Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-30
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Forward and backward substitution are correct, unique, and quadratic in scalar operations

Statement

Let F be a field, let n≥1, and let b∈Fn.

  1. If L∈Mn(F) is lower triangular with nonzero diagonal, the forward substitution recursion of Forward and backward substitution for triangular linear systems produces the unique solution of Lx=b.
  2. If U∈Mn(F) is upper triangular with nonzero diagonal, the backward substitution recursion produces the unique solution of Ux=b.

Each solve uses exactly n divisions, n(n−1)2 multiplications, and n(n−1)2 additions or subtractions, hence O(n2) scalar operations.

Facts & Assumptions

Given: A field F, a natural number n≥1, a vector b∈Fn, and a triangular matrix T∈Mn(F) with nonzero diagonal; when T=L it is lower triangular, and when T=U it is upper triangular.

[L1]

Forward substitution defines xi from the earlier coordinates by xi=ℓii−1(bi−∑j<iℓijxj),and backward substitution defines xi from the later coordinates byxi=uii−1(bi−∑j>iuijxj) (Forward and backward substitution for triangular linear systems).

Proof

Proof technique: induction on the row index, then reverse induction for the upper-triangular case.

1.1

Forward substitution. For i=0, [L1] gives x0=b0/ℓ00, so the first row equation is satisfied. Assume the first i rows are satisfied. Then the (i+1)-st row of a lower-triangular system is ∑j≤i+1ℓi+1,jxj=bi+1, and solving for xi+1 gives exactly the recursion in [L1]. Because ℓi+1,i+1≠0, this value is forced and is unique. Induction proves that the forward recursion solves Lx=b and that no other vector can differ in any coordinate.

L1induction
1.2

Backward substitution is the same argument from the last row upward: the last row determines xn−1=bn−1/un−1,n−1, and once xi+1,…,xn−1 are fixed, the i-th row forces xi=uii−1(bi−∑j>iuijxj). Hence the backward recursion solves Ux=b and the solution is unique.

L1induction
2.1

In either triangular case, the i-th step forms one sum with i products and i additions in the lower-triangular solve, or with n−1−i products and n−1−i additions in the upper-triangular solve, followed by one division. Summing over all rows gives n divisions and ∑i=0n−1i=∑i=0n−1(n−1−i)=n(n−1)/2 multiplications and the same number of additions.

step 1.1step 1.2algebra
3.1

Step 1.1 proves claim 1, step 1.2 proves claim 2, and step 2.1 gives the operation counts.

step 1.1step 1.2step 2.1discharge-induction∎

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Sources