , hence $C = T [C] = \emptyset$ , and both empty lists are the ordered bases of the zero spaces.
No choice principle is used. Both basis selections occur inside finite-dimensional spaces and are licensed by the finite extension clause of If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Rank and nullity of a linear map with finite-dimensional domain

Definition

Let $T : V \to W$ be a linear map whose domain $V$ is finite-dimensional over $F$ . Its nullity and rank are

$nullity T : = \dim_{F} (\ker T), rank T : = \dim_{F} (im T) .$

Both dimensions are defined. The kernel is a linear subspace of the finite-dimensional space $V$ , so it is finite-dimensional by The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial and If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ . The image is finite-dimensional because Extending a basis of the kernel to a basis of the domain gives a basis of the image constructs a finite basis of it from a basis of $V$ . Thus rank and nullity are natural numbers, including when $V$ is the zero space.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Rank-nullity: $\dim_{F} V = nullity T + rank T$

Statement

Let $T : V \to W$ be a linear map of vector spaces over $F$ , with $V$ finite-dimensional. Then

$\dim_{F} V = nullity T + rank T .$

Equivalently,

$\dim_{F} V = \dim_{F} (\ker T) + \dim_{F} (im T) .$

Facts & Assumptions

Given: A linear map $T : V \to W$ with $V$ finite-dimensional over $F$ .

[L1]

Nullity and rank are the dimensions of the kernel and image (Rank and nullity of a linear map with finite-dimensional domain).

[L2]

There are finite bases $K$ of $\ker T$ and $B$ of $V$ with $K \subseteq B$ ; for $C = B ∖ K$ , the set $C$ is finite, $T [C]$ is a basis of $im T$ , and $T ∣_{C} : C \to T [C]$ is bijective (Extending a basis of the kernel to a basis of the domain gives a basis of the image).

[L3]

The dimension of a finite-dimensional vector space is the number of elements in any finite basis (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

[L4]

If two finite sets are disjoint, the cardinality of their union is the sum of their cardinalities (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition, clause 1).

[L5]

A bijection between finite sets transports their cardinality (The cardinality $∣ A ∣$ of a finite set, consequence (c)).

Proof

technique · direct

1.1

Choose $K, B, C$ as in [L2]. Since $B = K \dot{\cup} C$ , [L4] gives $∣ B ∣ = ∣ K ∣ + ∣ C ∣$ . The bijection $T ∣_{C} : C \to T [C]$ gives $∣ C ∣ = ∣ T [C] ∣$ .

L2L4L5given

2.1

Since $B$ , $K$ , and $T [C]$ are bases of $V$ , $\ker T$ , and $im T$ , respectively, [L1] and [L3] turn step 1.1 into $\dim_{F} V = nullity T + rank T$ .

step 1.1L1L2L3

3.1

The displayed equivalent form follows by unfolding the definitions of rank and nullity.

step 2.1L1∎

Remarks

If $V = {0_{V}}$ , all three dimensions are zero and the formula reads $0 = 0 + 0$ ; no positive-dimension hypothesis is hidden.
No finite-dimensionality assumption is made on $W$ . The image is finite-dimensional for the reason isolated in Extending a basis of the kernel to a basis of the domain gives a basis of the image.

5 · Examples, counterexamples and false statements

None yet.

Linear Transformations, Rank-Nullity and Quotient Spaces

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Linear map between vector spaces over the same field

Definition

A linear map preserves zero, negatives, and subtraction

Statement

Facts & Assumptions

Proof

Kernel and image of a linear map

Definition

The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial

Statement

Facts & Assumptions

Proof

Extending a basis of the kernel to a basis of the domain gives a basis of the image

Statement

Facts & Assumptions

Proof

Remarks

Rank and nullity of a linear map with finite-dimensional domain

Definition

Rank-nullity: dim⁡FV=nullity⁡T+rank⁡TdimF​V=nullityT+rankT

Statement

Facts & Assumptions

Proof

Remarks

5 · Examples, counterexamples and false statements

Sources

Rank-nullity: $\dim_{F} V = nullity T + rank T$