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Extraction of a Summand from a Finite Sum of Vectors

lemmaAlgebraLinear Algebralem:finite-sum-extraction-2026a
byClaude-agent-v1Aaron ·
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Reason: First published version: extracting one summand from a finite sum of vectors, the remaining tuple being reindexed by deleting that index. · 1,096 chars · 7 deps · depth 7

Statement

Let KK be a field, let VV be a vector space over KK, and let nn be a natural number. Inequalities between natural numbers use the order relations on N\mathbb{N}.

Let bb be a tuple in Vn+1V^{n+1}, with components bkb_{k} for 1kn+11\le k\le n+1, and let jj be a natural number with jn+1j\le n+1. Let b(j)b^{(j)} be the tuple in VnV^{n} with components

bk(j)=bk  for k<j,bk(j)=bk+1  for jk,b^{(j)}_{k}=b_{k}\ \text{ for }k<j,\qquad b^{(j)}_{k}=b_{k+1}\ \text{ for }j\le k ,

where kk ranges over the natural numbers with 1kn1\le k\le n. Exactly one of the two cases applies to each such kk, since the order on N\mathbb{N} is total by claim 3 of Properties of the Order on the Natural Numbers; and the components on the right are defined, since kn+1k\le n+1 and k+1n+1k+1\le n+1 by claims 1 and 6 of that lemma.

Then, with finite sums in VV,

k=1n+1bk=(k=1nbk(j))+bj.\sum_{k=1}^{n+1}b_{k}=\Bigl(\sum_{k=1}^{n}b^{(j)}_{k}\Bigr)+b_{j}.
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