4.6 Bases and Dimension

Transcription

1 46 Bases and Dimension (a) Show that {1,x,x 2,x 3 } is linearly independent on every interval (b) If f k (x) = x k for k = 0, 1,,n, show that {f 0,f 1,,f n } is linearly independent on every interval for all fixed n 41 (a) Show that the functions f 1 (x) = e r1x, f 2 (x) = e r2x, f 3 (x) = e r 3x have Wronskian W [f 1,f 2,f 3 ](x) = e (r 1+r 2 +r 3 )x r 1 r 2 r 3 r1 2 r2 2 r2 3 = e (r 1+r 2 +r 3 )x (r 3 r 1 )(r 3 r 2 )(r 2 r 1 ), and hence determine the conditions on r 1,r 2,r 3 such that {f 1,f 2,f 3 } is linearly independent on every interval (b) More generally, show that the set of functions {e r1x,e r2x,,e rnx } is linearly independent on every interval if and only if all of the r i are distinct [Hint: Show that the Wronskian of the given functions is a multiple of the n n Vandermonde determinant, and then use Problem 21 in Section 33] 42 Let {v 1, v 2 } be a linearly independent set in a vector space V, and let v = αv 1 + v 2, w = v 1 + αv 2, where α is a constant Use Definition 454 to determine all values of α for which {v, w} is linearly independent 43 If v 1 and v 2 are vectors in a vector space V, and u 1, u 2, u 3 are each linear combinations of them, prove that {u 1, u 2, u 3 } is linearly dependent 44 Let v 1, v 2,,v m be a set of linearly independent vectors in a vector space V and suppose that the vectors u 1, u 2,,u n are each linear combinations of them It follows that we can write m u k = a ik v i, k = 1, 2,,n, i=1 for appropriate constants a ik (a) If n>m, prove that {u 1, u 2,,u n } is linearly dependent on V (b) If n = m, prove that {u 1, u 2,,u n } is linearly independent in V if and only if det[a ij ] 0 (c) If n<m, prove that {u 1, u 2,,u n } is linearly independent in V if and only if rank(a) = n, where A =[a ij ] (d) Which result from this section do these results generalize? 45 Prove from the definition of linearly independent that if {v 1, v 2,,v n } is linearly independent and if A is an invertible n n matrix, then the set {Av 1,Av 2,,Av n } is linearly independent 46 Prove that if {v 1, v 2 } is linearly independent and v 3 is not in span{v 1, v 2 }, then {v 1, v 2, v 3 } is linearly independent 47 Generalizing the previous exercise, prove that if {v 1, v 2,,v k } is linearly independent and v k+1 is not in span{v 1, v 2,,v k }, then {v 1, v 2,,v k+1 } is linearly independent 48 Prove Theorem Prove Proposition Prove that if {v 1, v 2,,v k } spans a vector space V, then for every vector v in V, {v, v 1, v 2,,v k } is linearly dependent 51 Prove that if V = P n and S ={p 1,p 2,,p k } is a set of vectors in V each of a different degree, then S is linearly independent [Hint: Assume without loss of generality that the polynomials are ordered in descending degree: deg(p 1 )>deg(p 2 )> > deg(p k ) Assuming that c 1 p 1 + c 2 p 2 + +c k p k = 0, first show that c 1 is zero by examining the highest degree Then repeat for lower degrees to show successively that c 2 = 0, c 3 = 0, and so on] 46 Bases and Dimension The results of the previous section show that if a minimal spanning set exists in a (nontrivial) vector space V, it cannot be linearly dependent Therefore if we are looking for minimal spanning sets for V, we should focus our attention on spanning sets that are linearly independent One of the results of this section establishes that every spanning set for V that is linearly independent is indeed a minimal spanning set Such a set will be

2 282 CHAPTER 4 Vector Spaces called a basis This is one of the most important concepts in this text and a cornerstone of linear algebra DEFINITION 461 A set of vectors {v 1, v 2,,v k } in a vector space V is called a basis 4 for V if (a) The vectors are linearly independent (b) The vectors span V Notice that if we have a finite spanning set for a vector space, then we can always, in principle, determine a basis for V by using the technique of Corollary 4512 Furthermore, the computational aspects of determining a basis have been covered in the previous two sections, since all we are really doing is combining the two concepts of linear independence and linear span Consequently, this section is somewhat more theoretically oriented than the preceding ones The reader is encouraged not to gloss over the theoretical aspects, as these really are fundamental results in linear algebra There do exist vector spaces V for which it is impossible to find a finite set of linearly independent vectors that span V The vector space C n (I), n 1, is such an example (Example 4619) Such vector spaces are called infinite-dimensional vector spaces Our primary interest in this text, however, will be vector spaces that contain a finite spanning set of linearly independent vectors These are known as finite-dimensional vector spaces, and we will encounter numerous examples of them throughout the remainder of this section We begin with the vector space R n InR 2, the most natural basis, denoted {e 1, e 2 }, consists of the two vectors e 1 = (1, 0), e 2 = (0, 1), (461) and in R 3, the most natural basis, denoted {e 1, e 2, e 3 }, consists of the three vectors e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1) (462) The verification that the sets (461) and (462) are indeed bases of R 2 and R 3, respectively, is straightforward and left as an exercise 5 These bases are referred to as the standard basis on R 2 and R 3, respectively In the case of the standard basis for R 3 given in (462), we recognize the vectors e 1, e 2, e 3 as the familiar unit vectors i, j, k pointing along the positive x-, y-, and z-axes of the rectangular Cartesian coordinate system More generally, consider the set of vectors {e 1, e 2,,e n } in R n defined by e 1 = (1, 0,,0), e 2 = (0, 1,,0),, e n = (0, 0,,1) These vectors are linearly independent by Corollary 4515, since det([e 1, e 2,,e n ]) = det(i n ) = 1 0 Furthermore, the vectors span R n, since an arbitrary vector v = (x 1,x 2,,x n ) in R n can be written as v = x 1 (1, 0,,0) + x 2 (0, 1,,0) + +x n (0, 0,,1) = x 1 e 1 + x 2 e 2 + +x n e n 4 The plural of basis is bases 5 Alternatively, the verification is a special case of that given shortly for the general case of R n

3 46 Bases and Dimension 283 Consequently, {e 1, e 2,,e n } is a basis for R n We refer to this basis as the standard basis for R n The general vector in R n has n components, and the standard basis vectors arise as the n vectors that are obtained by sequentially setting one component to the value 1 and the other components to 0 In general, this is how we obtain standard bases in vector spaces whose vectors are determined by the specification of n independent constants We illustrate with some examples Example 462 Determine the standard basis for M 2 (R) Solution: The general matrix in M 2 (R) is [ ] ab cd Consequently, there are four independent parameters that give rise to four special vectors in M 2 (R) Sequentially setting one of these parameters to the value 1 and the others to 0 generates the following four matrices: A 1 = [ ], A 2 = [ ], A 3 = [ ] 00, A 10 4 = We see that {A 1,A 2,A 3,A 4 } is a spanning set for M 2 (R) Furthermore, holds if and only if c 1 [ c 1 A 1 + c 2 A 2 + c 3 A 3 + c 4 A 4 = 0 2 ] [ ] [ ] [ ] c 2 + c c 10 4 = 01 [ ] [ ] that is, if and only if c 1 = c 2 = c 3 = c 4 = 0 Consequently, {A 1,A 2,A 3,A 4 } is a linearly independent spanning set for M 2 (R), hence it is a basis This is the standard basis for M 2 (R) Remark More generally, consider the vector space of all m n matrices with real entries, M m n (R)IfweletE ij denote the m n matrix with value 1 in the (i, j)-position and zeros elsewhere, then we can show routinely that {E ij : 1 i m, 1 j n} is a basis for M m n (R), and it is the standard basis Example 463 Determine the standard basis for P 2 Solution: We have P 2 ={a 0 + a 1 x + a 2 x 2 : a 0,a 1,a 2 R}, so that the vectors in P 2 are determined by specifying values for the three parameters a 0,a 1, and a 2 Sequentially setting one of these parameters to the value 1 and the other two to the value 0 yields the following vectors in P 2 : p 0 (x) = 1, p 1 (x) = x, p 2 (x) = x 2

4 284 CHAPTER 4 Vector Spaces We have shown in Example 446 that {p 0,p 1,p 2 } is a spanning set for P 2 Furthermore, W[p 0,p 1,p 2 ](x) = 1 xx 2 012x 00 2 = 2 0, which implies that {p 0,p 1,p 2 } is linearly independent on any interval 6 Consequently, {p 0,p 1,p 2 } is a basis for P 2 This is the standard basis for P 2 Remark More generally, the reader can check that the standard basis for the vector space of all polynomials of degree n or less, P n,is {1,x,x 2,,x n } Dimension of a Finite-Dimensional Vector Space The reader has probably realized that there can be many different bases for a given vector space V In addition to the standard basis {e 1, e 2, e 3 } on R 3, for example, it can be checked 7 that {(1, 2, 3), (4, 5, 6), (7, 8, 8)} and {(1, 0, 0), (1, 1, 0), (1, 1, 1)} are also bases for R 3 And there are countless others as well Despite the multitude of different bases available for a vector space V, they all share one common feature: the number of vectors in each basis for V is the same This fact will be deduced as a corollary of our next theorem, a fundamental result in the theory of vector spaces Theorem 464 If a finite-dimensional vector space has a basis consisting of m vectors, then any set of more than m vectors is linearly dependent Proof Let {v 1, v 2,,v m } be a basis for V, and consider an arbitrary set of vectors in V, say, {u 1, u 2,,u n }, with n>mwewishtoprovethat{u 1, u 2,,u n } is necessarily linearly dependent Since {v 1, v 2,,v m } is a basis for V, it follows that each u j can be written as a linear combination of v 1, v 2,,v m Thus, there exist constants a ij such that u 1 = a 11 v 1 + a 21 v a m1 v m, u 2 = a 12 v 1 + a 22 v a m2 v m, u n = a 1n v 1 + a 2n v a mn v m To prove that {u 1, u 2,,u n } is linearly dependent, we must show that there exist scalars c 1,c 2,,c n, not all zero, such that c 1 u 1 + c 2 u 2 + +c n u n = 0 (463) Inserting the expressions for u 1, u 2,,u n into Equation (463) yields c 1 (a 11 v 1 + a 21 v 2 + +a m1 v m ) + c 2 (a 12 v 1 + a 22 v 2 + +a m2 v m ) + +c n (a 1n v 1 + a 2n v 2 + +a mn v m ) = 0 6 Alternatively, we can start with the equation c 0 p 0 (x) + c 1 p 1 (x) + c 2 p 2 (x) = 0 for all x in R and show readily that c 0 = c 1 = c 2 = 0 7 The reader desiring extra practice at the computational aspects of verifying a basis is encouraged to pause here to check these examples

5 46 Bases and Dimension 285 Rearranging terms, we have (a 11 c 1 + a 12 c 2 + +a 1n c n )v 1 + (a 21 c 1 + a 22 c 2 + +a 2n c n )v 2 + +(a m1 c 1 + a m2 c 2 + +a mn c n )v m = 0 Since {v 1, v 2,,v m } is linearly independent, we can conclude that a 11 c 1 + a 12 c a 1n c n = 0, a 21 c 1 + a 22 c a 2n c n = 0, a m1 c 1 + a m2 c a mn c n = 0 Thisisanm n homogeneous system of linear equations with m<n, and hence, from Corollary 2511, it has nontrivial solutions for c 1,c 2,,c n It therefore follows from Equation (463) that {u 1, u 2,,u n } is linearly dependent Corollary 465 All bases in a finite-dimensional vector space V contain the same number of vectors Proof Suppose {v 1, v 2,,v n } and {u 1, u 2,,u m } are two bases for V From Theorem 464 we know that we cannot have m>n(otherwise {u 1, u 2,,u m } would be a linearly dependent set and hence could not be a basis for V ) Nor can we have n>m (otherwise {v 1, v 2,,v n } would be a linearly dependent set and hence could not be a basis for V ) Thus, it follows that we must have m = n We can now prove that any basis provides a minimal spanning set for V Corollary 466 If a finite-dimensional vector space V has a basis consisting of n vectors, then any spanning set must contain at least n vectors Proof If the spanning set contained fewer than n vectors, then there would be a subset of less than n linearly independent vectors that spanned V ; that is, there would be a basis consisting of less than n vectors But this would contradict the previous corollary The number of vectors in a basis for a finite-dimensional vector space is clearly a fundamental property of the vector space, and by Corollary 465 it is independent of the particular chosen basis We call this number the dimension of the vector space DEFINITION 467 The dimension of a finite-dimensional vector space V, written dim[v ], is the number of vectors in any basis for V IfV is the trivial vector space, V ={0}, then we define its dimension to be zero Remark We say that the dimension of the world we live in is three for the very reason that the maximum number of independent directions that we can perceive is three If a vector space has a basis containing n vectors, then from Theorem 464, the maximum number of vectors in any linearly independent set is n Thus, we see that the terminology dimension used in an arbitrary vector space is a generalization of a familiar idea Example 468 It follows from our examples earlier in this section that dim[r 3 ]=3, dim[m 2 (R)] =4, and dim[p 2 ]=3

6 286 CHAPTER 4 Vector Spaces More generally, the following dimensions should be remembered: dim[r n ]=n, dim[m m n (R)] =mn, dim[m n (R)] =n 2, dim[p n ]=n + 1 These values have essentially been established previously in our discussion of standard bases The standard basis for R n is {e 1, e 2,,e n }, where e i is the n-tuple with value 1 in the ith position and value 0 elsewhere Thus, this basis contains n vectors The standard basis for M m n (R) is the set of matrices E ij (1 i m, 1 j n) with value1inthe(i, j) position and value 0 elsewhere There are mn such matrices in this standard basis The case of M n (R) is just a special case of M m n (R) in which m = n Finally, the standard basis for P n is {1,x,x 2,,x n }, a set of n + 1 vectors Next, let us return once more to Example 1216 to cast its results in terms of the basis concept Example 469 Determine a basis for the solution space to the differential equation y + y = 0 on any interval I Solution: Our results from Example 1216 tell us that all solutions to the given differential equation are of the form y(x) = c 1 cos x + c 2 sin x Consequently, {cos x,sin x} is a linearly independent spanning set for the solution space of the differential equation and therefore is a basis More generally, we will show in Chapter 6 that all solutions to the differential equation y + a 1 (x)y + a 2 (x)y = 0 on the interval I have the form y(x) = c 1 y 1 (x) + c 2 y 2 (x), where {y 1,y 2 } is any linearly independent set of solutions to the differential equation Using the terminology introduced in this section, it will therefore follow that: The set of all solutions to y + a 1 (x)y + a 2 (x)y = 0onan interval I is a vector space of dimension two If a vector space has dimension n, then from Theorem 464, the maximum number of vectors in any linearly independent set is n On the other hand, from Corollary 466, the minimum number of vectors that can span V is also n Thus, a basis for V must be a linearly independent set of n vectors Our next theorem establishes that any set of n linearly independent vectors is a basis for V Theorem 4610 If dim[v ]=n, then any set of n linearly independent vectors in V is a basis for V

7 46 Bases and Dimension 287 Proof Let v 1, v 2,,v n be n linearly independent vectors in V We need to show that they span V Todothis,letv be an arbitrary vector in V From Theorem 464, the set of vectors {v, v 1, v 2,,v n } is linearly dependent, and so there exist scalars c 0,c 1,,c n, not all zero, such that c 0 v + c 1 v 1 + +c n v n = 0 (464) If c 0 = 0, then the linear independence of {v 1, v 2,,v n } and (464) would imply that c 0 = c 1 = =c n = 0, a contradiction Hence, c 0 0, and so, from Equation (464), v = 1 c 0 (c 1 v 1 + c 2 v 2 + +c n v n ) Thus v, and hence any vector in V, can be written as a linear combination of v 1, v 2,,v n, and hence, {v 1, v 2,,v n } spans V, in addition to being linearly independent Hence it is a basis for V, as required Theorem 4610 is one of the most important results of the section In Chapter 6, we will explicitly construct a basis for the solution space to the differential equation y (n) + a 1 (x)y (n 1) + +a n 1 (x)y + a n (x)y = 0 consisting of n vectors That is, we will show that the solution space to this differential equation is n-dimensional It will then follow immediately from Theorem 4610 that every solution to this differential equation is of the form y(x) = c 1 y 1 (x) + c 2 y 2 (x) + +c n y n (x), where {y 1,y 2,,y n } is any linearly independent set of n solutions to the differential equation Therefore, determining all solutions to the differential equation will be reduced to determining any linearly independent set of n solutions A similar application of the theorem will be used to develop the theory for systems of differential equations in Chapter 7 More generally, Theorem 4610 says that if we know in advance that the dimension of the vector space V is n, then n linearly independent vectors in V are already guaranteed to form a basis for V without the need to explicitly verify that these n vectors also span V This represents a useful reduction in the work required to verify a basis Here is an example: Example 4611 Verify that {1 + x,2 2x + x 2, 1 + x 2 } is a basis for P 2 Solution: Since dim[p 2 ]=3, Theorem 4610 will guarantee that the three given vectors are a basis, once we confirm only that they are linearly independent The polynomials have Wronskian p 1 (x) = 1 + x, p 2 (x) = 2 2x + x 2, p 3 (x) = 1 + x 2 W[p 1,p 2,p 3 ](x) = 1 + x 2 2x + x x x 2x = 6 0 Since the Wronskian is nonzero, the given set of vectors is linearly independent on any interval Consequently, {1 + x,2 2x + x 2, 1 + x 2 } is indeed a basis for P 2

8 288 CHAPTER 4 Vector Spaces There is a notable parallel result to Theorem 4610 which can also cut down the work required to verify that a set of vectors in V is a basis for V, provided that we know the dimension of V in advance Theorem 4612 If dim[v ]=n, then any set of n vectors in V that spans V is a basis for V Proof Let v 1, v 2,,v n be n vectors in V that span V To confirm that {v 1, v 2,,v n } is a basis for V, we need only show that this is a linearly independent set of vectors Suppose, to the contrary, that {v 1, v 2,,v n } is a linearly dependent set By Corollary 4512, there is a linearly independent subset of {v 1, v 2,,v n }, with fewer than n vectors, which also spans V But this implies that V contains a basis with fewer than n vectors, a contradiction Putting the results of Theorems 4610 and 4612 together, the following result is immediate Corollary 4613 If dim[v ] = n and S = {v 1, v 2,,v n } is a set of n vectors in V, the following statements are equivalent: 1 S is a basis for V 2 S is linearly independent 3 S spans V We emphasize once more the importance of this result It means that if we have a set S of dim[v ] vectors in V, then to determine whether or not S is a basis for V,we need only check if S is linearly independent or if S spans V, not both We next establish another corollary to Theorem 4610 Corollary 4614 Let S be a subspace of a finite-dimensional vector space V If dim[v ]=n, then Furthermore, if dim[s] =n, then S = V dim[s] n Proof Suppose that dim[s] >n Then any basis for S would contain more than n linearly independent vectors, and therefore we would have a linearly independent set of more than n vectors in V This would contradict Theorem 464 Thus, dim[s] n Now consider the case when dim[s] =n = dim[v ] In this case, any basis for S consists of n linearly independent vectors in S and hence n linearly independent vectors in V Thus, by Theorem 4610, these vectors also form a basis for V Hence, every vector in V is spanned by the basis vectors for S, and hence, every vector in V lies in S Thus, V = S Example 4615 Give a geometric description of the subspaces of R 3 of dimensions 0, 1, 2, 3 Solution: Zero-dimensional subspace: This corresponds to the subspace {(0, 0, 0)}, and therefore it is represented geometrically by the origin of a Cartesian coordinate system One-dimensional subspace: These are subspaces generated by a single (nonzero) basis vector Consequently, they correspond geometrically to lines through the origin Two-dimensional subspace: These are the subspaces generated by any two noncollinear vectors and correspond geometrically to planes through the origin

9 46 Bases and Dimension 289 Three-dimensional subspace: Since dim[r 3 ]=3, it follows from Corollary 4614 that the only three-dimensional subspace of R 3 is R 3 itself Example 4616 Theorem 4617 Determine a basis for the subspace of R 3 consisting of all solutions to the equation x 1 + 2x 2 x 3 = 0 Solution: We can solve this problem geometrically The given equation is that of a plane through the origin and therefore is a two-dimensional subspace of R 3 In order to determine a basis for this subspace, we need only choose two linearly independent (ie, noncollinear) vectors that lie in the plane A simple choice of vectors is 8 v 1 = (1, 0, 1) and v 2 = (2, 1, 0) Thus, a basis for the subspace is {(1, 0, 1), (2, 1, 0)} Corollary 4614 has shown that if S is a subspace of a finite-dimensional vector space V with dim[s] =dim[v ], then S = V Our next result establishes that, in general, a basis for a subspace of a finite-dimensional vector space V can be extended to a basis for V This result will be required in the next section and also in Chapter 5 Let S be a subspace of a finite-dimensional vector space V Any basis for S is part of a basis for V Proof Suppose dim[v ]=n and dim[s] =k By Corollary 4614, k n Ifk = n, then S = V, so that any basis for S is a basis for V Suppose now that k<n, and let {v 1, v 2,,v k } be a basis for S These basis vectors are linearly independent, but they fail to span V (otherwise they would form a basis for V, contradicting k<n) Thus, there is at least one vector, say v k+1,inv that is not in span{v 1, v 2,,v k } Hence, {v 1, v 2,,v k, v k+1 } is linearly independent If k+1 = n, then we have a basis for V by Theorem 4610, and we are done Otherwise, we can repeat the procedure to obtain the linearly independent set {v 1, v 2,,v k, v k+1, v k+2 } The process will terminate when we have a linearly independent set containing n vectors, including the original vectors v 1, v 2,,v k in the basis for S This proves the theorem Remark a basis The process used in proving the previous theorem is referred to as extending Example 4618 Let S denote the subspace of M 2 (R) consisting of all symmetric 2 2 matrices Determine a basis for S, and find dim[s] Extend this basis for S to obtain a basis for M 2 (R) Solution: We first express S in set notation as S ={A M 2 (R) : A T = A} In order to determine a basis for S, we need to obtain the element form of the matrices in S We can write {[ ] } ab S = : a,b,c R bc Since [ ] [ ] [ ] [ ] ab = a + b + c, bc it follows that 8 There are many others, of course {[ 10 S = span 00 ], [ ] 01, 10 [ ]} 00 01

10 290 CHAPTER 4 Vector Spaces Furthermore, it is easily shown that the matrices in this spanning set are linearly independent Consequently, a basis for S is {[ ] [ ] [ ]} ,,, so that dim[s] =3 Since dim[m 2 (R)] =4, in order to extend the basis for S to a basis for M 2 (R), we need to add one additional matrix from M 2 (R) such that the resulting set is linearly independent We must choose a nonsymmetric matrix, for any symmetric matrix can be expressed as a linear combination of the three basis vectors for S, and this would create a linear dependency among the matrices A simple choice of nonsymmetric matrix (although this is certainly not the only choice) is [ ] Adding this vector to the basis for S yields the linearly independent set {[ ] [ ] [ ] [ ]} ,,, (465) Since dim[m 2 (R)] =4, Theorem 4610 implies that (465) is a basis for M 2 (R) It is important to realize that not all vector spaces are finite dimensional Some are infinite-dimensional In an infinite-dimensional vector space, we can find an arbitrarily large number of linearly independent vectors We now give an example of an infinitedimensional vector space that is of primary importance in the theory of differential equations, C n (I) Example 4619 Show that the vector space C n (I) is an infinite-dimensional vector space Solution: Consider the functions 1,x,x 2,,x k in C n (I) Of course, each of these functions is in C k (I) as well, and for each fixed k, the Wronskian of these functions is nonzero (the reader can check that the matrix involved in this calculation is upper triangular, with nonzero entries on the main diagonal) Hence, the functions are linearly independent on I by Theorem 4521 Since we can choose k arbitrarily, it follows that there are an arbitrarily large number of linearly independent vectors in C n (I), hence C n (I) is infinite-dimensional In this example we showed that C n (I) is an infinite-dimensional vector space Consequently, the use of our finite-dimensional vector space theory in the analysis of differential equations appears questionable However, the key theoretical result that we will establish in Chapter 6 is that the solution set of certain linear differential equations is a finite-dimensional subspace of C n (I), and therefore our basis results will be applicable to this solution set Exercises for 46 Key Terms Basis, Standard basis, Infinite-dimensional, Finitedimensional, Dimension, Extension of a subspace basis Skills Be able to determine whether a given set of vectors forms a basis for a vector space V Be able to construct a basis for a given vector space V Be able to extend a basis for a subspace of V to V itself Be familiar with the standard bases on R n, M m n (R), and P n

11 46 Bases and Dimension 291 Be able to give the dimension of a vector space V Be able to draw conclusions about the properties of a set of vectors in a vector space (ie, spanning or linear independence) based solely on the size of the set Understand the usefulness of Theorems 4610 and 4612 True-False Review For Questions 1 11, decide if the given statement is true or false, and give a brief justification for your answer If true, you can quote a relevant definition or theorem from the text If false, provide an example, illustration, or brief explanation of why the statement is false 1 A basis for a vector space V is a set S of vectors that spans V 2 If V and W are vector spaces of dimensions n and m, respectively, and if n>m, then W is a subspace of V 3 A vector space V can have many different bases 4 dim[p n ]=dim[r n ] 5 If V is an n-dimensional vector space, then any set S of m vectors with m>nmust span V 6 Five vectors in P 3 must be linearly dependent 7 Two vectors in P 3 must be linearly independent 8 Ten vectors in M 3 (R) must be linearly dependent 9 If V is an n-dimensional vector space, then every set S with fewer than n vectors can be extended to a basis for V 10 Every set of vectors that spans a finite-dimensional vector space V contains a subset which forms a basis for V 11 The set of all 3 3 upper triangular matrices forms a three-dimensional subspace of M 3 (R) Problems For Problems 1 5, determine whether the given set of vectors is a basis for R n 1 {(1, 1), ( 1, 1)} 2 {(1, 2, 1), (3, 1, 2), (1, 1, 1)} 3 {(1, 1, 1), (2, 5, 2), (3, 11, 5)} 4 {(1, 1, 1, 2), (1, 0, 1, 1), (2, 1, 1, 1)} 5 {(1, 1, 0, 2), (2, 1, 3, 1), ( 1, 1, 1, 2), (2, 1, 1, 2)} 6 Determine all values of the constant k for which the set of vectors {(0, 1, 0,k), (1, 0, 1, 0), (0, 1, 1, 0), (k, 0, 2, 1)} is a basis for R 4 7 Determine a basis S for P 3, and hence, prove that dim[p 3 ]=4 Be sure to prove that S is a basis 8 Determine a basis S for P 3 whose elements all have the same degree Be sure to prove that S is a basis For Problems 9 12, find the dimension of the null space of the given matrix A [ ] A = A = A = A = Let S be the subspace of R 3 that consists of all solutions to the equation x 3y + z = 0 Determine a basis for S, and hence, find dim[s] 14 Let S be the subspace of R 3 consisting of all vectors of the form (r, r 2s, 3s 5r), where r and s are real numbers Determine a basis for S, and hence, find dim[s] 15 Let S be the subspace of M 2 (R) consisting of all 2 2 upper triangular matrices Determine a basis for S, and hence, find dim[s] 16 Let S be the subspace of M 2 (R) consisting of all 2 2 matrices with trace zero Determine a basis for S, and hence, find dim[s] 17 Let S be the subspace of R 3 spanned by the vectors v 1 = (1, 0, 1), v 2 = (0, 1, 1), v 3 = (2, 0, 2) Determine a basis for S, and hence, find dim[s]

12 292 CHAPTER 4 Vector Spaces 18 Let S be the vector space consisting of the set of all linear combinations of the functions f 1 (x) = e x,f 2 (x) = e x,f 3 (x) = sinh(x) Determine a basis for S, and hence, find dim[s] 19 Determine a basis for the subspace of M 2 (R) spanned by [ ] [ ] [ ] [ ] ,,, Let v 1 = (1, 1) and v 2 = ( 1, 1) (a) Show that {v 1, v 2 } spans R 2 (b) Show that {v 1, v 2 } is linearly independent (c) Conclude from (a) or (b) that {v 1, v 2 } is a basis for R 2 What theorem in this section allows you to draw this conclusion from either (a) or (b), without proving both? 21 Let v 1 = (2, 1) and v 2 = (3, 1) (a) Show that {v 1, v 2 } spans R 2 (b) Show that {v 1, v 2 } is linearly independent (c) Conclude from (a) or (b) that {v 1, v 2 } is a basis for R 2 What theorem in this section allows you to draw this conclusion from either (a) or (b), without proving both? 22 Let v 1 = (0, 6, 3), v 2 = (3, 0, 3), and v 3 = (6, 3, 0) Show that {v 1, v 2, v 3 } is a basis for R 3 [Hint: You need not show that the set is both linearly independent and a spanning set for P 2 Use a theorem from this section to shorten your work] 23 Determine all values of the constant α for which {1 + αx 2, 1 + x + x 2, 2 + x} is a basis for P 2 24 Let p 1 (x) = 1 + x,p 2 (x) = x(x 1), p 3 (x) = 1+2x 2 Show that {p 1,p 2,p 3 } is a basis for P 2 [Hint: You need not show that the set is both linearly independent and a spanning set for P 2 Use a theorem from this section to shorten your work] 25 The Legendre polynomial of degree n, p n (x),isdefined to be the polynomial solution of the differential equation (1 x 2 )y 2xy + n(n + 1)y = 0, which has been normalized so that p n (1) = 1 The first three Legendre polynomials are p 0 (x) = 1,p 1 (x) = x, and p 2 (x) = 1 2 (3x2 1) Show that {p 0,p 1,p 2 } is a basis for P 2 [The hint for the previous problem applies again] 26 Let 27 Let [ 11 A 1 = 01 [ 10 A 3 = 12 ], A 2 = ], A 4 = [ ] 13, 1 0 [ ] (a) Show that {A 1,A 2,A 3,A 4 } is a basis for M 2 (R) [The hint on the previous problems applies again] (b) Express the vector [ ] as a linear combination of the basis in (a) A = , and let v 1 = ( 2, 7, 5, 0) and v 2 = (3, 8, 0, 5) (a) Show that {v 1, v 2 } is a basis for the null space of A (b) Using the basis in part (a), write an expression for an arbitrary vector (x,y,z,w) in the null space of A 28 Let V = M 3 (R) and let S be the subset of all vectors in V such that the sum of the entries in each row and in each column is zero (a) Find a basis and the dimension of S (b) Extend the basis in (a) to a basis for V 29 Let V = M 3 (R) and let S be the subset of all vectors in V such that the sum of the entries in each row and in each column is the same (a) Find a basis and the dimension of S (b) Extend the basis in (a) to a basis for V For Problems 30 31, Sym n (R) and Skew n (R) denote the vector spaces consisting of all real n n matrices that are symmetric and skew-symmetric, respectively 30 Find a basis for Sym 2 (R) and Skew 2 (R), and show that dim[sym 2 (R)]+dim[Skew 2 (R)] =dim[m 2 (R)]

13 47 Change of Basis Determine the dimensions of Sym n (R) and Skew n (R), and show that dim[sym n (R)]+dim[Skew n (R)] =dim[m n (R)] For Problems 32 34, a subspace S of a vector space V is given Determine a basis for S and extend your basis for S to obtain a basis for V 32 V = R 3, S is the subspace consisting of all points lying on the plane with Cartesian equation x + 4y 3z = 0 33 V = M 2 (R), S is the subspace consisting of all matrices of the form [ ] ab ba 34 V = P 2, S is the subspace consisting of all polynomials of the form (2a 1 +a 2 )x 2 +(a 1 +a 2 )x +(3a 1 a 2 ) 35 Let S be a basis for P n 1 Prove that S {x n } is a basis for P n 36 Generalize the previous problem as follows Let S be a basis for P n 1, and let p be any polynomial of degree n Prove that S {p} is a basis for P n 37 (a) What is the dimension of C n as a real vector space? Determine a basis (b) What is the dimension of C n as a complex vector space? Determine a basis 47 Change of Basis Throughout this section, we restrict our attention to vector spaces that are finite-dimensional If we have a (finite) basis for such a vector space V, then, since the vectors in a basis span V, any vector in V can be expressed as a linear combination of the basis vectors The next theorem establishes that there is only one way in which we can do this Theorem 471 If V is a vector space with basis {v 1, v 2,,v n }, then every vector v V can be written uniquely as a linear combination of v 1, v 2,,v n Proof Since v 1, v 2,,v n span V, every vector v V can be expressed as for some scalars a 1,a 2,,a n Suppose also that v = a 1 v 1 + a 2 v 2 + +a n v n, (471) v = b 1 v 1 + b 2 v 2 + +b n v n, (472) for some scalars b 1,b 2,,b n We will show that a i = b i for each i, which will prove the uniqueness assertion of this theorem Subtracting Equation (472) from Equation (471) yields (a 1 b 1 )v 1 + (a 2 b 2 )v 2 + +(a n b n )v n = 0 (473) But {v 1, v 2,,v n } is linearly independent, and so Equation (473) implies that a 1 b 1 = 0, a 2 b 2 = 0,, a n b n = 0 That is, a i = b i for each i = 1, 2,,n Remark The converse of Theorem 471 is also true That is, if every vector v in a vector space V can be written uniquely as a linear combination of the vectors in {v 1, v 2,,v n }, then {v 1, v 2,,v n } is a basis for V The proof of this fact is left as an exercise (Problem 38) Up to this point, we have not paid particular attention to the order in which the vectors of a basis are listed However, in the remainder of this section, this will become