Lecture 4: Linear independence, span, and bases (1)


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1 Lecture 4: Linear independence, span, and bases (1) Travis Schedler Tue, Sep 20, 2011 (version: Wed, Sep 21, 6:30 PM) Goals (2) Understand linear independence and examples Understand span and examples Understand basis and examples Preview dimension Prove the main theorem on linear independence and span, for the finitedimensional situation Warmup exercise 1 (3) What are all of the subspaces of R 2? Try to define them also using set notation; e.g., {(a, b) R 2 ab = 1} or {(a, 1 a ) a R, a 0} would describe the hyperbola xy = 1. Answer: The subspaces are {0} (otherwise denoted by 0), R 2 itself, and all of the lines through the origin, L λ,µ = {(a, b) R 2 µa + λb = 0} = {(λt, µt) t R}, where λ and µ are not both zero. (Note that L λ,µ = L cλ,cµ for nonzero c R). Warmup exercise 2 (4) True or false: If true, explain why. If false, give a counterexample: Suppose V = V 1 + V 2 + V 3. Then, V = V 1 V 2 V 3 if and only if 0 = V 1 V 2 = V 1 V 3 = V 2 V 3. Answer: False: one counterexample is V = R 2, with V 1 = the xaxis, V 2 = the yaxis, and V 3 = the line x = y. Then, V = R 2 = V 1 + V 2 + V 3, but V i V j = 0 for all i j. 1
2 Linear (in)dependence (5) Recall from the book: Definition 1. A list (v 1,..., v m ) is linearly independent if the only choice of a 1,..., a m F such that a 1 v a m v m = 0 (0.1) is a 1 = = a m = 0. The list is linearly dependent if there exists a 1,..., a m F, not all zero, such that (0.1) holds. Examples (6) Any sublist of a linearly independent list is still linearly independent. For any nonzero vector v V, the list (v), just containing the vector v only, is linearly independent. Any list containing zero is linearly dependent. A list (v, w) of two elements is linearly independent if and only if v 0 and w is not a scalar multiple of v. The case of R 3 : The list ((1, 0, 0), (0, 1, 0), (0, 0, 1)) is linearly independent. The list ((1, 1, 2), (2, 2, 4)) is linearly dependent. The list ((1, 0, 0), (0, 1, 1), (1, 2, 2)) is linearly dependent. In general, (u, v, w) is linearly independent if and only if they do not all lie in the same plane through the origin. Any list of length four or more turns out to be linearly dependent (why?) Span (7) Definition 2. The span of a list (v 1,..., v m ) of vectors in V is the vector subspace Span(v 1,..., v m ) := {a 1 v a m v m a 1,..., a m F}. (0.2) Given a vector space V, a list (v 1,..., v m ) of elements of V is a spanning list for V if V = Span(v 1,..., v m ). Note that, in a definition, if really means if and only if, since we are making a definition. Exercises: 2
3 Show that the set Span(v 1,..., v m ) is always a vector subspace. Show that Span(v 1,..., v m ) = Span(v 1 ) + + Span(v m ). This relates span to sum! Show that (v 1,..., v m ) is linearly independent if and only if Span(v 1,..., v m ) = Span(v 1 ) Span(v m ) and all the v i are nonzero. Examples (8) If we append any vectors to a spanning list, the result is still spanning. Span(0) = {0}. In general, Span(v 1,..., v m ) = {0} if and only if all of v 1,..., v m are zero. For nonzero v, Span(v) = {λv λ F} is the line through v and the origin. For nonzero v and for w not a multiple of v, Span(v, w) = {av +bw a, b F} is the plane through the origin containing v and w. Span(x, x 2,..., x n ) = the space of degree n polynomials which are multiples of x (i.e., which vanish at 0). The linear dependence lemma (9) We strengthen (a) from Lemma 2.4 and state it differently: Lemma 3. (a) A list (v 1,..., v m ) is linearly independent if and only if v 1 0 and, for all 2 j m, v j / Span(v 1,..., v j 1 ). (b) If the list is linearly dependent and v j Span(v 1,..., v j 1 ), then the span of the list (v 1,..., v j 1, v j+1,..., v m ) obtained by removing v j is the same as the span of (v 1,..., v m ). Intuitive idea: As we saw, a list (v 1,..., v m ) in R n is linearly independent if and only if v 1 spans a line, (v 1, v 2 ) spans a plane, etc., and (v 1,..., v j ) spans a jdimensional space for all 1 j m, i.e., each v j is not in the span of the previous v 1,..., v j 1. For part (b), the idea is: if v j Span(v 1,..., v j 1 ), then we can replace v j by a linear combination of v 1,..., v j 1, so the span is unchanged if we discard v j. Proof of lemma (10) Proof. Axler proves half of (a) and all of (b): If (v 1,..., v m ) is linearly dependent, then either v 1 = 0 or, for some j, v j Span(v 1,..., v j 1 ) [half of part (a)]. In the latter case, the span of (v 1,..., v m ) is the same as the span of the list obtained by removing v j [part (b)]. 3
4 Using this, it remains to prove the other half of part (a). Suppose that v 1 = 0. Then (0, v 2,..., v m ) is linearly dependent: 1 v 1 +0 v v m = 0. Suppose that j 2 and v j Span(v 1,..., v j 1 ). Write v j = a 1 v a j 1 v j 1. Then a 1 v a j 1 v j 1 + ( 1) v j = 0. So (v 1,..., v m ) is linearly dependent. Bases (11) Definition 4. A basis of V is a list (v 1,..., v m ) of vectors in V that is both linearly independent and a spanning list. Main (characterizing) property: Proposition 0.3 (Proposition 2.8). A list (v 1,..., v m ) in V is a basis if and only if, for every vector v V, there exist unique a 1,..., a m F such that v = a 1 v a m v m. Proof: Read it in Axler and understand it! Idea of proof: Break into parts: (a) The list (v 1,..., v m ) is spanning if and only if every vector v can be written in at least one way as v = a 1 v a m v m. (b) The list (v 1,..., v m ) is linearly independent if and only if every vector v can be written in at most one way as v = a 1 v a m v m. Exercise: Read the proof in Axler and use it to prove (a) and (b) separately. Examples (12) A basis of R 2 : ((1, 0), (0, 1)). More generally, (u, v) where u 0 and v is not a multiple of u. The standard basis of R n : (v 1,..., v n ), where v i = (0, 0,..., 0, 1, 0, 0,..., 0), with 1 in the ith place. Functions {1, 2,..., n} F: a basis is the collection of delta functions, (δ 1,..., δ n ), where δ i (j) = 0 if i j, and δ i (i) = 1. Deep result we will prove: For F n or the space of functions {1,..., n} F, a spanning list is a basis if and only if it has length n. Similarly, again for F n or the space of functions {1,..., n} F, a linearly independent list is a basis if and only if it has length n. 4
5 Preview on Dimension (13) The above deep properties are saying that F n and {1,..., n} F have dimension n. Definition 5. A vector space V is finitedimensional if and only if there is a (finite) spanning list (v 1,..., v m ). A deep result we will prove next time: Proposition 0.4. If V is finitedimensional, then the following numbers coincide: (a) The minimum length of a spanning list; (b) the maximum length of a linearly independent list; (c) the length of every basis. This number is called the dimension of V. A vector space which is not finitedimensional is called infinitedimensional. This includes P(F) and F. Main theorem on finite spanning and linearly independent lists (14) The key ingredient in the proof of the previous proposition on dimension is our main theorem: Theorem 6. In a finitedimensional vector space, the length of every linearly independent list is less than or equal to the length of every spanning list. Proof: Let (w 1,..., w n ) be a spanning list and (u 1,..., u m ) be a linearly independent list. We have to show m n. In fact, we will show that (u 1,..., u m ) can be extended to a spanning list of length n by adding some w j s, so in particular m n. To do so, we go backwards, beginning with (w 1,..., w n ), and at each step replacing a w j with a u i. Proof of main theorem continued (15) First, let us append u 1. Form the list (u 1, w 1,..., w m ). Since u 1 Span(w 1,..., w n ), this list is linearly dependent. By the linear dependence lemma, since u 1 0, for some j, w j Span(u 1, w 1,..., w j 1 ). Then the list (u 1, w 1,..., w j 1, w j+1,..., w m ) is also spanning. Inductively, suppose that we have a spanning list of length n, beginning with u 1,..., u i, and ending with n i of the w s. Append u i+1 to the list. Since u i+1 is in the span of the other vectors, the result is linearly dependent. 5
6 Reorder the list to begin with u 1,..., u i+1. By the linear dependence lemma, either u j Span(u 1,..., u j 1 ) (impossible since (u 1,..., u i+1 ) is linearly independent), or else one of the v s, say v k, is in the span of the previous elements in the list. Now discard v k, completing the induction. We get a spanning list of length n, beginning with u 1,..., u m. So m n. For next time: Read all of Chapter 2, and understand the proofs of Theorems 2.10 and 2.12; bring questions to class. 6
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