第 18726 題：對偶向量空間

原題與已知條件

有限維R^n中的線性泛函與明示基底座標。操作=dual-dual-basis-certificate；資料={"basis":[[1,0],[1,1]],"candidate_dual":[[1,-1],[0,1]]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 有限維R^n中的線性泛函與明示基底座標。操作=dual-dual-basis-certificate；資料={"basis":[[1,0],[1,1]],"candidate_dual":[[1,-1],[0,1]]}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-dual-vector-space

逐步推導

source operation=dual-dual-basis-certificate;original snapshot and mathematics retained

explicit ordered source output fields=[pairing_matrix, is_dual_basis, certificate]

dual original ordered basis:complete ordered matrix=[[1, 1], [0, 1]];rows=2;columns=2

use exact invertible ROW operations only;original COLUMN roles/order stay fixed

dual original ordered basis pivot0 column0 normalized entry0 := div(1, 1)

dual original ordered basis pivot0 column0 normalized entry0 = 1

dual original ordered basis pivot0 column0 normalized entry1 := div(1, 1)

dual original ordered basis pivot0 column0 normalized entry1 = 1

dual original ordered basis:pivot row=0;column=0;nonzero pivot=1;old row=[1, 1]

normalized full row=[1, 1]

dual original ordered basis pivot1 column1 normalized entry0 := div(0, 1)

dual original ordered basis pivot1 column1 normalized entry0 = 0

dual original ordered basis pivot1 column1 normalized entry1 := div(1, 1)

dual original ordered basis pivot1 column1 normalized entry1 = 1

dual original ordered basis:pivot row=1;column=1;nonzero pivot=1;old row=[0, 1]

normalized full row=[0, 1]

dual original ordered basis eliminate row0 pivot1 entry0 product := mul(1, 0)

dual original ordered basis eliminate row0 pivot1 entry0 product = 0

dual original ordered basis eliminate row0 pivot1 entry0 difference := sub(1, 0)

dual original ordered basis eliminate row0 pivot1 entry0 difference = 1

dual original ordered basis eliminate row0 pivot1 entry1 product := mul(1, 1)

dual original ordered basis eliminate row0 pivot1 entry1 product = 1

dual original ordered basis eliminate row0 pivot1 entry1 difference := sub(1, 1)

dual original ordered basis eliminate row0 pivot1 entry1 difference = 0

dual original ordered basis:target row=0;old=[1, 1];factor=1;pivot row=[0, 1]

new full row=[1, 0]

dual original ordered basis:complete exact reduced matrix=[[1, 0], [0, 1]];pivot columns=[0, 1]

rank=2;pivot columns independent and all original column relations preserved;zero rows retained

dual original ordered basis:complete ORIGINAL independent columns=[[1, 0], [1, 1]]

declared subspace dimension=2;basis of its full real span;empty basis means zero subspace

dual candidate row0 basis column0 coordinate0 product := mul(1, 1)

dual candidate row0 basis column0 coordinate0 product = 1

dual candidate row0 basis column0 coordinate1 product := mul(-1, 0)

dual candidate row0 basis column0 coordinate1 product = 0

dual candidate row0 basis column0 complete sum := sum(1, 0)

dual candidate row0 basis column0 complete sum = 1

dual candidate row0 basis column1 coordinate0 product := mul(1, 1)

dual candidate row0 basis column1 coordinate0 product = 1

dual candidate row0 basis column1 coordinate1 product := mul(-1, 1)

dual candidate row0 basis column1 coordinate1 product = -1

dual candidate row0 basis column1 complete sum := sum(1, -1)

dual candidate row0 basis column1 complete sum = 0

dual candidate row1 basis column0 coordinate0 product := mul(0, 1)

dual candidate row1 basis column0 coordinate0 product = 0

dual candidate row1 basis column0 coordinate1 product := mul(1, 0)

dual candidate row1 basis column0 coordinate1 product = 0

dual candidate row1 basis column0 complete sum := sum(0, 0)

dual candidate row1 basis column0 complete sum = 0

dual candidate row1 basis column1 coordinate0 product := mul(0, 1)

dual candidate row1 basis column1 coordinate0 product = 0

dual candidate row1 basis column1 coordinate1 product := mul(1, 1)

dual candidate row1 basis column1 coordinate1 product = 1

dual candidate row1 basis column1 complete sum := sum(0, 1)

dual candidate row1 basis column1 complete sum = 1

ORIGINAL basis columns=[[1, 0], [1, 1]];candidate functional ROWS=[[1, -1], [0, 1]];ALL phi_i(v_j)=[[1, 0], [0, 1]]

complete Kronecker delta identity=[[1, 0], [0, 1]];candidate is dual basis=是;not just sampled pairings

逐步反向核對

• S028 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S027 · premise_and_result_audit — 以精確有理數重算 dual candidate row1 basis column1 complete sum，不以小數近似替代。
• S026 · premise_and_result_audit — 以精確有理數重算 dual candidate row1 basis column1 coordinate1 product，不以小數近似替代。
• S025 · premise_and_result_audit — 以精確有理數重算 dual candidate row1 basis column1 coordinate0 product，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 dual candidate row1 basis column0 complete sum，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 dual candidate row1 basis column0 coordinate1 product，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 dual candidate row1 basis column0 coordinate0 product，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 dual candidate row0 basis column1 complete sum，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 dual candidate row0 basis column1 coordinate1 product，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 dual candidate row0 basis column1 coordinate0 product，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 dual candidate row0 basis column0 complete sum，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 dual candidate row0 basis column0 coordinate1 product，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 dual candidate row0 basis column0 coordinate0 product，不以小數近似替代。
• S015 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S014 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S013 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S012 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis eliminate row0 pivot1 entry1 difference，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis eliminate row0 pivot1 entry1 product，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis eliminate row0 pivot1 entry0 difference，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis eliminate row0 pivot1 entry0 product，不以小數近似替代。
• S008 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S007 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis pivot1 column1 normalized entry1，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis pivot1 column1 normalized entry0，不以小數近似替代。
• S005 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S004 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis pivot0 column0 normalized entry1，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 dual original ordered basis pivot0 column0 normalized entry0，不以小數近似替代。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

([[1, 0], [0, 1]], 是, phi_i_v_j_equals_Kronecker_delta)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。