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# d_concept2 - Lecture 6 Linear Independence Spanning Basis...

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Lecture 6: Linear Independence, Spanning, Basis and Dimension. The homogeneous system A x = 0 can be studied from a different perspective by writing them as vector equations; (1) 1 2 - 3 3 5 9 5 9 3 x 1 x 2 x 3 = 0 0 0 x 1 1 3 5 + x 2 2 5 9 + x 3 - 3 9 3 = 0 0 0 A set of vectors { v 1 , . . . , v p } in R n is said to be linearly independent if x 1 v 1 + · · · + x p v p = 0 has only the trivial solution x 1 = · · · = x p = 0. The set { v 1 , . . . , v p } is said to be linearly dependent if there are weights λ 1 , . . . , λ p not all 0, such that λ 1 v 1 + · · · + λ p v p = 0 . Ex 1 Determine if v 1 = 1 3 5 , v 2 = 2 5 9 , v 3 = - 3 9 3 are linearly independent? Sol They are linearly dependent if (1) has a nontrivial solution. Augmented: 1 2 - 3 0 3 5 9 0 5 9 3 0 1 2 - 3 0 0 - 1 18 0 0 - 1 18 0 1 2 - 3 0 0 - 1 18 0 0 0 0 0 1 0 33 0 0 - 1 18 0 0 0 0 0 Since x 3 is free there are nontrivial solutions x 1 = - 33 x 3 , x 2 =18 x 3 , x 3 is free. If we e.g. let x 3 =1 then x 1 = - 33 and x 2 =18 so we have the linear dependence relation - 33 1 3 5 + 18 2 5 9 + 1 - 3 9 3 = 0 0 0 1 2 - 3 3 5 9 5 9 3 - 33 18 1 = 0 0 0 The columns of A are linearly independent A x = 0 has only the trivial solution. Th n vectors { v 1 , · · · , v n } in R m are linearly dependent if n > m . Pf A =[ v 1 · · · v n ] is a m × n matrix. If n>m then the homogeneous system A x = 0 has more unkonowns than equations so there must be free variables. Hence A x = 0 has a nontrivial solution x 6 = 0 so the columns of A are linearly dependent.

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