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Linear Dependence and Linear Independence 267 i i “main” 2007/2/16 page 267 i i 4.5 Linear Dependence and Linear Independence 267 3 32. {v1, v2}, where v1, v2 are collinear vectors in R . 34. Prove that 33. Prove that if S and S are subsets of a vector space V span{v1, v2, v3}=span{v1, v2} such that S is a subset of S, then span(S) is a subset of span(S). if and only if v3 can be written as a linear combination of v1 and v2. 4.5 Linear Dependence and Linear Independence As indicated in the previous section, in analyzing a vector space we will be interested in determining a spanning set. The reader has perhaps already noticed that a vector space V can have many such spanning sets. Example 4.5.1 Observe that {(1, 0), (0, 1)}, {(1, 0), (1, 1)}, and {(1, 0), (0, 1), (1, 2)} are all spanning sets for R2. As another illustration, two different spanning sets for V = M2(R) were given in Exam- ple 4.4.5 and the remark that followed. Given the abundance of spanning sets available for a given vector space V , we are faced with a natural question: Is there a “best class of” spanning sets to use? The answer, to a large degree, is “yes”. For instance, in Exam- ple 4.5.1, the spanning set {(1, 0), (0, 1), (1, 2)} contains an “extra” vector, (1, 2), which seems to be unnecessary for spanning R2, since {(1, 0), (0, 1)} is already a spanning set. In some sense, {(1, 0), (0, 1)} is a more efficient spanning set. It is what we call a mini- mal spanning set, since it contains the minimum number of vectors needed to span the vector space.3 But how will we know if we have found a minimal spanning set (assuming one exists)? Returning to the example above, we have seen that span{(1, 0), (0, 1)}=span{(1, 0), (0, 1), (1, 2)}=R2. Observe that the vector (1, 2) is already a linear combination of (1, 0) and (0, 1), and therefore it does not add any new vectors to the linear span of {(1, 0), (0, 1)}. As a second example, consider the vectors v1 = (1, 1, 1), v2 = (3, −2, 1), and v3 = 4v1 + v2 = (7, 2, 5). It is easily verified that det([v1, v2, v3]) = 0. Consequently, the three vectors lie in a plane (see Figure 4.5.1) and therefore, since they are not collinear, the linear span of these three vectors is the whole of this plane. Furthermore, the same plane is generated if we consider the linear span of v1 and v2 alone. As in the previous example, the reason that v3 does not add any new vectors to the linear span of {v1, v2} is that it is already a linear combination of v1 and v2. It is not possible, however, to generate all vectors in the plane by taking linear combinations of just one vector, as we could generate only a line lying in the plane in that case. Consequently, {v1, v2} is a minimal spanning set for the subspace of R3 consisting of all points lying on the plane. As a final example, recall from Example 1.2.16 that the solution space to the differ- ential equation y + y = 0 3Since a single (nonzero) vector in R2 spans only the line through the origin along which it points, it cannot span all of R2; hence, the minimum number of vectors required to span R2 is 2. i i i i i i “main” 2007/2/16 page 268 i i 268 CHAPTER 4 Vector Spaces z ϭ ϩ (7, 2, 5) v3 4v1 v2 (3,Ϫ2, 1) v v1 2 (1, 1, 1) y (3,Ϫ2, 0) (1, 1, 0) x (7, 2, 0) Figure 4.5.1: v3 = 4v1 + v2 lies in the plane through the origin containing v1 and v2, and so, span{v1, v2, v3}=span{v1, v2}. can be written as span{y1,y2}, where y1(x) = cos x and y2(x) = sin x.However,ifwe let y3(x) = 3 cos x − 2sinx, for instance, then {y1,y2,y3} is also a spanning set for the solution space of the differential equation, since span{y1,y2,y3}={c1 cos x + c2 sin x + c3(3 cos x − 2sinx) : c1,c2,c3 ∈ R} ={(c1 + 3c3) cos x + (c2 − 2c3) sin x : c1,c2,c3 ∈ R} ={d1 cos x + d2 sin x : d1,d2 ∈ R} = span{y1,y2}. The reason that {y1,y2,y3} is not a minimal spanning set for the solution space is that y3 is a linear combination of y1 and y2, and therefore, as we have just shown, it does not add any new vectors to the linear span of {cos x,sin x}. More generally, it is not too difficult to extend the argument used in the preceding examples to establish the following general result. Theorem 4.5.2 Let {v1, v2,...,vk} be a set of at least two vectors in a vector space V . If one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors and the linear span of the resulting set of vectors will be the same as the linear span of {v1, v2,...,vk}. Proof The proof of this result is left for the exercises (Problem 48). For instance, if v1 is a linear combination of v2, v3,...,vk, then Theorem 4.5.2 says that span{v1, v2,...,vk}=span{v2, v3,...,vk}. In this case, the set {v1, v2,...,vk} is not a minimal spanning set. To determine a minimal spanning set, the problem we face in view of Theorem 4.5.2 is that of determining when a vector in {v1, v2,...,vk} can be expressed as a linear combination of the remaining vectors in the set. The correct formulation for solving this problem requires the concepts of linear dependence and linear independence, which we are now ready to introduce. First we consider linear dependence. i i i i i i “main” 2007/2/16 page 269 i i 4.5 Linear Dependence and Linear Independence 269 DEFINITION 4.5.3 A finite nonempty set of vectors {v1, v2,...,vk} in a vector space V issaidtobe linearly dependent if there exist scalars c1, c2, ..., ck, not all zero, such that c1v1 + c2v2 +···+ckvk = 0. Such a nontrivial linear combination of vectors is sometimes referred to as a linear dependency among the vectors v1, v2,...,vk. A set of vectors that is not linearly dependent is called linearly independent. This can be stated mathematically as follows: DEFINITION 4.5.4 A finite, nonempty set of vectors {v1, v2,...,vk} in a vector space V is said to be linearly independent if the only values of the scalars c1,c2,...,ck for which c1v1 + c2v2 +···+ckvk = 0 are c1 = c2 =···=ck = 0. Remarks 1. It follows immediately from the preceding two definitions that a nonempty set of vectors in a vector space V is linearly independent if and only if it is not linearly dependent. 2. If {v1, v2,...,vk} is a linearly independent set of vectors, we sometimes informally say that the vectors v1, v2,...,vk are themselves linearly independent. The same remark applies to the linearly dependent condition as well. Consider the simple case of a set containing a single vector v.Ifv = 0, then {v} is linearly dependent, since for any nonzero scalar c1, c10 = 0. On the other hand, if v = 0, then the only value of the scalar c1 for which c1v = 0 is c1 = 0. Consequently, {v} is linearly independent. We can therefore state the next theorem. Theorem 4.5.5 A set consisting of a single vector v in a vector space V is linearly dependent if and only if v = 0. Therefore, any set consisting of a single nonzero vector is linearly independent. We next establish that linear dependence of a set containing at least two vectors is equivalent to the property that we are interested in—namely, that at least one vector in the set can be expressed as a linear combination of the remaining vectors in the set. i i i i i i “main” 2007/2/16 page 270 i i 270 CHAPTER 4 Vector Spaces Theorem 4.5.6 Let {v1, v2,...,vk} be a set of at least two vectors in a vector space V . Then {v1, v2,...,vk} is linearly dependent if and only if at least one of the vectors in the set can be expressed as a linear combination of the others. Proof If {v1, v2,...,vk} is linearly dependent, then according to Definition 4.5.3, there exist scalars c1,c2,...,ck, not all zero, such that c1v1 + c2v2 +···+ckvk = 0. Suppose that ci = 0. Then we can express vi as a linear combination of the other vectors as follows: 1 vi =− (c1v1 + c2v2 +···+ci−1vi−1 + ci+1vi+1 +···+ckvk). ci Conversely, suppose that one of the vectors, say, vj , can be expressed as a linear combi- nation of the remaining vectors. That is, vj = c1v1 + c2v2 +···+cj−1vj−1 + cj+1vj+1 +···+ckvk. Adding (−1)vj to both sides of this equation yields c1v1 + c2v2 +···+cj−1vj−1 − vj + cj+1vj+1 +···+ckvk = 0. Since the coefficient of vj is −1 = 0, the set of vectors {v1, v2,...,vk} is linearly dependent. As far as the minimal-spanning-set idea is concerned, Theorems 4.5.6 and 4.5.2 tell y us that a linearly dependent spanning set for a (nontrivial) vector space V cannot be a minimal spanning set.
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