Question 11258: A Zero Combination Must Also Be Nontrivial to Witness Dependence

Original question and givens

Use the explicitly supplied finite integer-coordinate vector family in a real coordinate space. Preserve vector order and coefficient pairing. Rank is computed by exact rational elimination; a basis requires both independence and spanning the declared space, not just one condition. vectors=[[1, 0], [0, 1], [1, -5]], coefficients=[-1, 5, 1]. Give (vectors,coefficients,weighted sum,some coefficient nonzero?,sum is zero?,both witness conditions?).

• G001 — Use the explicitly supplied finite integer-coordinate vector family in a real coordinate space. Preserve vector order and coefficient pairing. Rank is computed by exact rational elimination; a basis requires both independence and spanning the declared space, not just one condition. vectors=[[1, 0], [0, 1], [1, -5]], coefficients=[-1, 5, 1]. Give (vectors,coefficients,weighted sum,some coefficient nonzero?,sum is zero?,both witness conditions?).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

• mko-linear-independence-basis

Step-by-step derivation

dependence witness:ordered vectors=[[1, 0], [0, 1], [1, -5]];aligned weights=[-1, 5, 1]

dimension=2;preserve duplicates and order

dependence witness coordinate0 term0 := mul(-1, 1)

dependence witness coordinate0 term0 = -1

dependence witness coordinate0 term1 := mul(5, 0)

dependence witness coordinate0 term1 = 0

dependence witness coordinate0 term2 := mul(1, 1)

dependence witness coordinate0 term2 = 1

dependence witness coordinate0 sum := sum(-1, 0, 1)

dependence witness coordinate0 sum = 0

dependence witness coordinate1 term0 := mul(-1, 0)

dependence witness coordinate1 term0 = 0

dependence witness coordinate1 term1 := mul(5, 1)

dependence witness coordinate1 term1 = 5

dependence witness coordinate1 term2 := mul(1, -5)

dependence witness coordinate1 term2 = -5

dependence witness coordinate1 sum := sum(0, 5, -5)

dependence witness coordinate1 sum = 0

given coefficients=[-1, 5, 1];weighted sum=[0, 0]

nontrivial=Yes;zero_sum=Yes;dependence_witness=Yes

Reverse checks

• S010 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S009 · premise_and_result_audit — Recompute dependence witness coordinate1 sum with exact rational arithmetic, not rounded decimals.
• S008 · premise_and_result_audit — Recompute dependence witness coordinate1 term2 with exact rational arithmetic, not rounded decimals.
• S007 · premise_and_result_audit — Recompute dependence witness coordinate1 term1 with exact rational arithmetic, not rounded decimals.
• S006 · premise_and_result_audit — Recompute dependence witness coordinate1 term0 with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Recompute dependence witness coordinate0 sum with exact rational arithmetic, not rounded decimals.
• S004 · premise_and_result_audit — Recompute dependence witness coordinate0 term2 with exact rational arithmetic, not rounded decimals.
• S003 · premise_and_result_audit — Recompute dependence witness coordinate0 term1 with exact rational arithmetic, not rounded decimals.
• S002 · premise_and_result_audit — Recompute dependence witness coordinate0 term0 with exact rational arithmetic, not rounded decimals.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

([[1, 0], [0, 1], [1, -5]], [-1, 5, 1], [0, 0], Yes, Yes, Yes)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.