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2 3 40. (a) Show that {1,x,x ,x } is linearly independent (a) If n>m, prove that {u1, u2,...,un} is linearly on every interval. dependent on V . k (b) If fk(x) = x for k = 0, 1,...,n, show that (b) If n = m, prove that {u1, u2,...,un} is linearly {f0,f1,...,fn} is linearly independent on every independent in V if and only if det[aij ] = 0. interval for all fixed n. (c) If n

4.6 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

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called a . This is one of the most important concepts in this text and a cornerstone of .

DEFINITION 4.6.1 4 A set of vectors {v1, v2,...,vk} in a vector space V is called a basis 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 4.5.12. Fur- thermore, 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 and . Consequently, this section is somewhat more the- oretically 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 Cn(I), n ≥ 1, is such an example (Example 4.6.19). Such vector spaces are called infinite-dimensional vector spaces. Our primary interest in this text, however, will be vector spaces that contain a finite span- ning 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. n 2 We begin with the vector space R .InR , the most natural basis, denoted {e1, e2}, consists of the two vectors

e1 = (1, 0), e2 = (0, 1), (4.6.1)

3 and in R , the most natural basis, denoted {e1, e2, e3}, consists of the three vectors

e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). (4.6.2)

The verification that the sets (4.6.1) and (4.6.2) are indeed bases of R2 and R3, respec- tively, is straightforward and left as an exercise.5 These bases are referred to as the on R2 and R3, respectively. In the case of the standard basis for R3 given in (4.6.2), we recognize the vectors e1, e2, e3 as the familiar unit vectors i, j, k pointing along the positive x-, y-, and z-axes of the rectangular Cartesian . n More generally, consider the set of vectors {e1, e2,...,en} in R defined by

e1 = (1, 0,...,0), e2 = (0, 1,...,0), ..., en = (0, 0,...,1).

These vectors are linearly independent by Corollary 4.5.15, since

det([e1, e2,...,en]) = det(In) = 1 = 0.

n n Furthermore, the vectors span R , since an arbitrary vector v = (x1,x2,...,xn) in R can be written as

v = x1(1, 0,...,0) + x2(0, 1,...,0) +···+xn(0, 0,...,1) = x1e1 + x2e2 +···+xnen.

4 The plural of basis is bases. 5Alternatively, the verification is a special case of that given shortly for the general case of Rn.

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n Consequently, {e1, e2,...,en} is a basis for R . We refer to this basis as the standard basis for Rn. The general vector in Rn 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 4.6.2 Determine the standard basis for M2(R).

Solution: The general matrix in M2(R) is   ab . cd

Consequently, there are four independent parameters that give rise to four special vectors in M2(R). Sequentially setting one of these parameters to the value 1 and the others to 0 generates the following four matrices:         10 01 00 00 A = ,A= ,A= ,A= . 1 00 2 00 3 10 4 01

We see that {A1,A2,A3,A4} is a spanning set for M2(R). Furthermore,

c1A1 + c2A2 + c3A3 + c4A4 = 02

holds if and only if           10 01 00 00 00 c + c + c + c = 1 00 2 00 3 10 4 01 00

—that is, if and only if c1 = c2 = c3 = c4 = 0. Consequently, {A1,A2,A3,A4} is a linearly independent spanning set for M2(R), hence it is a basis. This is the standard basis for M2(R). 

Remark More generally, consider the vector space of all m × n matrices with real entries, Mm×n(R).IfweletEij denote the m×n matrix with value 1 in the (i, j)- and zeros elsewhere, then we can show routinely that

{Eij : 1 ≤ i ≤ m, 1 ≤ j ≤ n}

is a basis for Mm×n(R), and it is the standard basis.

Example 4.6.3 Determine the standard basis for P2. Solution: We have

2 P2 ={a0 + a1x + a2x : a0,a1,a2 ∈ R},

so that the vectors in P2 are determined by specifying values for the three parameters a0,a1, and a2. Sequentially setting one of these parameters to the value 1 and the other two to the value 0 yields the following vectors in P2:

2 p0(x) = 1,p1(x) = x, p2(x) = x .

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We have shown in Example 4.4.6 that {p0,p1,p2} is a spanning set for P2. Furthermore, 1 xx2 W[p0,p1,p2](x) = 012x = 2 = 0, 00 2

6 which implies that {p0,p1,p2} is linearly independent on any interval. Consequently, {p0,p1,p2} is a basis for P2. This is the standard basis for P2. 

Remark More generally, the reader can check that the standard basis for the vector space of all polynomials of degree n or less, Pn,is {1,x,x2,...,xn}.

Dimension of a Finite-Dimensional Vector Space The reader has probably realized that there can be many different bases for a given 3 vector space V . In addition to the standard basis {e1, e2, e3} on R , for example, it can be checked7 that {(1, 2, 3), (4, 5, 6), (7, 8, 8)} and {(1, 0, 0), (1, 1, 0), (1, 1, 1)} are also bases for R3. 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 4.6.4 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 {v1, v2,...,vm} be a basis for V , and consider an arbitrary set of vectors in V , say, {u1, u2,...,un}, with n>m.Wewishtoprovethat{u1, u2,...,un} is necessarily linearly dependent. Since {v1, v2,...,vm} is a basis for V , it follows that each uj can be written as a of v1, v2,...,vm. Thus, there exist constants aij such that

u1 = a11v1 + a21v2 + ··· + am1vm, u2 = a12v1 + a22v2 + ··· + am2vm, . . un = a1nv1 + a2nv2 + ··· + amnvm.

To prove that {u1, u2,...,un} is linearly dependent, we must show that there exist scalars c1,c2,...,cn, not all zero, such that

c1u1 + c2u2 +···+cnun = 0. (4.6.3)

Inserting the expressions for u1, u2,...,un into Equation (4.6.3) yields

c1(a11v1 + a21v2 +···+am1vm) + c2(a12v1 + a22v2 +···+am2vm) +···+cn(a1nv1 + a2nv2 +···+amnvm) = 0.

6 Alternatively, we can start with the equation c0p0(x) + c1p1(x) + c2p2(x) = 0 for all x in R and show readily that c0 = c1 = c2 = 0. 7The reader desiring extra practice at the computational aspects of verifying a basis is encouraged to pause here to check these examples.

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Rearranging terms, we have

(a11c1 + a12c2 +···+a1ncn)v1 + (a21c1 + a22c2 +···+a2ncn)v2 +···+(am1c1 + am2c2 +···+amncn)vm = 0.

Since {v1, v2,...,vm} is linearly independent, we can conclude that

a11c1 + a12c2 + ··· + a1ncn = 0, a21c1 + a22c2 + ··· + a2ncn = 0, . . am1c1 + am2c2 + ··· + amncn = 0. Thisisanm × n homogeneous system of linear equations with m

Corollary 4.6.5 All bases in a finite-dimensional vector space V contain the same number of vectors.

Proof Suppose {v1, v2,...,vn} and {u1, u2,...,um} are two bases for V . From The- orem 4.6.4 we know that we cannot have m>n(otherwise {u1, u2,...,um} would be a linearly dependent set and hence could not be a basis for V ). Nor can we have n>m (otherwise {v1, v2,...,vn} 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 4.6.6 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 4.6.5 it is independent of the particular chosen basis. We call this number the dimension of the vector space.

DEFINITION 4.6.7 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 4.6.4, 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.

3 Example 4.6.8 It follows from our examples earlier in this section that dim[R ]=3, dim[M2(R)]=4, and dim[P2]=3. 

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More generally, the following dimensions should be remembered:

n 2 dim[R ]=n, dim[Mm×n(R)]=mn, dim[Mn(R)]=n , dim[Pn]=n + 1.

These values have essentially been established previously in our discussion of standard n bases. The standard basis for R is {e1, e2,...,en}, where ei is the n-tuple with value 1intheith position and value 0 elsewhere. Thus, this basis contains n vectors. The standard basis for Mm×n(R) is the set of matrices Eij (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 Mn(R) is just a special case of Mm×n(R) in which m = n. 2 n Finally, the standard basis for Pn is {1,x,x ,...,x }, a set of n + 1 vectors. Next, let us return once more to Example 1.2.16 to cast its results in terms of the basis concept.

Example 4.6.9 Determine a basis for the solution space to the differential equation

y + y = 0

on any interval I. Solution: Our results from Example 1.2.16 tell us that all solutions to the given differential equation are of the form

y(x) = c1 cos x + c2 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 + a1(x)y + a2(x)y = 0 on the interval I have the form

y(x) = c1y1(x) + c2y2(x),

where {y1,y2} 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 + a1(x)y + a2(x)y = 0onan interval I is a vector space of dimension two.

If a vector space has dimension n, then from Theorem 4.6.4, the maximum number of vectors in any linearly independent set is n. On the other hand, from Corollary 4.6.6, 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 4.6.10 If dim[V ]=n, then any set of n linearly independent vectors in V is a basis for V .

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Proof Let v1, v2,...,vn 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 4.6.4, the set of vectors {v, v1, v2,...,vn} is linearly dependent, and so there exist scalars c0,c1,...,cn, not all zero, such that

c0v + c1v1 +···+cnvn = 0. (4.6.4)

If c0 = 0, then the linear independence of {v1, v2,...,vn} and (4.6.4) would imply that c0 = c1 =···=cn = 0, a contradiction. Hence, c0 = 0, and so, from Equation (4.6.4), 1 v =− (c1v1 + c2v2 +···+cnvn). c0

Thus v, and hence any vector in V , can be written as a linear combination of v1, v2,...,vn, and hence, {v1, v2,...,vn} spans V , in addition to being linearly independent. Hence it is a basis for V , as required. Theorem 4.6.10 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

(n) (n−1) y + a1(x)y +···+an−1(x)y + an(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 4.6.10 that every solution to this differential equation is of the form

y(x) = c1y1(x) + c2y2(x) +···+cnyn(x),

where {y1,y2,...,yn} 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 4.6.10 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:

2 2 Example 4.6.11 Verify that {1 + x,2 − 2x + x , 1 + x } is a basis for P2.

Solution: Since dim[P2]=3, Theorem 4.6.10 will guarantee that the three given vectors are a basis, once we confirm only that they are linearly independent. The poly- nomials

2 2 p1(x) = 1 + x, p2(x) = 2 − 2x + x ,p3(x) = 1 + x

have Wronskian

1 + x 2 − 2x + x2 1 + x2 W[p1,p2,p3](x) = 1 −2 + 2x 2x =−6 = 0. 02 2

Since the Wronskian is nonzero, the given set of vectors is linearly independent on any 2 2 interval. Consequently, {1 + x,2 − 2x + x , 1 + x } is indeed a basis for P2. 

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There is a notable parallel result to Theorem 4.6.10 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 4.6.12 If dim[V ]=n, then any set of n vectors in V that spans V is a basis for V .

Proof Let v1, v2,...,vn be n vectors in V that span V . To confirm that {v1, v2,...,vn} is a basis for V , we need only show that this is a linearly independent set of vectors. Suppose, to the contrary, that {v1, v2,...,vn} is a linearly dependent set. By Corol- lary 4.5.12, there is a linearly independent subset of {v1, v2,...,vn}, 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 4.6.10 and 4.6.12 together, the following result is immediate.

Corollary 4.6.13 If dim[V ]=n and S ={v1, v2,...,vn} 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 4.6.10.

Corollary 4.6.14 Let S be a subspace of a finite-dimensional vector space V . If dim[V ]=n, then

dim[S]≤n.

Furthermore, if dim[S]=n, then S = V .

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 4.6.4. 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 4.6.10, 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 4.6.15 Give a geometric description of the subspaces of R3 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.

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Three-dimensional subspace: Since dim[R3]=3, it follows from Corollary 4.6.14 that the only three-dimensional subspace of R3 is R3 itself. 

Example 4.6.16 Determine a basis for the subspace of R3 consisting of all solutions to the equation x1 + 2x2 − x3 = 0. Solution: We can solve this problem geometrically. The given equation is that of a through the origin and therefore is a two-dimensional subspace of R3. In order to determine a basis for this subspace, we need only choose two linearly independent (i.e., 8 noncollinear) vectors that lie in the plane. A simple choice of vectors is v1 = (1, 0, 1) and v2 = (2, −1, 0). Thus, a basis for the subspace is {(1, 0, 1), (2, −1, 0)}.  Corollary 4.6.14 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.

Theorem 4.6.17 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 4.6.14, k ≤ n.Ifk = n, then S = V , so that any basis for S is a basis for V . Suppose now that k

Remark The process used in proving the previous theorem is referred to as extending a basis.

Example 4.6.18 Let S denote the subspace of M2(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 M2(R). Solution: We first express S in set notation as

T S ={A ∈ M2(R) : A = 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 10 01 00 = a + b + c , bc 00 10 01 it follows that       10 01 00 S = span , , . 00 10 01

8There are many others, of course.

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Furthermore, it is easily shown that the matrices in this spanning set are linearly inde- pendent. Consequently, a basis for S is       10 01 00 , , , 00 10 01

so that dim[S]=3. Since dim[M2(R)]=4, in order to extend the basis for S to a basis for M2(R), we need to add one additional matrix from M2(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   01 . 00 Adding this vector to the basis for S yields the linearly independent set         10 01 00 01 , , , . (4.6.5) 00 10 01 00

Since dim[M2(R)]=4, Theorem 4.6.10 implies that (4.6.5) is a basis for M2(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 infinite- dimensional vector space that is of primary importance in the theory of differential equations, Cn(I).

Example 4.6.19 Show that the vector space Cn(I) is an infinite-dimensional vector space. Solution: Consider the functions 1,x,x2,...,xk in Cn(I). Of course, each of these functions is in Ck(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 4.5.21. Since we can choose k arbitrarily, it follows that there are an arbitrarily large number of linearly independent vectors in Cn(I), hence Cn(I) is infinite-dimensional.  In this example we showed that Cn(I) is an infinite-dimensional vector space. Con- sequently, the use of our finite-dimensional vector space theory in the analysis of differ- ential 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 Cn(I), and therefore our basis results will be applicable to this solution set.

Exercises for 4.6 Key Terms • Be able to construct a basis for a given vector space V Basis, Standard basis, Infinite-dimensional, Finite- . dimensional, Dimension, Extension of a subspace basis. • Be able to extend a basis for a subspace of V to V Skills itself. n • Be able to determine whether a given set of vectors • Be familiar with the standard bases on R , Mm×n(R), forms a basis for a vector space V . and Pn.

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• Be able to give the dimension of a vector space V . 4. {(1, 1, −1, 2), (1, 0, 1, −1), (2, −1, 1, −1)}.

• Be able to draw conclusions about the properties of a 5. {(1, 1, 0, 2), (2, 1, 3, −1), (−1, 1, 1, −2), (2, −1, 1, 2)}. set of vectors in a vector space (i.e., spanning or linear independence) based solely on the size of the set. 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), • Understand the usefulness of Theorems 4.6.10 and (k, 0, 2, 1)} is a basis for R4. 4.6.12. 7. Determine a basis S for P3, and hence, prove that True-False Review dim[P3]=4. Be sure to prove that S is a basis. For Questions 1–11, decide if the given statement is true or 8. Determine a basis S for P3 whose elements all have false, and give a brief justification for your answer. If true, the same degree. Be sure to prove that S is a basis. you can quote a relevant definition or theorem from the text. If false, provide an example, illustration, or brief explanation For Problems 9–12, find the dimension of the null space of of why the statement is false. the given matrix A.   1. A basis for a vector space V is a set S of vectors that 13 9. A = . spans V . −2 −6   2. If V and W are vector spaces of dimensions n and m, 000 respectively, and if n>m, then W is a subspace of V . 10. A =  000. 3. A vector space V can have many different bases. 010   n 4. dim[Pn]=dim[R ]. 1 −14 11. A =  23−2 . 5. If V is an n-dimensional vector space, then any set S 12−2 of m vectors with m>nmust span V .   1 −123 6. Five vectors in P3 must be linearly dependent.    2 −134 12. A =  . 7. Two vectors in P3 must be linearly independent. 1011 3 −145 8. Ten vectors in M3(R) must be linearly dependent. 13. S R3 9. If V is an n-dimensional vector space, then every set Let be the subspace of that consists of all solu- x − y + z = S with fewer than n vectors can be extended to a basis tions to the equation 3 0. Determine a S [S] for V . basis for , and hence, find dim . R3 10. Every set of vectors that spans a finite-dimensional 14. Let S be the subspace of consisting of all vectors − − vector space V contains a subset which forms a basis of the form (r, r 2s,3s 5r), where r and s are for V . real numbers. Determine a basis for S, and hence, find dim[S]. 11. The set of all 3 × 3 upper triangular matrices forms a R × three-dimensional subspace of M3(R). 15. Let S be the subspace of M2( ) consisting of all 2 2 upper triangular matrices. Determine a basis for S, and Problems hence, find dim[S].

For Problems 1–5, determine whether the given set of vectors 16. Let S be the subspace of M2(R) consisting of all 2 ×2 is a basis for Rn. matrices with trace zero. Determine a basis for S, and hence, find dim[S]. 1. {(1, 1), (−1, 1)}. R3 2. {(1, 2, 1), (3, −1, 2), (1, 1, −1)}. 17. Let S be the subspace of spanned by the vectors v1 = (1, 0, 1), v2 = (0, 1, 1), v3 = (2, 0, 2). Deter- 3. {(1, −1, 1), (2, 5, −2), (3, 11, −5)}. mine a basis for S, and hence, find dim[S].

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18. Let S be the vector space consisting of the set of 26. Let all linear combinations of the functions f1(x) =     x −x −11 13 e ,f2(x) = e ,f3(x) = sinh(x). Determine a basis A = ,A= , 1 01 2 −10 for S, and hence, find dim[S].     10 0 −1 19. M (R) A = ,A= . Determine a basis for the subspace of 2 spanned 3 12 4 23 by         13 00 −14 5 −6 , , , . −12 00 11 −51 (a) Show that {A1,A2,A3,A4} is a basis for M2(R). [The hint on the previous problems applies again.] = = − 20. Let v1 (1, 1) and v2 ( 1, 1). (b) Express the vector (a) Show that {v , v } spans R2.   1 2 56 { } (b) Show that v1, v2 is linearly independent. 78 (c) Conclude from (a) or (b) that {v1, v2} is a ba- sis for R2. What theorem in this section allows as a linear combination of the basis in (a). you to draw this conclusion from either (a) or (b), 27. Let without proving both?   11−11 =  − −  21. Let v1 = (2, 1) and v2 = (3, 1). A 2 356 , 502−3 2 (a) Show that {v1, v2} spans R . and let v1 = (−2, 7, 5, 0) and v2 = (3, −8, 0, 5). (b) Show that {v1, v2} is linearly independent.

(c) Conclude from (a) or (b) that {v1, v2} is a ba- (a) Show that {v1, v2} is a basis for the null space of sis for R2. What theorem in this section allows A. you to draw this conclusion from either (a) or (b), (b) Using the basis in part (a), write an expression for without proving both? an arbitrary vector (x,y,z,w) in the null space

22. Let v1 = (0, 6, 3), v2 = (3, 0, 3), and v3 = of A. (6, −3, 0). Show that {v , v , v } is a basis for R3. 1 2 3 28. Let V = M (R) and let S be the subset of all vectors [Hint: You need not show that the set is both linearly 3 in V such that the sum of the entries in each row and independent and a spanning set for P . Use a theorem 2 in each column is zero. from this section to shorten your work.] 23. Determine all values of the constant α for which (a) Find a basis and the dimension of S. 2 2 {1 + αx , 1 + x + x , 2 + x} is a basis for P2. (b) Extend the basis in (a) to a basis for V . = + = − = 24. Let p1(x) 1 x,p2(x) x(x 1), p3(x) 29. Let V = M (R) and let S be the subset of all vectors + 2 { } 3 1 2x . Show that p1,p2,p3 is a basis for P2.[Hint: in V such that the sum of the entries in each row and You need not show that the set is both linearly inde- in each column is the same. pendent and a spanning set for P2. Use a theorem from this section to shorten your work.] (a) Find a basis and the dimension of S.

25. The Legendre of degree n, pn(x),isde- (b) Extend the basis in (a) to a basis for V . fined to be the polynomial solution of the differential R R equation For Problems 30–31, Symn( ) and Skewn( ) denote the vector spaces consisting of all real n × n matrices that are (1 − x2)y − 2xy + n(n + 1)y = 0, symmetric and skew-symmetric, respectively.

which has been normalized so that pn(1) = 1. The first 30. Find a basis for Sym2(R) and Skew2(R), and show three Legendre polynomials are p0(x) = 1,p1(x) = that 1 2 x, and p2(x) = (3x − 1). Show that {p0,p1,p2} 2 [ R ]+ [ R ]= [ R ] is a basis for P2. [The hint for the previous problem dim Sym2( ) dim Skew2( ) dim M2( ) . applies again.]

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31. Determine the dimensions of Symn(R) and 34. V = P2, S is the subspace consisting of all polynomi- 2 Skewn(R), and show that als of the form (2a1 +a2)x +(a1 +a2)x +(3a1 −a2).

dim[Symn(R)]+dim[Skewn(R)]=dim[Mn(R)]. n 35. Let S be a basis for Pn−1. Prove that S ∪{x } is a basis For Problems 32–34, a subspace S of a vector space V is for Pn. given. Determine a basis for S and extend your basis for S to obtain a basis for V . 36. Generalize the previous problem as follows. Let S be a 32. V = R3, S is the subspace consisting of all points basis for Pn−1, and let p be any polynomial of degree lying on the plane with Cartesian equation n. Prove that S ∪{p} is a basis for Pn. x + 4y − 3z = 0. 37. (a) What is the dimension of Cn as a real vector 33. V = M2(R), S is the subspace consisting of all ma- space? Determine a basis. trices of the form   n ab (b) What is the dimension of C as a complex vector . ba space? Determine a basis.

4.7 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 4.7.1 If V is a vector space with basis {v1, v2,...,vn}, then every vector v ∈ V can be written uniquely as a linear combination of v1, v2,...,vn.

Proof Since v1, v2,...,vn span V , every vector v ∈ V can be expressed as

v = a1v1 + a2v2 +···+anvn, (4.7.1)

for some scalars a1,a2,...,an. Suppose also that

v = b1v1 + b2v2 +···+bnvn, (4.7.2)

for some scalars b1,b2,...,bn. We will show that ai = bi for each i, which will prove the uniqueness assertion of this theorem. Subtracting Equation (4.7.2) from Equation (4.7.1) yields

(a1 − b1)v1 + (a2 − b2)v2 +···+(an − bn)vn = 0. (4.7.3)

But {v1, v2,...,vn} is linearly independent, and so Equation (4.7.3) implies that

a1 − b1 = 0,a2 − b2 = 0, ..., an − bn = 0.

That is, ai = bi for each i = 1, 2,...,n.

Remark The converse of Theorem 4.7.1 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 {v1, v2,...,vn}, then {v1, v2,...,vn} 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

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