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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
294 CHAPTER 4 Vector Spaces
an important consideration. By an ordered basis for a vector space, we mean a basis in which we are keeping track of the order in which the basis vectors are listed.
DEFINITION 4.7.
If B = { v 1 , v 2 ,... , v n} is an ordered basis for V and v is a vector in V , then the scalars c 1 , c 2 ,... , c (^) n in the unique n-tuple (c 1 , c 2 ,... , c (^) n ) such that
v = c 1 v 1 + c 2 v 2 + · · · + cn v n
are called the components of v relative to the ordered basis B = { v 1 , v 2 ,... , v n}. We denote the column vector consisting of the components of v relative to the ordered basis B by [ v ]B , and we call [ v ]B the component vector of v relative to B.
Example 4.7.3 Determine the components of the vector v = ( 1 , 7 ) relative to the ordered basis B = {( 1 , 2 ), ( 3 , 1 )}. Solution: If we let v 1 = ( 1 , 2 ) and v 2 = ( 3 , 1 ), then since these vectors are not collinear, B = { v 1 , v 2 } is a basis for R^2. We must determine constants c 1 , c 2 such that
c 1 v 1 + c 2 v 2 = v.
We write c 1 ( 1 , 2 ) + c 2 ( 3 , 1 ) = ( 1 , 7 ). This requires that c 1 + 3 c 2 = 1 and 2 c 1 + c 2 = 7. The solution to this system is ( 4 , − 1 ), which gives the components of v relative to the ordered basis B = { v 1 , v 2 }. (See Figure 4.7.1.) Thus,
v = 4 v 1 − v 2.
Therefore, we have
[ v ]B =
[
]
x
y (4, 8) (1, 7) (^4) v 1
v 1 v 2
v 2
v = 4 v 1 v 2
(1, 2) (3, 1)
Figure 4.7.1: The components of the vector v = ( 1 , 7 ) relative to the basis {( 1 , 2 ), ( 3 , 1 )}.
Remark In the preceding example, the component vector of v = ( 1 , 7 ) relative to the ordered basis B′^ = {( 3 , 1 ), ( 1 , 2 )} is
[ v ]B′ =
[
]
Thus, even though the bases B and B′^ contain the same vectors, the fact that the vectors are listed in different order affects the components of the vectors in the vector space.
Example 4.7.4 In P 2 , determine the component vector of p(x) = 5 + 7 x − 3 x^2 relative to the following:
(a) The standard (ordered) basis B = { 1 , x, x^2 }.
(b) The ordered basis C = { 1 + x, 2 + 3 x, 5 + x + x^2 }.
296 CHAPTER 4 Vector Spaces
components of a vector relative to one basis to components relative to another basis. The tool we need in order to do this efficiently is the change-of-basis matrix. Before we describe this matrix, we pause to record the linearity properties satisfied by the components of a vector. These properties will facilitate the discussion that follows.
Lemma 4.7.5 Let V be a vector space with ordered basis B = { v 1 , v 2 ,... , v n}, let x and y be vectors in V , and let c be a scalar. Then we have
(a) [ x + y ]B = [ x ]B + [ y ]B.
(b) [c x ]B = c[ x ]B.
Proof Write
x = a 1 v 1 + a 2 v 2 + · · · + an v n and y = b 1 v 1 + b 2 v 2 + · · · + bn v n ,
so that x + y = (a 1 + b 1 ) v 1 + (a 2 + b 2 ) v 2 + · · · + (an + bn ) v n.
Hence,
[ x + y ]B =
a 1 + b 1 a 2 + b 2 .. . an + bn
a 1 a 2 .. . an
b 1 b 2 .. . bn
= [ x ]B + [ y ]B ,
which establishes (a). The proof of (b) is left as an exercise (Problem 37).
DEFINITION 4.7.
Let V be an n-dimensional vector space with ordered bases B and C given in (4.7.4). We define the change-of-basis matrix from B to C by
PC←B =
[
[ v 1 ]C , [ v 2 ]C ,... , [ v n]C
]
In words, we determine the components of each vector in the “old basis” B with respect the “new basis” C and write the component vectors in the columns of the change-of-basis matrix.
Remark Of course, there is also a change-of-basis matrix from C to B, given by
PB←C =
[
[ w 1 ]B , [ w 2 ]B ,... , [ w n]B
]
We will see shortly that the matrices PB←C and PC←B are intimately related. Our first order of business at this point is to see why the matrix in (4.7.5) converts the components of a vector relative to B into components relative to C. Let v be a vector in V and write v = a 1 v 1 + a 2 v 2 + · · · + an v n.
4.7 Change of Basis 297
Then
[ v ]B =
a 1 a 2 .. . an
Hence, using Theorem 2.2.9 and Lemma 4.7.5, we have
PC←B [ v ]B = a 1 [ v 1 ]C +a 2 [ v 2 ]C +· · ·+an[ v n]C = [a 1 v 1 +a 2 v 2 +· · ·+an v n]C = [ v ]C.
This calculation shows that premultiplying the component vector of v relative to B by the change of basis matrix PC←B yields the component vector of v relative to C:
[ v ]C = PC←B [ v ]B. (4.7.6)
Example 4.7.7 Let V = R^2 , B = {( 1 , 2 ), ( 3 , 4 )}, C = {( 7 , 3 ), ( 4 , 2 )}, and v = ( 1 , 0 ). It is routine to verify that B and C are bases for V.
(a) Determine [ v ]B and [ v ]C.
(b) Find PC←B and PB←C.
(c) Use (4.7.6) to compute [ v ]C , and compare your answer with (a).
Solution:
(a) Solving ( 1 , 0 ) = a 1 ( 1 , 2 ) + a 2 ( 3 , 4 ), we find a 1 = −2 and a 2 = 1. Hence,
[ v ]B =
[
]
Likewise, setting ( 1 , 0 ) = b 1 ( 7 , 3 ) + b 2 ( 4 , 2 ), we find b 1 = 1 and b 2 = − 1 .5. Hence, [ v ]C =
[
]
(b) A short calculation shows that
[( 1 , 2 )]C =
[
]
and [( 3 , 4 )]C =
[
]
Thus, we have PC←B =
[
]
Likewise, another short calculation shows that
[( 7 , 3 )]B =
[
]
and [( 4 , 2 )]B =
[
]
Hence, PB←C =
[
]
4.7 Change of Basis 299
In much the same way that we showed above that the matrices PC←B and PB←C are inverses of one another, we can make the following observation.
Theorem 4.7.9 Let V be a vector space with ordered bases A, B, and C. Then
PC←A = PC←B PB←A. (4.7.8)
Proof Using (4.7.6), for every v ∈ V , we have
PC←B PB←A[ v ]A = PC←B [ v ]B = [ v ]C = PC←A[ v ]A ,
so that premultiplication of [ v ]A by either matrix in (4.7.8) yields the same result. Hence, the matrices on either side of (4.7.8) are the same. We conclude this section by using Theorem 4.7.9 to show how an arbitrary change- of-basis matrix PC←B in Rn^ can be expressed as a product of change-of-basis matrices involving the standard basis E = { e 1 , e 2 ,... , e n} of Rn. Let B = { v 1 , v 2 ,... , v n} and C = { w 1 , w 2 ,... , w n} be arbitrary ordered bases for Rn. Since [ v ]E = v for all column vectors v in Rn, the matrices
PE←B = [[ v 1 ]E , [ v 2 ]E ,... , [ v n]E ] = [ v 1 , v 2 ,... , v n]
and PE←C = [[ w 1 ]E , [ w 2 ]E ,... , [ w n]E ] = [ w 1 , w 2 ,... , w n] can be written down immediately. Using these matrices, together with Theorem 4.7.9, we can compute the arbitrary change-of-basis matrix PC←B with ease:
PC←B = PC←E PE←B = (PE←C )−^1 PE←B.
Exercises for 4.
Key Terms
Ordered basis, Components of a vector relative to an ordered basis, Change-of-basis matrix.
Skills
- Be able to find the components of a vector relative to a given ordered basis for a vector space V.
- Be able to compute the change-of-basis matrix for a vector space V from one ordered basis B to another ordered basis C.
- Be able to use the change-of-basis matrix from B to C to determine the components of a vector relative to C from the components of the vector relative to B.
- Be familiar with the relationship between the two change-of-basis matrices PC←B and PB←C.
True-False Review
For Questions 1–8, 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. Every vector in a finite-dimensional vector space V can be expressed uniquely as a linear combination of vectors comprising a basis for V. 2. The change-of-basis matrix PB←C acts on the com- ponent vector of a vector v relative to the basis C and produces the component vector of v relative to the ba- sis B. 3. A change-of-basis matrix is always a square matrix. 4. A change-of-basis matrix is always invertible.
300 CHAPTER 4 Vector Spaces
5. For any vectors v and w in a finite-dimensional vector space V with basis B, we have [ v − w ]B = [ v ]B −[ w ]B. 6. If the bases B and C for a vector space V contain the same set of vectors, then [ v ]B = [ v ]C for every vector v in V. 7. If B and C are bases for a finite-dimensional vector space V , and v and w are in V such that [ v ]B = [ w ]C , then v = w. 8. The matrix PB←B is the identity matrix for any basis B for V.
Problems
For Problems 1–13, determine the component vector of the given vector in the vector space V relative to the given or- dered basis B.
1. V = R^2 ; B = {( 2 , − 2 ), ( 1 , 4 )}; v = ( 5 , − 10 ). 2. V = R^2 ; B = {(− 1 , 3 ), ( 3 , 2 )}; v = ( 8 , − 2 ). 3. V = R^3 ; B = {( 1 , 0 , 1 ), ( 1 , 1 , − 1 ), ( 2 , 0 , 1 )}; v = (− 9 , 1 , − 8 ). 4. V = R^3 ; B = {( 1 , − 6 , 3 ), ( 0 , 5 , − 1 ), ( 3 , − 1 , − 1 )}; v = ( 1 , 7 , 7 ). 5. V = R^3 ; B = {( 3 , − 1 , − 1 ), ( 1 , − 6 , 3 ), ( 0 , 5 , − 1 )}; v = ( 1 , 7 , 7 ). 6. V = R^3 ; B = {(− 1 , 0 , 0 ), ( 0 , 0 , − 3 ), ( 0 , − 2 , 0 )}; v = ( 5 , 5 , 5 ). 7. V = P 2 ; B = {x^2 + x, 2 + 2 x, 1 }; p(x) = − 4 x^2 + 2 x + 6. 8. V = P 2 ; B = { 5 − 3 x, 1 , 1 + 2 x^2 }; p(x) = 15 − 18 x − 30 x^2. 9. V = P 3 ; B = { 1 , 1 + x, 1 + x + x^2 , 1 + x + x^2 + x^3 }; p(x) = 4 − x + x^2 − 2 x^3. 10. V = P 3 ; B = {x^3 + x^2 , x^3 − 1 , x^3 + 1 , x^3 + x}; p(x) = 8 + x + 6 x^2 + 9 x^3. 11. V = M 2 (R); B =
{[
]
[
]
[
]
[
]}
A =
[
]
12. V = M 2 (R);
B =
{[
]
[
]
[
]
[
]}
A =
[
]
13. V = M 2 (R);
B =
{[
]
[
]
[
]
[
]}
A =
[
]
14. Let v 1 = ( 0 , 6 , 3 ), v 2 = ( 3 , 0 , 3 ), and v 3 = ( 6 , − 3 , 0 ). Determine the component vector of an ar- bitrary vector v = (x, y, z) relative to the ordered basis { v 1 , v 2 , v 3 }. 15. Let p 1 (x) = 1 + x, p 2 (x) = x(x − 1 ), and p 3 (x) = 1 + 2 x^2. Determine the component vector of an arbi- trary polynomial p(x) = a 0 + a 1 x + a 2 x^2 relative to the ordered basis {p 1 , p 2 , p 3 }.
For Problems 16–25, find the change-of-basis matrix PC←B from the given ordered basis B to the given ordered basis C of the vector space V.
16. V = R^2 ; B = {( 9 , 2 ), ( 4 , − 3 )}; C = {( 2 , 1 ), (− 3 , 1 )}. 17. V = R^2 ; B = {(− 5 , − 3 ), ( 4 , 28 )}; C = {( 6 , 2 ), ( 1 , − 1 )}. 18. V = R^3 ; B = {( 2 , − 5 , 0 ), ( 3 , 0 , 5 ), ( 8 , − 2 , − 9 )}; C = {( 1 , − 1 , 1 ), ( 2 , 0 , 1 ), ( 0 , 1 , 3 )}. 19. V = R^3 ; B = {(− 7 , 4 , 4 ), ( 4 , 2 , − 1 ), (− 7 , 5 , 0 )}; C = {( 1 , 1 , 0 ), ( 0 , 1 , 1 ), ( 3 , − 1 , − 1 )}. 20. V = P 1 ; B = { 7 − 4 x, 5 x}; C = { 1 − 2 x, 2 + x}. 21. V = P 2 ; B = {− 4 +x − 6 x^2 , 6 + 2 x^2 , − 6 − 2 x + 4 x^2 }; C = { 1 − x + 3 x^2 , 2 , 3 + x^2 }. 22. V = P 3 ; B = {− 2 + 3 x+ 4 x^2 −x^3 , 3 x+ 5 x^2 + 2 x^3 , − 5 x^2 − 5 x^3 , 4 + 4 x + 4 x^2 }; C = { 1 − x^3 , 1 + x, x + x^2 , x^2 + x^3 }. 23. V = P 2 ; B = { 2 +x^2 , − 1 − 6 x + 8 x^2 , − 7 − 3 x − 9 x^2 }; C = { 1 + x, −x + x^2 , 1 + 2 x^2 }. 24. V = M 2 (R); B =
{[
]
[
]
[
]
[
]}
C =
{[
]
[
]
[
]
[
]}
