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

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The determinant of a triangular matrix is the product of its diagonal entries

Statement

If $A = (a_{i j}) \in M_{n} (R)$ is upper or lower triangular over a commutative ring, with $n \geq 1$ , then $\det (A) = \prod_{i < n} a_{i i} .$

Facts & Assumptions

Given: A triangular $n \times n$ matrix $A$ over a commutative ring.

[L1]

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

[L2]

Determinant is the sum of the signed products $\prod_{i} a_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L3]

The identity permutation has sign $1$ (Inversions, inversion number, the sign $sgn (σ) = (- 1)^{inv (σ)}$ , and even and odd permutations).

Proof

technique · direct

1.1L1L2L3

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

2.1step 1.1L2L3

The identity term has sign $1$ and equals $\prod_{i < n} a_{i i}$ , so it is the whole determinant.

3.1step 2.1L1L2∎

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

The determinant of a triangular matrix is the product of its diagonal entries

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources