a. Since S(a = (ca CER} and since al = (1,2,3,4), we see that 1 1: B S(al) if and only if there exists a scalar c such that

Transcription

1 3.2.1(L) a. Since S(a = (ca CER} and since al = (1,2,3,4), we see that 1 1: B S(al) if and only if there exists a scalar c such that Hence, by (11, Hence, if and only if there exist real numbers cl and c2 such that The main problem with (2) is that it may not seem apparent how, for example, the last two components depend on the first two. That is, we know that once two of the four components of (2) are given, cl and c2 are determined, whereupon the other two components are uniquely determined. One thing that we might do is replace the first two components of (2) by the single symbols, say, x1 and x2. That is, we could make the substitutions

2 3.2.1(L) continued and from which it follows that and In terms of row-reduced matrices, we obtain (4) from (3) by From (4), it is readily seen that while

3 3.2.1 (L) continued Thus, in terms of xl and x2, (2) becomes [In terms of cl and c2, (5) says that 3cl + 7c2 = (cl + 2c2) + (2cl + 5c2) and 4cl + 7c2 = 6(cl + 2c2) - (2cl + 5c2), so that the last two components of (2) are then expressed in terms of the first two components.] The advantage of (5) over (2) is that we can now tell by inspection whether (x1,x2,x3,x4) E S (alra2). Namely 3 and 1 ="+x2 c. Letting a3 = ( ). we see from (6) that a3 4 S(al,a2). Namely, in this case, xl = 3, x2 = 7, and x3 = 8; hence, X3 # X1 + X2' d. Knowing that fl = (3,7,y,z), we see from (6) that E s(a1,a2) * y = and z = 6(3) - 7. Hence, for f3 to belong to S(al,a2), it must be that We shall revisit this exercise as a note to Exercise (L) The main aim of this exercise is to illustrate some "tricks of the trade" in finding the space spanned by a set of vectors. We have chosen some special cases, each of which shall be expanded within

4 3.2.2 (L) continued the context of our solution, to show how we can replace vectors in the given set by other vectors which span the same space. This new technique will supply us with a "neater" way of obtaining the same results as those of the type obtained in the previous exercise. Thus, B E W means there exist real numbers cl, c2, c3, and c4 such that But since vector addition is associative and commutative, we have from (1) that - Note (i) The significance of this part of the exercise is to show that the space spanned by any set of vectors, (a1,..., an} does not depend on the order in which the vectors are listed. True, we have taken the special case n = 4 and the particular reordering given by (2), but in the same way we derived (21, we could have shown how other rearrangements could be made, and the case n = 4 was chosen only to get away from the usual geometric association of ideas. Any other value of n would work just as well.

5 k2a2 Solutions b. Let (L) continued (ii) In terms of linear structure, this exercise is analogous to the statement that the solution set of a system of algebraic equa- tions does not depend on the order in which the equations are written. wl = s(alla2,a3) = {clal + c2a2 + c3a3: c1,c2.c3 E RI and Then, if f3 E Wlr 1 or letting kl = 3,k2 = c2, and k3 = c3, That is, $ E Wl+ B E W Hencel 2' W1 C w2. Similarly, 7EW2+ y=3ka k3a3' so letting cl = 3kl, c2 = k21 C3 = k31 S.3.2.5

6 3.2.2(L) continued Y = C1al + c2a2 + c3a3 E S(alIa2,a3) = W1. Therefore, Hence, (6) and (7), together, imply that W1 = W2. Note - (i) Again what is important here is the fact that the space spanned by {al,..., an) is unchanged if one of the vectors ai is replaced by a non-zero multiple ciai. In our example, we took i = 1, but by part (a) this was no loss of generality. Namely, whatever ai was replaced by ciai, we could have rearranged (al,..., an) so that ai appeared first in the listing. Moreover, we picked ci = 3. All that was important was that ci # 0. For if ci = 0, we see from (5) that we would have to divide by 0, which is not permitted. (ii) The analogy here to systems of equations is that if we re- place one equation by a constant multiple of that equation, we do not change the solution set of the system. Therefore,

7 3.2.2 (L) continued Similarly, Therefore, So comparing (8) and (9), we have that W2 = W1. Note (i) What we have illustrated here is that if we are studying the space spanned by {a1,..., an), we may replace any ai by itself plus any multiple of another, say, ai by ai + k.a (where in our I j example i = 1, j = 2 and for computational simplicity, k = 1), j without changing the space spanned by the vectors. (ii) Since the three properties developed in parts (a), (b) and (c) are precisely the properties that one needs in order to use row-reduced matrix techniques, it should be clear that this matrix technique may be used to find the space spanned by a given set of vectors. To illustrate this idea, we shall revisit Exercise in the form of a note. NOTE ON MATRIX CODING SYSTEM Suppose we want to determine the space spanned by al = (1,2,3,4) and a2 = (2,5,7,7). We may use a 2 by 4 matrix in which the first row represents the components of al and the second row, the components of a2. We then have

8 3.2.2 (L) continued Now we have seen in this exercise that the space spanned by al and a is not altered if we replace al by a non-zero multiple of a 2 1' say, -2a1 [which is precisely the multiple of the first row of (1) that must be added to the second row when we use the usual rowreduction technique]. Thus, the matrix which is row-equivalent to (11, tells us that the space spanned by a1 and a2 is the same as that spanned by -2a1 and a2. In fact, proceeding quite mechanically, we have In terms of our code, (1), (21, (3), and (4) say that *We usually obtain (4) from (1) by the single step of replacing the second row of (1) by the second minus twice the first. We have included (2) and (3) to emphasize the validity our claim that (1) and (4) code the same space in terms of the basic properties described in this exercise.

9 3.2.2(L) continued In still other words, if we let B2 = a2-2al = (0,1,1,-l), then If we now complete the row-reduction of (1), we see from (4) that Using our code, (5) tells us that where and Now, B1 and B2 have a very nice form which help us express S(a1,a2) more conveniently. Namely, which checks with our result in Exercise

10 3.2.2 (L) continued From another perspective, we are saying that so that IB1,B2) seems to be a natural coordinate system (basis) for S(a1,a2). Again, with respect to Exercise 3.2.1, observe that which checks with our result in Exercise 3.2.2(d). Notice also that we may use the augmented matrix technique to con- vert from the a's to the B's; namely, we may write from which we see that

11 3.2.2 (L) continued We shall continue to reinforce this idea in the remaining exercises (L) a. Using our matrix coding system, we have From (1), we see that SIa1,a2,a3) = S(BllB2,B3) where

12 3.2.3(L) continued Therefore, B1,B2,B3 belong to S(a1,a2,a3) and (xlfx2,x3,x4) E V belongs to b. From (2), Combining the results of (3) and (4), we see that ~x1,x2,x3~x4)e S(alfa2~a3I if and only if [As a check of (5), observe that (1,2,3,4), (2,5,7,7), and (3,7,8,9) each obey (5). Namely, with xl = 1, x2 = 2, x3 = 3, we have from (5) that x4 = = 4; with xl= 2, x2 = 5, x3 = 7, we have x4 = = 7; and with x 1 = 3, x2 = 7, x 3 = 8, wehave x4 = = 9.1 At any rate, we have from (5) that

13 3.2.3 (L) continued Hence, in particular,. Here we emphasize the salient feature of the vectors obtained by our row-reduced matrix technique. Namely, the very form of B1, 62, and B3 guarantees that IB11f32,B3} is linearly independent. Namely, suppose Then Hence, Therefore, so that [and 5x1-2x2 + x3 = 01. Since xl = x2 = x3 = 0, we see from (6) that 1B11f32,B31 is linearly independent. More generally, when we employ the row-reduced matrix idea to compute S(ul,..., un), we look at our final reduced matrix and

14 3.2.3 (L) continued delete the rows which consist entirely of zeroes. The number of remaining rows will turn out to be the dimension of S(ul,..., un). In particular, if we name the non-zero rows of the reduced matrix by v1 I..., and vm (where m < n) then and the set {vl,..., v 1 is linearly independent. m (L) a. To find the space spanned by {al,a2,a3,a4) where al = (1,2,3,4), a2 = (2,3,5,5), a3 = (2,4,7,6) and a4 = (-1,2,3,4), we row-reduce the matrix i : Namely,

15 3.2.4 (L) continued From (3), we see that s(al, a2, a3, a4) = S(B1,B2,B3, B4) where But the 0-vector is redundant in any spanning set. Namely, clal + czaz + c3a = clal + c2a2. etc. Hence, where B1, B2, and B3 are as in (4). Consequently, That is, for (x1,x2,x3,x4) to belong to S(a1,a2,a3,a4) it is necessary and sufficient that x4 = 5x2-2x3. a4 all have this property. Notice that alra2,a3, - For example, with (1,2,3,4), x2 = 2, x3 = 3 whence 5x2-2x3-10-6=4=x 4 '

16 3.2.4 (L) continued b. We now use the augmented matrix technique to see why {al,a2,a3,a4) spanned less than 4-dimensions. We have From (61, we may conclude at once that - and

17 3.2.4 (L) continued That is, In other words, we see from (7) how to express B1, B2, and B3 as linear combinations of al, a2, and a3. From equation (81, we see that u4 is "redundant" in the sense that it is a linear cornbination of al, a2, and a3. Check + "2 "3 = (2,3,5,5) + (2,4,7,6) NOW, from (8), Hence, for any real number c, (10) implies that 3cal - 4ca2 + 2ca3 - ca4 = 0, Combining (9) and (ll), we have *Notice how we "pick off" this information simply by inspection.

18 3.2.4 (L) continued Thus, from (12), we see that there are as many ways of expressing (4,7,12,11) as a linear combination of alr a2, a3, and a4 as there are ways of choosing a value of c. In other words, since (alra2,a3,a41 is a linearly dependent set, every member of S(a1,a2,a3,a4) may be expressed as a linear combination of air a2, "31 and a4 in infinitely many ways. - Note This exercise is a concrete illustration of a more general result; namely, any finite set of vectors contains a linearly independent subset which generates (spans) the same subspace. This linearly independent subset can always be obtained by our row-reduced matrix technique. a. Using our matrix code for al, a2, a3, a4, and as, we have

19 3.2.5 (L) continued From (3), we see that S (al,a2,a3,a4,a5) is the same space as that spanned by B1 and B2 where B1 = (1,0,5) and B2 = (0,1,-1). [The last three rows of (3) all represent the 0-vector (0,O,0) and this adds nothing to the space spanned by B1 and B2.1 We next observe that From (4), we see that each element (x1,x2,x3) E S (B1,B2) can be expressed in one and only one way as a linear combination of B1 and B2. In particular, Hence, dim S(a1,a2,a3,a4,a5) = dim S(BlIB2) = 2. b. Using the augmented matrix idea in which our last five columns denote al, a2, a3, a4, and a5, respectively, we have

20 3.2.5 (L) continued Hence, B1 = (1,0,5) = 7al - 2a3 B2 = (0,1,-1) = -3a1 + a3 and Note -+ Geometrically speaking, if we view each 3-tuple as a vector in i, f , and k components originating at the origin, then i + 2j + 3k, t ; + 6;, ; + 8k, i + 3j + 2k, and i - 2j + 7k all lie

21 3.2.5 (L) continued in the same plane. This plane is determined by the pair of vec b -b tors al = i + 2j + 3k and a3 = 3i + 7j + 8k mined by the pair of vectors B1 = i + 5k and e2 = - k b -+ In terms of B plane -++ It is also deter- and B2-components, x i + x j + x k lies in the so just as in equation (4), we see that the equation of the plane is (or in more common notation, z = 5x - y). a. In a manner of speaking, this is the most important result in the first three units. What it says is that if we have a set of m linearly independent vectors in V, then no fewer than m vectors can span V. To make this a bit more concrete, we are saying, for example, that if vl, v2, v3, and v4 all belong to a vector space V, and if (v1,v2,v3,v4) is linearly independent, then no set of three, or less, elements in V can span V. As a still more cont -? + crete illustration, since 1, 1, and k are linearly independent, no fewer than three vectors can span xyz-space. Before we prove the theorem, let us explain why the result (assuming its true) is so important in the discussion of the dimension of a vector space. To begin with, let us notice that the con- structive techniques used in the lecture seem the answer might depend on how we choose our vectors. to indicate that In other words, suppose we carry out the construction described in the lec- ture and find two sets of linearly independent vectors that span V. How do we know that these two sets have the same number of elements? And if they don't have the same number, then it is ambiguous to define the dimension of a vector space to be the number of elements in a linearly independent set which spans V.

22 3.2.6 (L) continued If the theorem is true, let us suppose that (ul,..., u 1 is one r set of linearly independent vectors which span V and that (vlf..., vs} is another such set. Then by the theorem, since..., v 1 is linearly independent, S (ul,..., ur) span V and Cvl, r b s. Reversing the roles of the two sets we have that (vlf..., vsl span V and that (ul,..., u 1 is linearly indepen- dent. Hence, s b r. But since r 3 s and s b r, it can only be that r = s. In particu- lar, this proves that if {ul,..., url is one set of linearly independent vectors which span V, then any other set of linearly independent vectors which span V must also have r elements. can also be shown in this case that any set of r linearly indepen- dent vectors span V. linear combinations of ul,..., u r' etc.) r (Namely, we can express these r vectors as We then define a basis for V to be any set of linearly independent vectors which span V, whereupon we may then unambiguously define the dimension of V to be the number of elements in any basis of V since all bases have the same number of elements. In other words, when we say that the dimension of V is r (written dim V = r), we mean that there exists a set of r linearly indepen- dent vectors which span V. we write If (ul,..., ur1 is such a set, then It or more simply, Returning to the proof of the theorem, we let Certainly, S1 spans V since {a1,..., an) already spans V and S1 is linearly dependent since B1 is a linear combination of al,..., a. Hence, in the order given by (I),at least one of the members n of S is a linear combination of the preceding ones. It can't be

23 3.2.6 (L) continued B1, so it must be one of the a's. This a can be deleted from S1 without changing the space spanned by S1. Let us delete such an a and renumber the remaining a's to be al,..., "n-1' Thus, spans V also. Now let Since (BIIalI..., a n-1 I spans V, so also does S2 since {BIIalI..., a n- 1I c {B11B21a1, an-l I and since B2 is a linear combination of B1,al,... a n-1' we have that S2 is a linearly dependent set. Hence, in the order given in (2), one of the elements of S2 can be expressed as a linear combination of the preceding ones. But, since the 6's are linearly independent, B2 cannot be a scalar multiple of B1; hence, it must again be one of the a's which is expendable. conclude that Deleting this a, let's again renumber the a's and also spans V. We now let and apply the same argument as before. Since S3 is linearly dependent,the B's are linearly independent it must be one of the a's which is redundant. Continuing inductively in this manner, each time we tack on a 8, we must be able to delete an a. In particular. then, there are at least as many a's as there are B's.

24 3.2.6 (L) continued As a final note to this part of the exercise, let us observe that the procedure outlined here tells us how to augment any set of linearly independent vectors into a basis. Namely, we use the procedure outlined in our proof-whereby we list one of our independent vectors followed by a set of spanning vectors and then using some computational technique such as row-reducing matrices, we find that one of the spanning vectors can be deleted. We then augment our set of vectors by listing the next member of the lin- early independent set and finding another of the spanning vectors to delete. We continue in this way until the last member of the linearly independent set is added and we may then use the rowreduced technique to prune out the remaining redundancies. we are then left with is a basis which includes the originally given set of linearly independent vectors. What This technique will be employed in more concrete form in the fol- lowing units of this block, but for now, we close with the example given in (6).. The main aim here is to add concreteness to part (a) and at the same time, to show one way of augmenting a set of linearly inde- pendent vectors to form a basis. one at a time by u1,u2,u3, E 5. Thus, we begin by row-reducing What we do is augment (a1,a2,a3}, etc., until we have a basis for to make sure that {al,a2,a3) is linearly independent. to This leads so that {a1,a2,a31 is linearly independent. From (11, we see that S(a1,a2,a3) = S(B1,B2,B3) where 8, = (1,o,o,-1,-11, B~ = (0,1,0,3,0), B3 = (0,0,1,-1,2). We

25 3.2.6 (L) continued augment S(a1,a2,a3) [= S(B1rB2,B3)1 by u1 to obtain From (21, we see that S(al,a2,a3,u1) is a 4-dimensional, so we next look at S(a1,a2,a3,u1,u2) by row-reducing We see from (3) that E~ = S(al,a2,a3,ul,u2). Had u 2 been a linear combination of al,a2,a3, and u1' we would have deleted it and used row-reduction on S(alra2,a3,u1,u3) etc., until we wound up with the five elements which span V.

26 3.2.7 continued From (11, we see that S(a1,a2,a3) = S(B1,B2,B3) where BL = (1,0,0,g), B2 = (0,1,0,g), and B3 =(0,0,1,7). Moreover, Hence,

27 3.2.7 continued b. The dimension of W = S(al,a2,a3) = 3. Any three linearly independent members of W span W and conversely any set of three elements of W which span W are linearly independent. The "natural" basis for W is (f31,f32ff33~ since ( x ~ ~ x ~ ~ E x W ~ *, x ~ ) From (1), we conclude that and

28 3.2.8 continued Hence, a3 = 7al + 6a2-4a4. [Of course, if we so desire, we may use (3) to obtain From (4') and (2), we could express (1,0,0), (0,1,0), and (0,0,1) as linear combinations of al, a2, and a3. Equation (4') is more conventional in the sense that for linear dependence, we like to express a vector as a linear combination of its predecessors.] (L) The main aim of this exercise is to illustrate the existence of an infinite-dimensional vector space. First of all, since every polynomial is continuous and since the set of all continuous func- tions is a vector space (with respect to the usual meanings of the sum of two functions and the product of a scalar and a function), we know that the set of all polynomials is at least a sub& of the space of continuous functions. To prove that this subset is a subspace, we need only know that the sum of two members of this subset belongs to the subset as does any scalar multiple of a mem- ber of the subset. Clearly, the sum of two polynomials is a polynomial and a scalar multiple of a polynomial is also a polynomial. Therefore, the set of all polynomials is itself a vector space. But we know from our treatment of elementary calculus that the powers of x are linearly

29 3.2.9 (L) continued independent. That is, x n cannot be expressed as a linear combinan- 1 tion of 1, x,..., and x. In other words, for any value of n no matter how large (1, x,..., xn) is a linearly independent set. Yet, the space spanned by (1, x,..., x n I? can never yield the entire space P since x E P but xn+l S(1, x,..., x n ). n-1 Hence, in the language of today's lecture with an = x that for each n, we see but for no n will S(al,..., an ) yield all of our vector space. That is, the constructive device described at the end of the lecture never terminates and accordingly, we refer to the space of all polynomials in x as an infinite-dimensional vector space. As a final note to this exercise, recall that since every analytic function may be represented as a power series, the space spanned by the finite set (1, x, x 2,..., x n,...i is the space of all analytic functions.

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