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Unformatted text preview: Notes on linear algebra (Monday 17th October, 2016, 23:10) page 215 4.6. <DRAFT> More on subspaces If U is a subspace of a vector space V, then the sum of any two elements of LI must belong to U (by the definition of a subspace). The same holds for sums of any (finite) number of elements of LI: Proposition 4.47. Let V be a vector space. Let U be a subspace of V. (a) If 111,112,... .,uk are elements of U, then H1 + Liz -I- ' ' ' + “k 6 LL (b) If 111, 1,12,. . ., ak are elements of LI, then every linear combination of H1, H2, . . . , ”k also lies in U. (c) Let a1, a2,. . ., ak be elements of U. Then, (”LHZu-uuk) Q U. The claim of Proposition 4.47 (b) is often expressed as the following short slogan: ”A subspace U of a vector space V is closed under linear combination”. Proof of Proposition 4 47 (a) Roughly speaking, this is just a matter of applying the ”closed under addition” axiom several times But there IS a subtlety involved (the sum of 0 elements of U 15 not obtained by addition but rather defined as 0 ) and I want to illustrate the principle of mathematical induction once again so I am going to present the proof in full detail. ...
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