Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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The determinant of a triangular matrix is the product of its diagonal entries

Statement

If A=(aij)Mn(R)A=(a_{ij})\in M_n(R) is upper or lower triangular over a commutative ring, with n1n\ge1, then det(A)=i<naii.\det(A)=\prod_{i<n}a_{ii}.

Facts & Assumptions

Given: A triangular n×nn\times n matrix AA over a commutative ring.

[L1]

Upper triangular means aij=0a_{ij}=0 for i>ji>j, while lower triangular means aij=0a_{ij}=0 for i<ji<j (Upper triangular, lower triangular and diagonal square matrices over a commutative ring).

[L2]

Determinant is the sum of the signed products iaσ(i),i\prod_i a_{\sigma(i),i} (For n1n\ge1, the determinant over a commutative ring by the Leibniz formula, and detA|\det A| for a real matrix).

Proof

technique · direct
1.1

For an upper triangular matrix, a nonzero Leibniz term requires σ(i)i\sigma(i)\le i for every ii. A permutation with these inequalities is the identity: starting at i=0i=0 and proceeding upward, bijectivity forces σ(i)=i\sigma(i)=i. Thus every nonidentity term contains a zero entry.

L1L2L3
2.1

The identity term has sign 11 and equals i<naii\prod_{i<n}a_{ii}, so it is the whole determinant.

step 1.1L2L3
3.1

For a lower triangular matrix the same argument uses σ(i)i\sigma(i)\ge i and proceeds downward, again forcing the identity. The argument includes n=1n=1 and makes no cancellation assumption, so zero divisors and the zero ring cause no exception.

step 2.1L1L2

Depends on

Used by

Dependency tree · next 3 levels

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