MATA23H3 Lecture Notes - Lecture 11: Linear Independence, Solution Set, Scalar Multiplication

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11 Feb 2016
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Mata23 - lecture 11 - homogeneousness, linear dependance, and bases. Linear dependence: let s = {(cid:126)v1, (cid:126)v2, . , (cid:126)vk} be a set of vectors in rn and r1, r2, . , rk r. if r1(cid:126)v1 + r2(cid:126)v2 + + rk(cid:126)vk = (cid:126)0 (*) is true with at least one ri (cid:54)= 0, then s is a linearly dependant set of vectors. 1: example 3, (cid:126)u = [1, 0], (cid:126)v = [2, 0, 2 (cid:126)u (cid:126)v = (cid:126)0, 2(cid:126)u + ( 1)(cid:126)v = (cid:126)0 (linear relationship) Let r(cid:126)u + s(cid:126)v = (cid:126)0, r, s r. 0: let s = {(cid:126)v1, (cid:126)v2, . , (cid:126)vk} be a set of vectors in rn. R1 r2 rk: r1(cid:126)v1 + r2(cid:126)v2 + + rk(cid:126)vk = (cid:126)0 a(cid:126)x = (cid:126)0 where (cid:126)x , (cid:126)v1 + (cid:126)v2 + + (cid:126)vk is linearly independant. A(cid:126)x = (cid:126)0 has the trivial solution only. The row-echelon form of a has k pivots. (cid:126)v1, (cid:126)v2, (cid:126)v3 are linearly dependent.

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