{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-linear-combination-span"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-linear-combination-span-01","title":{"zh-Hant":"線性組合與生成：以有限係數 witness 判定 span 成員","en":"Linear combinations and span: deciding membership using finite coefficient lists as witnesses"},"summary":{"zh-Hant":"（一）在 ℝ² 中，v₁=(1,2)、v₂=(3,−1)，判定 b=(7,0) 是否屬於 span{v₁,v₂}：先證明兩個向量的生成集恰為 {c₁v₁+c₂v₂}，再解係數、證明唯一，並由 witness 重建 b。（二）在 ℝ³ 中，u₁=(1,0,1)、u₂=(0,1,1)，證明 t=(2,3,4) 不屬於 span{u₁,u₂}，而 t'=(2,3,5) 屬於：同為三維並不決定成員資格。","en":"(1) In ℝ², with v₁=(1,2) and v₂=(3,−1), decide whether b=(7,0) lies in span{v₁,v₂}: first prove that the span of two vectors is exactly {c₁v₁+c₂v₂}, then solve for the coefficients, prove they are unique, and rebuild b from the witness. (2) In ℝ³, with u₁=(1,0,1) and u₂=(0,1,1), prove that t=(2,3,4) is not in span{u₁,u₂} while t'=(2,3,5) is: having three coordinates does not decide membership."},"mko_ids":["mko-linear-combination-span"]}],"title":"Concept derivation guides"}