Linear Algebra

100 Chapter Two. Vector Spaces1.9 Example In any vector space, any subset containing the zero vector is linearlydependent. For example, in the space P 2 of quadratic polynomials, consider thesubset {1 + x, x + x 2 , 0}.One way to see that this subset is linearly dependent is to use Lemma 1.3: wehave 0 ·⃗v 1 + 0 ·⃗v 2 + 1 ·⃗0 = ⃗0, and this is a nontrivial relationship as not all of thecoefficients are zero. Another way to see that this subset is linearly dependent isto go straight to Definition 1.2: we can express the third member of the subsetas a linear combination of the first two, namely, we can satisfy c 1 ⃗v 1 + c 2 ⃗v 2 = ⃗0by taking c 1 = 0 and c 2 = 0 (in contrast to the lemma, the definition allows allof the coefficients to be zero).There is subtler way to see that this subset is dependent. The zero vector isequal to the trivial sum, the sum of the empty set. So a set containing the zerovector has an element that is a combination of a subset of other vectors fromthe set, specifically, the zero vector is a combination of the empty subset.1.10 Remark [Velleman] Definition 1.2 says that when we decide whether some Sis linearly independent, we must consider it as a multiset. Here is an exampleshowing that we can need multiset rather than set (recall that in a set repeatedelements collapse so that the set {0, 1, 0} equals the set {0, 1}, whereas in amultiset they do not collapse so that the multiset {0, 1, 0} contains the element 0twice). In the next chapter we will look at functions. Let the function f: P 1 → Rbe f(a + bx) = a; for instance, f(1 + 2x) = 1. Consider the subset B = {1, 1 + x}of the domain. The images of the elements are f(1) = 1 and f(1 + x) = 1.Because in a set repeated elements collapse to be a single element these imagesform the one-element set {1}, which is linearly independent. But in a multisetrepeated elements do not collapse so these images form a linearly dependentmultiset {1, 1}. The second case is the correct one: B is linearly independent butits image under f is linearly dependent.Most of the time we won’t need the set-multiset distinction and we willtypically follow the standard convention of referring to a linearly independentor dependent “set.”This section began with a discussion and an example about when a setcontains “repeat” elements, ones that we can omit without shrinking the span.The next result characterizes when this happens. And, it supports the definitionof linear independence because it says that such a set is a minimal spanning setin that we cannot omit any element without changing its span.1.11 Lemma If ⃗v is a member of a vector space V and S ⊆ V then [S − {⃗v}] ⊆ [S].Also: (1) if ⃗v ∈ S then [S − {⃗v}] = [S] if and only if ⃗v ∈ [S − {⃗v}] and (2) thecondition that removal of any ⃗v ∈ S shrinks the span [S − {⃗v}] ≠ [S] holds if andonly if S is linearly independent.Proof First, [S−{⃗v}] ⊆ [S] because an element of [S−{⃗v}] is a linear combinationof elements of S − {⃗v}, and so is a linear combination of elements of S, and so isan element of [S].