LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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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

1.1L1L2L7given

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 $B = K \dot{\cup} C$ .

1.2L5L6discharge-induction

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.

2.1step 1.1L3L5L7L8construct

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 $V$ .

3.1step 2.1step 1.2L1L3L4L6

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 $T (c_{0}), \dots, T (c_{r - 1})$ .

3.2step 2.1step 1.2L1L3L4L6

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 $b_{j} = 0_{F}$ .

4.1step 3.1step 3.2L3L8

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.

5.1step 1.1step 4.1∎

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

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 $U = V$ .

Depends on

Dependency tree · 2 levels

Sources

Axler, Linear Algebra Done Right, 4th ed., Chapter 3