TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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.

A square matrix is invertible exactly when its multiplication map is a linear isomorphism; matrices preserve inverses of linear isomorphisms

Statement

For $A \in M_{n} (F)$ , let $L_{A} : M_{n \times 1} (F) \to M_{n \times 1} (F)$ be $L_{A} (x) = A x$ . Then $A$ is invertible if and only if $L_{A}$ is a linear isomorphism.

More generally, if $T : V \to W$ is a linear isomorphism between finite-dimensional spaces and $B, C$ are ordered bases, then $[T]_{B}^{C}$ is invertible and

$[T^{- 1}]_{C}^{B} = ([T]_{B}^{C})^{- 1} .$

Facts & Assumptions

Given: A field $F$ , a natural $n$ , and the matrix multiplication map $L_{A}$ ; for the general claim, an isomorphism $T$ and ordered bases $B, C$ .

[L1]

An invertible matrix has a two-sided matrix inverse, and a linear isomorphism has a two-sided linear inverse (Invertible matrices and the general linear group ${GL}_{n} (F)$ , Invertible linear maps, linear isomorphisms, and inverse linear maps).

[L2]

Matrix representation sends a composite of linear maps to the product of their matrices ( $[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$ ).

[L3]

Matrix multiplication distributes over addition and is compatible with scalar multiplication (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[L4]

Every linear map between finite-dimensional coordinate spaces has a matrix in chosen ordered bases ( $T \mapsto [T]_{B}^{C}$ is a vector-space isomorphism $L (V, W) ≅ M_{m \times n} (F)$ ).

[L5]

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

Proof

technique · direct

1.1

By [L3], $L_{A}$ and $L_{B}$ are linear. If $A B = B A = I_{n}$ , then $L_{A} L_{B} = L_{A B} = id$ and $L_{B} L_{A} = L_{B A} = id$ , so $L_{A}$ is a linear isomorphism with inverse $L_{B}$ . This includes $n = 0$ , where the unique empty matrix and the unique zero-space map are their own inverses.

givenL1L3

2.1

Conversely, if $L_{A}$ has a linear inverse $S$ , [L4] represents $S$ by a matrix $B$ in the standard coordinate basis. The two identity composites and [L2] give $A B = I_{n} = B A$ , so $A$ is invertible.

step 1.1L1L2L4

3.1

For a general linear isomorphism $T$ , the list $T (B)$ is an ordered basis of $W$ : $T^{- 1}$ transfers both linear independence and spanning back to $B$ . Hence [L5] shows that $B$ and $C$ have the same length, so the two displayed representation matrices are square. Representing $T^{- 1} \circ T$ and $T \circ T^{- 1}$ in the chosen bases, [L2] gives both inverse equations for $[T]_{B}^{C}$ and $[T^{- 1}]_{C}^{B}$ .

step 2.1L1L2L5∎

Depends on

Used by

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

S. Axler, Linear Algebra Done Right, 4th ed., Theorem 3.86 (standard reference, not scraped)
S. Schiavone, MIT 18.700 Day 9, invertibility section (standard reference, not scraped)