Alphabeta MathHow 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.
For a positive-dimensional finite-dimensional operator,
Statement
Let be a linear operator on a finite-dimensional vector space over a field, with . Then
Facts & Assumptions
Given: as in the statement and an ordered basis .
For a positive-sized square matrix , (For every positive-sized square matrix over a commutative ring, ).
The matrix-representation map is injective ( is a vector-space isomorphism ).
Proof
Put . By [F1] and [L2], the matrices of and are respectively and .
By [L1], both matrices in step 1.1 equal , which equals by [L3].
Equality of representing matrices gives both asserted operator identities.
Depends on
- The adjugate of an operator on a positive-dimensional finite-dimensional vector space, defined by the adjugate matrix in any basis
- For every positive-sized square matrix over a commutative ring, $A\operatorname{adj}(A)=\operatorname{adj}(A)A=\det(A)I$
- The determinant of a linear operator is independent of the chosen ordered basis
- $[S\circ T]_{\mathcal B}^{\mathcal D}=[S]_{\mathcal C}^{\mathcal D}[T]_{\mathcal B}^{\mathcal C}$
- $T\mapsto[T]_{\mathcal B}^{\mathcal C}$ is a vector-space isomorphism $\mathcal L(V,W)\cong M_{m\times n}(F)$
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
23 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
- Sheldon Axler, Linear Algebra Done Right, 4th ed. (standard reference, not scraped)