ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13

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.

One operator has matrices $diag (2, 3)$ and $(\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array})$ in two bases, both with determinant $6$

Example

Let $T (x, y) = (2 x, 3 y)$ on $R^{2}$ . In the standard ordered basis $E$ , its matrix is $diag (2, 3)$ . In the ordered basis $B = ((1, 1), (0, 1))$ , its matrix is $(\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array})$ . Both determinants are $6$ .

Facts & Assumptions

Given: $T, E, B$ as in the example.

[F1]

$R$ is a field (The reals form a field).

[L1]

If $P = P_{E \leftarrow B}$ , then $[T]_{B} = P^{- 1} [T]_{E} P$ ( $[T]_{B^{'}}^{C^{'}} = P_{C^{'} \leftarrow C} [T]_{B}^{C} P_{B \leftarrow B^{'}}$ ).

[F2]

The $2 \times 2$ determinant is the two-term Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L2]

The operator determinant is independent of the ordered basis (The determinant of a linear operator is independent of the chosen ordered basis).

Verification

technique · direct

1.1

Direct evaluation gives $[T]_{E} = A = (\begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array})$ and $P = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array})$ , whose inverse is $P^{- 1} = (\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array})$ .

F1givenalgebra

2.1

Matrix multiplication in [L1] gives $[T]_{B} = P^{- 1} A P = (\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array}) .$

step 1.1L1algebra

3.1

By the two-term determinant formula, $\det (A) = 2 \cdot 3 = 6$ and $\det ([T]_{B}) = 2 \cdot 3 - 0 \cdot 1 = 6$ .

step 2.1F2algebra

4.1

The explicit computations in steps 3.1 and 2.1 illustrate the equality asserted abstractly by [L2].

step 3.1step 2.1L2∎

Depends on

Used by

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Dependency tree · two levels

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One operator has matrices diag⁡(2,3)diag(2,3) and (2013)(21​03​) in two bases, both with determinant 66