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Span and Independence Math 130 Linear Algebra Theorem 3 Theorem 2. The span of a set S is a subspace of V . You can also describe span(S) as the smallest subspace of V that contains all of S. Span and independence Math 130 Linear Algebra Theorem 3. The span of a set S is the intersection D Joyce, Fall 2013 of all subspaces of V that contain S. \ We're looking at bases of vector spaces. Recall span(S) = fW; a subspace of V j S ⊆ W g that a basis β of a vector space V is a set of vectors Proof. First note that span(S) is a vector space β = (b ; b ;:::; b ) such that each vector v in V 1 2 n that contains all of S, so it's one of spaces W in can be uniquely represented as a linear combination the intersection. Second, span(S) only has linear of vectors from β combinations of vectors in S, so every vector in span(S) has to be in every vector space W that v = v1b1 + v2b2 + ··· + vnbn: contains all of S. Therefore span(S) is a subset How can a set S of vectors fail to be a basis for of all the spaces W in the intersection, so it's the a vector space V ? There are two ways. It might be smallest one, and, therefore, equals the intersection that some vectors aren't linear combinations of S, of all of them. q.e.d. that is, there aren't enough vectors to span all of Some examples. A single nontrivial vector in Rn V . It might be that some vectors can be expressed spans the line through the origin that contains it. as linear combinations of S but in more than one Two vectors in R3 that don't both lie in the same way, that is, there are too many vectors in S. We'll line span a plane. The functions sin t and cos t span study these two phenomena next. the solution space of the differential equation y00 = −y. The span of a set of vectors. Since vector spaces are closed under linear combinations, we Definition 4. We say a set S of vectors in a vector should have a name for the set of all linear com- space V spans V if V = span(S). Equivalently, binations of a given set of vectors, and that will be every vector in V is a linear combination of vectors their span. in S. Note that this definition does not require that a Definition 1. Let S be a set of vectors in a vector vector can be a linear combination in only one way, space V . The span of S, written span(S), is the set just that there is at least one way. That's how this of all linear combinations of vectors in S. That is, definition differs from the definition for basis of a span(S) consists of all vectors of the form vector space. An example. The set S = f(1; 3); (2; 2); (3; 1)g c v + c v + ··· + c v 1 1 2 2 n k spans the vector space R2, but it's not a basis of it. (Why not?) where each ci is a scalar and each vi is a vector in S. The linear combination problem in Mat- The proof of the following theorem is left for you lab. Consider the question whether a particular to prove. It depends on showing that a linear com- vector v is a linear combination of given vectors bination of linear combinations is a linear combina- v1; v2;:::; vn. This is the same question as: Is v in tion. the span of the given vectors? 1 This question can be solved in Matlab. After of vectors in a vector space V is said to be linearly all, you're just looking to solve the vector equation dependent if there are scalars c1; c2; : : : ; ck not all 0, such that v = c1v1 + c2v2 + ··· + cnvk c1v1 + c2v2 + ··· + ckvk = 0: for the unknowns c1; c2; : : : ; cn, and that's just a system of linear equations. You can read this as saying that at least one of Here we'll determine if the vector v = (1; 4; 2; 6) the vectors is a linear combination of the rest, for if is a linear combination of the vectors x = c 6= 0, then v is a linear combination of the rest. (3; −1; 1; 0), y = (1; 1; 1; 1), and z = (1; 2; −1; 3). i i If the vectors aren't linearly dependent, then we Note that the system will have 4 equations (one for say they're linearly independent. In other words, no each coordinate) in three unknowns (being c , c , 1 2 vector in S is a linear combination of the others. and c ), so we don't expect it to have a solution. 3 A logically equivalent statement is that S is lin- Treat all the vectors as column vectors, place the early independent if the only way a linear combi- vectors as columns in an augmented matrix, and nation of vectors in S can equal 0, row reduce it using the function rref. (In fact, I'll enter them as rows, then transpose.) c1v1 + c2v2 + ··· + ckvk = 0; >> aug=[3 -1 1 0;1 1 1 1;1 2 -1 3;1 4 2 6]' is when each of the scalars c ; c ; : : : ; c are all 0. In aug = 1 2 k other words, 0 is not a nontrivial linear combination 3 1 1 1 of the vectors in S. -1 1 2 4 How do you know whether the vectors in S are 1 1 -1 2 linearly dependent or independent? Just solve the 0 1 3 6 vector equation c1v1 + c2v2 + ··· + ckvk = 0 for c ; c ; : : : ; c . This single vector equation is a sys- >> rref(aug) 1 2 k tem of homogeneous linear equations. If you only ans = get the trivial solution, then the vectors in S are 1 0 0 0 linearly independent. If you get any other solution, 0 1 0 0 then they're dependent. 0 0 1 0 For Rn, the standard unit vectors e ; e ;:::; e 0 0 0 1 1 2 n are linearly independent. You can see that each of th Thus the four equations are inconsistent since the them, ei, is the only one of them with a nonzero i last equation says 0 = 1. Thus, v is not a linear coordinate, therefore it is not a linear combination combination of the others. of the rest. So they're all independent. In general, two vectors v and w are linearly in- Linear independence. The question of span- dependent if and only if each is not a multiple of ning a vector space asks if you have enough vectors the other. Geometrically that means they do not in a set S to get all other vectors in a space as a lin- lie on the same line through the origin 0. ear combination of the vectors in S. The question of independence asks if you have too many, that is, Testing for linear independence using Mat- can you do without some of them because they're lab. In order to tell if vectors v1; v2;:::; vk are redundant. independent, check to see if the homogeneous sys- tem c v + c v + ··· + c v = 0 has any nontrivial Definition 5. A set 1 1 2 2 k k solutions. We can do that in Matlab with the S = fv1; v2;:::; vkg rref function. 2 For example, let's see see if the vectors (1; 3; 5; 7), of S. We have yet to show there's only one such (2; 0; 1; 3), (−1; 2; −1; 0), and (4; 3; 1; −5) are inde- linear combination. Suppose that a vector v can pendent. Place them in four columns of a coefficient be represented in two ways: matrix, and row reduce the matrix. v = c1v1 + c2v2 + ··· + ckvk >>A=[1 3 5 7;2 0 1 3; -1 2 -1 0;-5 17 9 15]' v = d1v1 + d2v2 + ··· + dkvk: A = Subtracting the second equation from the first 1 2 -1 -5 yields the equation 3 0 2 17 5 1 -1 9 0 = (c1 − d1)v1 + (c2 − d2)v2 + ··· + (ck − dk)vk: 7 3 0 15 But S is linearly independent, so 0 is only a triv- >> rref(A) ial linear combination of the basis vectors, that ans = is, ci − di = 0 for each index i. Therefore, each 1 0 0 3 ci = di. Hence the two representations were the 0 1 0 -2 same. q.e.d. 0 0 1 4 Theorem 7. Given a finite set of vectors spans a 0 0 0 0 vector space, then it has a subset which is a basis for that vector space. There are nontrivial solutions. The unknown c4 can be chosen freely, and the general solution is Proof. Let the finite set S span the vector space V . (c1; c2; c3; c4) = (−3c4; 2c4; −4c4; c4). Thus they are There are a couple of ways that you can find an not independent. Taking c4 = −1, we can write 0 independent subset of S that spans V . as a nontrivial combination of the four vectors by One way is to throw out redundant vectors in S. If S is already independent, you're done. If not, one 3v1 − 2v2 + 4v3 − v4 = 0: of the vectors v depends on the rest. Then S0 = S− fvg also spans V since, as v is a linear combination A basis is a linearly independent spanning of S0, and every vector is a linear combination of set.
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