Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

If χT(x)=i<n(xλi) in F[x], then χp(T)(y)=i<n(yp(λi)) for every pF[x]: the eigenvalues of p(T) are p(λi), counted with algebraic multiplicity

Statement

Let T act on an n-dimensional F-vector space and suppose

χT(x)=i<n(xλi).

For every pF[x],

χp(T)(y)=i<n(yp(λi)).

Consequently the eigenvalues of p(T) are the values p(λi), counted with the combined algebraic multiplicities shown by this product.

Facts & Assumptions

Given: T, the displayed split factorization of χT, and pF[x].

[L1]

Polynomial evaluation is p(T)=kakTk (Polynomial evaluation at an endomorphism: p(T)=kakTk). Matrix representation sends sums and scalar multiples to matrix sums and scalar multiples (T[T]BC is a vector-space isomorphism L(V,W)Mm×n(F)) and composites to matrix products ([ST]BD=[S]CD[T]BC).

[L2]

The characteristic polynomial of an operator is computed in any basis and is 1 in dimension zero (The basis-independent characteristic polynomial χT of an endomorphism of a finite-dimensional space, including χT=1 in dimension zero).

[L4]

An independent subset of a finite-dimensional space extends without Choice to a basis (If dimFV=n and U is a linear subspace of V, then U is finite-dimensional, dimFUn, and dimFU=n if and only if U=V, clause 3), whose representing matrix records the coordinate columns of the images (Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

[L5]

A block-triangular characteristic polynomial is the product of the characteristic polynomials of its diagonal blocks (The characteristic polynomial of a block upper- or lower-triangular matrix is the product of the characteristic polynomials of its diagonal blocks).

[L6]

The ring F[x] is a unique factorisation domain, hence an integral domain and admits cancellation of nonzero polynomials (For every field F, F[x] is a unique factorisation domain).

[L7]

Algebraic multiplicity is the exponent of a root's linear factor in the characteristic polynomial (Algebraic multiplicity as the exponent of xλ in χT, and geometric multiplicity as dimEλ(T)).

Proof

technique · induction on $n$
1.1

If n=0, both sides are the empty product 1 by [L2].

baseL2algebra
1.2

Assume n>0 and the result in dimension n1. The factor xλ0 shows that λ0 is a root of χT, so [L3] supplies a nonzero eigenvector v. Extend (v) by [L4] to a basis. In that basis [L4] gives [T]=(λ00B).

ihL3L4givenchoose
2.1

By [L5], χT(x)=(xλ0)χB(x). Comparing with the given factorization and cancelling xλ0 in the domain [L6] gives χB(x)=0<i<n(xλi).

step 1.2L5L6given
3.1

Powers, linear combinations, and [L1] preserve this block upper-triangular shape, so [p(T)]=(p(λ0)0p(B)). The induction hypothesis applied to B and step 2.1 gives χp(B)(y)=0<i<n(yp(λi)).

step 1.2step 2.1ihL1
4.1

Applying [L5] to step 3.1 yields χp(T)(y)=(yp(λ0))χp(B)(y)=i<n(yp(λi)).

step 3.1L2L5
5.1

The base case and induction step prove the polynomial identity for every finite dimension. By [L3] and [L7], its roots are the eigenvalues of p(T) and repeated equal values acquire their combined algebraic multiplicity.

step 1.1step 4.1L3L7discharge-induction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 129 results over 25 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