Abstract Algebra Theory and Applications - Computer Science ...

316 CHAPTER 18 VECTOR SPACESProposition 18.3 Let {v 1 , v 2 , . . . , v n } be a set of linearly independent vectorsin a vector space. Suppose thatv = α 1 v 1 + α 2 v 2 + · · · + α n v n = β 1 v 1 + β 2 v 2 + · · · + β n v n .Then α 1 = β 1 , α 2 = β 2 , . . . , α n = β n .Proof. Ifv = α 1 v 1 + α 2 v 2 + · · · + α n v n = β 1 v 1 + β 2 v 2 + · · · + β n v n ,then(α 1 − β 1 )v 1 + (α 2 − β 2 )v 2 + · · · + (α n − β n )v n = 0.Since v 1 , . . . , v n are linearly independent, α i − β i = 0 for i = 1, . . . , n.□The definition of linear dependence makes more sense if we consider thefollowing proposition.Proposition 18.4 A set {v 1 , v 2 , . . . , v n } of vectors in a vector space V islinearly dependent if and only if one of the v i ’s is a linear combination ofthe rest.Proof. Suppose that {v 1 , v 2 , . . . , v n } is a set of linearly dependent vectors.Then there exist scalars α 1 , . . . , α n such thatα 1 v 1 + α 2 v 2 + · · · + α n v n = 0,with at least one of the α i ’s not equal to zero. Suppose that α k ≠ 0. ThenThenv k = − α 1α kv 1 − · · · − α k−1Conversely, suppose thatα kv k−1 − α k+1α kv k+1 − · · · − α nα kv n .v k = β 1 v 1 + · · · + β k−1 v k−1 + β k+1 v k+1 + · · · + β n v n .β 1 v 1 + · · · + β k−1 v k−1 − v k + β k+1 v k+1 + · · · + β n v n = 0.The following proposition is a consequence of the fact that any system ofhomogeneous linear equations with more unknowns than equations will havea nontrivial solution. We leave the details of the proof for the end-of-chapterexercises.□