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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

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 $\ker T = {0_{V}}$ .

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]

[L3]

[L4]

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

technique · direct

1.1L1L2L3given

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 $a u + v \in \ker T$ .

1.2L1L2L3given

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.

1.3L1L3L4given

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.

1.4L1L3L4given

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 $u = v$ .

2.1step 1.1step 1.2L2

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

3.1step 1.3step 1.4∎

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

Depends on