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4.6 Bases and Dimension 281
40. (a) Show that { 1 , x, x^2 , x^3 } is linearly independent on every interval. (b) If fk (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 r^1 x^ , f 2 (x) = e r^2 x^ , f 3 (x) = e r^3 x have Wronskian
W f 1 , f 2 , f 3 = e (r^1 +r^2 +r^3 )x
r 1 r 2 r 3 r^21 r^22 r^23 = 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 r^1 x^ , e r^2 x^ ,... , e r^ n^ x^ } is linearly independent on every interval if and only if all of the ri 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 3.3.]
42. Let { v 1 , v 2 } be a linearly independent set in a vector space V , and let v = α v 1 + v 2 , w = v 1 + α v 2 , where α is a constant. Use Definition 4.5.4 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 vec- tors 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
u k =
∑^ m
i= 1
aik v i , k = 1 , 2 ,... , n,
for appropriate constants aik.
(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 = [aij ].
(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 {A v 1 , A v 2 ,... , A v 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 4.5.2. 49. Prove Proposition 4.5.7. 50. 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 lin- early dependent. 51. Prove that if V = Pn 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 descend- ing degree: deg(p 1 ) > deg(p 2 ) > · · · > deg(pk ). Assuming that c 1 p 1 + c 2 p 2 + · · · + ck pk = 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.]
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
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 4.6.
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 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 linear independence and linear span. 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 C n^ (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. We begin with the vector space Rn. In R^2 , the most natural basis, denoted { e 1 , e 2 }, consists of the two vectors
e 1 = ( 1 , 0 ), e 2 = ( 0 , 1 ), (4.6.1)
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 ). (4.6.2)
The verification that the sets (4.6.1) and (4.6.2) are indeed bases of R^2 and R^3 , respec- tively, 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 (4.6.2), 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 Rn^ defined by
e 1 = ( 1 , 0 ,... , 0 ), e 2 = ( 0 , 1 ,... , 0 ),... , e n = ( 0 , 0 ,... , 1 ).
These vectors are linearly independent by Corollary 4.5.15, since
det([ e 1 , e 2 ,... , e n]) = det(I (^) n ) = 1 = 0.
Furthermore, the vectors span Rn, since an arbitrary vector v = (x 1 , x 2 ,... , x (^) n ) in Rn 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 + · · · + xn 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 Rn.
284 CHAPTER 4 Vector Spaces
We have shown in Example 4.4.6 that {p 0 , p 1 , p 2 } is a spanning set for P 2. Furthermore,
W p 0 , p 1 , p 2 =
1 x x^2 0 1 2x 0 0 2
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, Pn, 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 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 { 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 > m. We wish to prove that { 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 aij such that u 1 = a 11 v 1 + a 21 v 2 + · · · + am 1 v m , u 2 = a 12 v 1 + a 22 v 2 + · · · + am 2 v m , .. . u n = a 1 n v 1 + a 2 n v 2 + · · · + amn 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 + · · · + cn u n = 0. (4.6.3)
Inserting the expressions for u 1 , u 2 ,... , u n into Equation (4.6.3) yields
c 1 (a 11 v 1 + a 21 v 2 + · · · + am 1 v m ) + c 2 (a 12 v 1 + a 22 v 2 + · · · + am 2 v m )
- · · · + cn (a 1 n v 1 + a 2 n v 2 + · · · + amn 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.
4.6 Bases and Dimension 285
Rearranging terms, we have
(a 11 c 1 + a 12 c 2 + · · · + a 1 n cn ) v 1 + (a 21 c 1 + a 22 c 2 + · · · + a 2 n cn ) v 2
- · · · + (am 1 c 1 + am 2 c 2 + · · · + amn cn ) v m = 0.
Since { v 1 , v 2 ,... , v m} is linearly independent, we can conclude that
a 11 c 1 + a 12 c 2 + · · · + a 1 n cn = 0 , a 21 c 1 + a 22 c 2 + · · · + a 2 n cn = 0 , .. . am 1 c 1 + am 2 c 2 + · · · + amn cn = 0.
This is an m × n homogeneous system of linear equations with m < n, and hence, from Corollary 2.5.11, it has nontrivial solutions for c 1 , c 2 ,... , c (^) n. It therefore follows from Equation (4.6.3) that { u 1 , u 2 ,... , u n} is linearly dependent.
Corollary 4.6.5 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 The- orem 4.6.4 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 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.
The dimension of a finite-dimensional vector space V , written dim[V ], is the number of vectors in any basis for V. If V 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.
Example 4.6.8 It follows from our examples earlier in this section that dim[R^3 ] = 3 , dim[M 2 (R)] = 4, and dim[P 2 ] = 3.
4.6 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. To do this, let v be an arbitrary vector in V. From Theorem 4.6.4, 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 + · · · + cn v n = 0. (4.6.4)
If c 0 = 0, then the linear independence of { v 1 , v 2 ,... , v n} and (4.6.4) would imply that c 0 = c 1 = · · · = cn = 0, a contradiction. Hence, c 0 = 0, and so, from Equation (4.6.4),
v = −
c 0
(c 1 v 1 + c 2 v 2 + · · · + cn 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 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
y (n)^ + a 1 (x)y (n−^1 )^ + · · · + 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) = c 1 y 1 (x) + c 2 y 2 (x) + · · · + cn yn (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 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:
Example 4.6.11 Verify that { 1 + x, 2 − 2 x + x^2 , 1 + x^2 } is a basis for P 2.
Solution: Since dim[P 2 ] = 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
p 1 (x) = 1 + x, p 2 (x) = 2 − 2 x + x^2 , p 3 (x) = 1 + x^2
have Wronskian
W p 1 , p 2 , p 3 =
1 + x 2 − 2 x + x^2 1 + x^2 1 − 2 + 2 x 2 x 0 2 2
Since the Wronskian is nonzero, the given set of vectors is linearly independent on any interval. Consequently, { 1 + x, 2 − 2 x + x^2 , 1 + x^2 } is indeed a basis for P 2.
288 CHAPTER 4 Vector Spaces
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 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 Corol- lary 4.5.12, 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 4.6.10 and 4.6.12 together, the following result is immediate.
Corollary 4.6.13 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 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 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.
290 CHAPTER 4 Vector Spaces
Furthermore, it is easily shown that the matrices in this spanning set are linearly inde- pendent. Consequently, a basis for S is {[ 1 0 0 0
]
[
]
[
]}
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 [ 0 1 0 0
]
Adding this vector to the basis for S yields the linearly independent set {[ 1 0 0 0
]
[
]
[
]
[
]}
Since dim[M 2 (R)] = 4, Theorem 4.6.10 implies that (4.6.5) 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 infinite- dimensional vector space that is of primary importance in the theory of differential equations, C n^ (I ).
Example 4.6.19 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 4.5.21. 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. 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 C n^ (I ), and therefore our basis results will be applicable to this solution set.
Exercises for 4.
Key Terms
Basis, Standard basis, Infinite-dimensional, Finite- dimensional, 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 Rn, Mm×n (R), and Pn.
4.6 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 (i.e., spanning or linear independence) based solely on the size of the set.
- Understand the usefulness of Theorems 4.6.10 and 4.6.12.
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[Pn] = dim[Rn]. 5. If V is an n-dimensional vector space, then any set S of m vectors with m > n must 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 Rn.
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 )}.
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.
9. A =
[
]
10. A =
11. A =
12. A =
13. Let S be the subspace of R^3 that consists of all solu- tions to the equation x − 3 y + 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 − 2 s, 3 s − 5 r), 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 ). Deter- mine a basis for S, and hence, find dim[S].
4.7 Change of Basis 293
31. Determine the dimensions of Symn (R) and Skewn (R), and show that dim[Symn (R)] + dim[Skew (^) n (R)] = dim[Mn (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 + 4 y − 3 z = 0. 33. V = M 2 (R), S is the subspace consisting of all ma- trices of the form (^) [ a b b a
]
34. V = P 2 , S is the subspace consisting of all polynomi- als of the form ( 2 a 1 +a 2 )x^2 +(a 1 +a 2 )x +( 3 a 1 −a 2 ). 35. Let S be a basis for Pn− 1. Prove that S ∪ {x n} is a basis for Pn. 36. Generalize the previous problem as follows. Let S be a basis for Pn− 1 , and let p be any polynomial of degree n. Prove that S ∪ {p} is a basis for Pn. 37. (a) What is the dimension of Cn^ as a real vector space? Determine a basis.
(b) What is the dimension of Cn^ as a complex vector 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 { 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
v = a 1 v 1 + a 2 v 2 + · · · + an v n , (4.7.1)
for some scalars a 1 , a 2 ,... , a (^) n. Suppose also that
v = b 1 v 1 + b 2 v 2 + · · · + bn v n , (4.7.2)
for some scalars b 1 , b 2 ,... , b (^) n. 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
(a 1 − b 1 ) v 1 + (a 2 − b 2 ) v 2 + · · · + (an − bn ) v n = 0. (4.7.3)
But { v 1 , v 2 ,... , v n} is linearly independent, and so Equation (4.7.3) implies that
a 1 − b 1 = 0 , a 2 − b 2 = 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 { 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
