Linear Algebra

Section II. Linear Independence 107So, in general, a linearly independent set may have a superset that is dependent.And, in general, a linearly independent set may have a superset that isindependent. We can characterize when the superset is one and when it is theother.1.16 Lemma Where S is a linearly independent subset of a vector space V ,S ∪ {⃗v} is linearly dependent if and only if ⃗v ∈ [S]for any ⃗v ∈ V with ⃗v ∉ S.Proof. One implication is clear: if ⃗v ∈ [S] then ⃗v = c 1 ⃗s 1 + c 2 ⃗s 2 + · · · + c n ⃗s nwhere each ⃗s i ∈ S and c i ∈ R, and so ⃗0 = c 1 ⃗s 1 + c 2 ⃗s 2 + · · · + c n ⃗s n + (−1)⃗v is anontrivial linear relationship among elements of S ∪ {⃗v}.The other implication requires the assumption that S is linearly independent.With S ∪ {⃗v} linearly dependent, there is a nontrivial linear relationship c 0 ⃗v +c 1 ⃗s 1 + c 2 ⃗s 2 + · · · + c n ⃗s n = ⃗0 and independence of S then implies that c 0 ≠ 0, orelse that would be a nontrivial relationship among members of S. Now rewritingthis equation as ⃗v = −(c 1 /c 0 )⃗s 1 − · · · − (c n /c 0 )⃗s n shows that ⃗v ∈ [S]. QED(Compare this result with Lemma 1.1. Both say, roughly, that ⃗v is a “repeat”if it is in the span of S. However, note the additional hypothesis here of linearindependence.)1.17 Corollary A subset S = {⃗s 1 , . . . , ⃗s n } of a vector space is linearly dependentif and only if some ⃗s i is a linear combination of the vectors ⃗s 1 , . . . , ⃗s i−1listed before it.Proof. Consider S 0 = {}, S 1 = { ⃗s 1 }, S 2 = {⃗s 1 , ⃗s 2 }, etc. Some index i ≥ 1 isthe first one with S i−1 ∪ {⃗s i } linearly dependent, and there ⃗s i ∈ [S i−1 ]. QEDLemma 1.16 can be restated in terms of independence instead of dependence:if S is linearly independent and ⃗v ∉ S then the set S ∪ {⃗v} is also linearlyindependent if and only if ⃗v ∉ [S]. Applying Lemma 1.1, we conclude that if Sis linearly independent and ⃗v ∉ S then S ∪ {⃗v} is also linearly independent ifand only if [S ∪ {⃗v}] ≠ [S]. Briefly, when passing from S to a superset S 1 , topreserve linear independence we must expand the span [S 1 ] ⊃ [S].Example 1.15 shows that some linearly independent sets are maximal — haveas many elements as possible — in that they have no supersets that are linearlyindependent. By the prior paragraph, a linearly independent sets is maximal ifand only if it spans the entire space, because then no vector exists that is notalready in the span.This table summarizes the interaction between the properties of independenceand dependence and the relations of subset and superset.S 1 ⊂ SS 1 ⊃ SS independent S 1 must be independent S 1 may be eitherS dependent S 1 may be either S 1 must be dependent