TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03

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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 $\operatorname{im}T$ 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 $\lambda u+v$ (One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $\lambda u + v \in W$ for all $\lambda \in F$ and $u, v \in W$ ).

[L3]

A linear map carries $au+v$ to $aT(u)+T(v)$ , 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 $T(au+v)=aT(u)+T(v)=0_W$ , so $au+v\in\ker T$ .

L1L2L3given

1.2

The image contains $0_W=T(0_V)$ . If $T(u),T(v)\in\operatorname{im}T$ and $a\in F$ , then $aT(u)+T(v)=T(au+v)$ 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 $\operatorname{im}T$ 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∎

Depends on

Used by

Cited to discharge well-definedness by Kernel and image of a linear map.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Axler, Linear Algebra Done Right, Chapter 3 (standard reference, not scraped)