4.3 Linear Combinations and Spanning Sets

Transcription

1 4.. LINEAR COMBINATIONS AND SPANNING SETS 5 4. Linear Combinations and Spanning Sets In the previous section, we looked at conditions under which a subset W of a vector space V was itself a vector space. In the next three section, we look at the following problem. If W is not a vector space, how can we build a vector space from it? Once we answer that, we will try to nd the most e cient way of doing it. We begin with some important de nitions. 4.. Linear Combinations De nition 7 Let V denote a vector space, and v V. We say that v is a linear combination of the vectors u ; u ; :::; u n if v can be written in the form where c ; c ; :::; c n are scalar. v = c u + c u + ::: + c n u n = c i u i Because V is a vector space, we know that given u ; u ; :::; u n in V, the linear combination c i u i is also in V. A more di cult question is: given v V and u ; u ; :::; u n in V, can we nd c ; c ; :::; c n such that v = c u +c u +:::+c n u n? To try to answer this question, we rst look at some example to familiarize ourselves with this concept. In the examples. Example 7 Let R be the underlying vector space. Is v = (; ) a linear combination of e = (; 0) and e = (0; )? It is easy to see that v = e + e. In fact, if v = (x; y), then v = xe + ye. So, any vector in R can be represented as a linear combination of e and e. Example 74 Let R be the underlying vector space. Is v = (; ) a linear combination of u = (; ) and u = (4; )? This time, it is a little bit more di cult. We must nd scalars c and c such that v = c u + c u. This amounts to solving the system c + 4c = c + c = Using any method learned in this class, we nd that the system has a unique solution 8 >< c = 7 >: c = Therefore, v = 7 u + u.

2 6 CHAPTER 4. VECTOR SPACES Example 75 Let R be the underlying vector space. Is v = (; 4; 6) a linear combination of u = (; ; ), u = (; ; ) and u = (; ; 4)? We need to solve the system c u + c u + c u = v. The augmented matrix of the system is We use Gaussian elimination to solve the system. Applying (R rr )! (R ) and (R R )! (R ), we obtain Then, applying (R R )! (R ), we obtain From the last line, we see that the system has no solution. expressed as a linear combination of u ; u and u. So, v cannot be In general, given v V and u ; u ; :::; u n in V, determining if v is a linear combination of the vectors u ; u ; :::; u n is simply a matter of solving the system of linear equations v = c u + c u + ::: + c n u n. We know that such systems can have no solution, a unique solution, or an in nite number of solutions. If the determinant of the coe cient matrix of the system is not zero, that is if the coe cient matrix is invertible, then the system has a unique solution. We will see later that it is important to know. However, if its determinant is zero, it does not mean there does not exist a linear combination. When the coe cient matrix of a system is not invertible, the system could have no solution, or an in nite number of solutions. Now, we are starting to get answers to the questions we asked in the abstract. Given u ; u ; :::; u n in V, it is possible that not every vector of V can be expressed as a linear combination of u ; u ; :::; u n. When it is the case that every vector in V can be expressed as a linear combination of u ; u ; :::; u n, then the vectors u ; u ; :::; u n are special. It is what we study next. 4.. Spanning Sets De nition 76 Let V denote a vector space and S = fu ; u ; :::; u n g a subset of V.. We say that S is a spanning set of V or that S spans V if for every vector v in V, v can be written as a linear combination of the vectors in S.

3 4.. LINEAR COMBINATIONS AND SPANNING SETS 7. The span of S, denoted span (S), is the set of all linear combinations of vectors in S. In other words, ( n ) X span (S) = c i u i jc i R and u i S Before we look at speci c examples, there are several important remarks to make and questions to ask. Remark 77. Since V is a vector space and S V, it follows that span (S) is a subset of V. Could span (S) equal V?. In the case that span (S) is a proper subset of V, what is span (S)?. Given S, how do we determine if span (S) = V? 4. Given S, how do we nd span (S)? We investigate the issues raised in the remark by rst looking at some examples. Example 78 We saw earlier that S = fe ; e g where e = (; 0) and e = (0; ) was a spanning set for R because every vector v = (x; y) in R could be written as a linear combination of e and e as follows: v = xe + ye. It is called the standard spanning set of R. There are others, as the next example suggests. Example 79 Show that S = fu ; u g where u = (; ) and u = (0; ) spans R Here, we have to show that any element of R can be expressed as a linear combination of vectors of S. A typical element of R is of the form v = (x; y). So, we have to show that the system c u + c u = v always has at least one solution. This can be done several ways. First, we can try to solve the system using Gaussian elimination. In this case, there is a faster and easier way. The 0 coe cient matrix of the system is. The determinant of this matrix is not zero, this means that the system always has a unique solution. The system c u + c u = v can always be solve, hence span (S) = R. Example 80 Similarly, S = fe ; e ; e g where e = (; 0; 0), e = (0; ; 0) and e = (0; 0; ) is a spanning set for R. It is called the standard spanning set of R. Example 8 Let P denote the set of polynomials of degree less than or equal to two. The set S = ; x; x is a spanning set for P. A typical polynomial of P is ax +bx+c, it is easy to see how this can be written as a linear combination of elements of S. It is called the standard spanning set of P.

4 8 CHAPTER 4. VECTOR SPACES Example 8 Does S = f(; ; ) ; ( ; ; ) ; (4; ; 4)g span R? A typical element of R is (x; y; z). The question becomes: can the system c (; ; )+c ( ; ; )+c (4; ; 4) = (x; y; z) always be solved? Be careful, here the unknowns are c ; c ; and c, not x; y; z. This system can also be written 8 < c c + 4c = x c c + c = y : c + c 4c = z The augmented matrix of the system is 4 y 4 x 4 z First, we perform R R! (R ) and (R + R )! (R ) to obtain x x + y x + z 7 5 ( For this system to have a solution, A vector of R must satisfy x + y = 0 x + z = 0 Obviously, not every vector of R satis es this. For example, (; ; ) does not. This means that not every vector of R can be written as a linear combination of vectors in S. Thus span (S) 6= R. As noted earlier, span (S) is always a subset of the underlying vector space V. If S is a spanning set, then span (S) = V, otherwise, span (S) is a proper subset. But, even in this case, span (S) has important properties, as the next theorem tells us. Theorem 8 Let V denote a vector space and S = fu ; u ; :::; u n g a subset of V.. span (S) is a subspace of V.. span (S) is the smallest subspace of V containing S. Proof. We prove each part separately.. We have already established that span (S) was a non-empty subset of V. We simply need to show that it is closed under both operations. Let v and v be two elements of span (S), let c be a scalar.

5 4.. LINEAR COMBINATIONS AND SPANNING SETS 9 (a) Closure under addition. We must show that v + v is also in S. Since v S, there exist scalars c ; c ; :::; c n such that v = c i u i. Similarly, v = d i u i for some scalars d ; d ; :::; d n. Therefore, v + v = = c i u i + d i u i (c i + d i ) u i And therefore, v + v is a linear combination of elements in S. (b) Closure under scalar multiplication. We must show that cv is also in S. Using the notation of the previous part of this proof, we have cv = c = c i u i cc i u i And therefore, cv is a linear combination of elements in S.. Let W be a subspace containing S. We need to show that W contains span (S). This is obvious. Since W is a subspace, it is closed under addition and scalar multiplication. Therefore, if W contains S, it must contain every linear combinations of elements of S. But the set of all possible linear combinations of elements of S is precisely span (S). There are still several unanswered questions. From the examples, we have seen that a space can have several spanning sets. Here are questions the reader should try to answer, especially as going through the next section.. If S is a spanning set of a vector space V, and one adds another vector of V to S, is the new set still a spanning set of V?. Same question in the case that we remove a vector from S. 4.. Problems Do # 7, 8, 9,, c, e on pages 9, 40.