6 Basis. 6.1 Introduction

Transcription

1 6 Basis 6 Introduction If x, e, and e 2 are as pictured, then using the geometrical rules for scaling and adding vectors we see that x = 7e +4e 2 We say that x has e -coordinate 7 and e 2 -coordinate 4 Similarly, we can find the coordinates of any vector with its tail at the origin by writing that vector as a linear combination of e and e 2 and using the scalar factors The vectors e and e 2 set up a coordinate system in the plane (It is this coordinate system we use when we write x = [7,4] T ) Other vectors in the plane can be used to set up a coordinate system as well For instance, let [ [ 3 2 = Then x = 2b +b 2, so x has b -coordinate 2 and b 2 -coordinate The vectors b and b 2 also set up a coordinate system in the plane

2 6 BASIS 2 There are restrictions on which vectors can set up a coordinate system: (i) The singlevectorb cannotsetup acoordinatesystem, since, forinstance, x cannot be written as a linear combination (= multiple) of b, so there is no way to get the coordinates of x (ii) The three vectors b = [ [ [ 3 2, b 2 =, b 3 = 0] cannotsetupacoordinatesystem, since, forinstancethevectorx = [7,4] T can be written as a linear combination of these vectors in two different ways, x = 2b +3b 2 +5b 3 and x = 6b +( )b 2 +5b 3, and we cannot tell whether the coordinates of x should be [ 2,3,5] T or [6,,5] T The problem in (i) is that the span of {b } is not all of R 2 The problem in (ii) is that the vectorsb, b 2, and b 3 are not linearly independent It turns out that if a collection of vectors spans the space we are trying to set up a coordinate system in and the vectors are linearly independent, then they can be used to set up a coordinate system Such a collection is called a basis for the space 62 Definition and examples Basis of subspace Let S be a subspace of R n and let b,b 2,,b s be vectors in S The set {b,b 2,,b s } is a basis for S if (i) Span{b,b 2,,b s } = S (ii) b,b 2,,b s are linearly independent 62 Example (Standard basis) Show that {e,e 2 } is a basis for R 2, where e = [ 0], e 2 = [ 0

3 6 BASIS 3 Solution Because of the wording is a basis for R 2, the role of S in the definition of basis is being played by R 2 We check the two properties: (i) (Span{e,e 2 } = R 2?) The way to show that two sets are equal is to show that each is a subset of the other Since linear combinations of vectors in R 2 are still vectors in R 2 we have Span{e,e 2 } R 2 For the other inclusion, we let x = [x,x 2 ] T be an arbitrary vector in R 2 (Must show that x is in Span{e,e 2 }) We need to show that x = α e + α 2 e 2 for some scalars α and α 2, that is, [ ] [ ] [ ] x 0 = α x +α We can let α = x and α 2 = x 2 (We used inspection here, but usually one has to solve a system) Therefore, R 2 Span{e,e 2 } and, since we showed the other inclusion earlier, we have Span{e,e 2 } = R 2 (ii) (e,e 2 linearlyindependent?) Neithervectorisalinearcombination(here, multiple) of the other, so e and e 2 are linearly independent Therefore, {e,e 2 } is a basis for R 2 More generally, the set of standard unit vectors {e,e 2,,e n } is a basis for R n, called the standard basis 622 Example Show that {b,b 2 } is a basis for R 2, where [ [ 3 2 = Solution We check the two properties (i) (Span{b,b 2 } = R 2?) Every linear combination of vectors in R 2 is a vector in R 2 so Span{b,b 2 } R 2 Let x = [x,x 2 ] T be an arbitrary vector in R 2 (Must show that x is in Span{b,b 2 }) We need to show that x = α b +α 2 b 2 for some scalars α and α 2, that is, [ x x 2 ] = α [ 3 ] [ ] +α 2 = 2 [ ] 3α +α 2 α +2α 2 This leads to a system with corresponding augmented matrix [ [ ] 3 x 3 x 2 x 2 ] x 3x 2 Since there is no pivot in the augmented column, a solution exists Therefore, R 2 Span{b,b 2 }, and, since we showed the other inclusion earlier, we have Span{b,b 2 } = R 2

4 6 BASIS 4 (ii) (b,b 2 linearly independent?) Neither vector is a linear combination (here, multiple) of the other, so b and b 2 are linearly independent Therefore, {b,b 2 } is a basis for R Example Show that {b,b 2 } is a basis for S = Span{b,b 2 }, where b =, b 2 = 0 0 Solution We check the two properties of basis: (i) (Span{b,b 2 } = S?) Since S is defined to be Span{b,b 2 } this property is automatically satisfied (ii) (b,b 2 linearly independent?) Neither vector is a linear combination (here, multiple) of the other, so b and b 2 are linearly independent Therefore, {b,b 2 } is a basis for S 624 Example Determine whether {e,e 2 } is a basis for R 3 (where e = [,0,0] T, e 2 = [0,,0] T ) Solution Span{e,e 2 }consistsofallvectorsoftheformα e +α 2 e 2 = [α,α 2,0] T (the third component must be zero) Since [0,0, T is a vector in R 3 that is not in Span{e,e 2 } we see that Span{e,e 2 } R 3 Therefore, {e,e 2 } is not a basis for R 3 (This counterexample was obtained by inspection Here is some possible scratch work Let x = [x,x 2,x 3 ] T be an arbitrary vector in R 3 Trying to write x as a linear combination of e and e 2 leads to a system with augmented matrix 0 x 0 x x 3 We see that there is no solution if x 3 0 This gives the idea for the counterexample above) 625 Example Determinewhether{b,b 2,b 3 }isabasisfors = Span{b,b 2,b 3 }, where b = 2, b 2 = 0 3, b 3 =

5 6 BASIS 5 Solution By inspection, the second vectoris a multiple ofthe third (b 2 = 3b 3 ), so the vectors are not linearly independent Therefore, the set is not a basis for S (If inspection such as this had been difficult, we would instead have applied row operations to the matrix having the vectors as columns, namely, [ b b 2 b 3 ] We would have found that this matrix does not have full rank, so the vectors are not linearly independent by the theorem before Example 523) 63 Coordinate vector Let b and b 2 be the vectors in R 2 given by [ [ 3 2 = Since {b,b 2 } is a basis for R 2 (see Example 622), any vector in R 2 can be written uniquely as a linear combination of these two vectors (a vector can be written at least one way as a linear combination due to the spanning property of basis and at most one way due to the linear independence property of basis) For instance, if x = [7,4] T, then x = 2b +b 2 We say that the coordinate vector of x relative to the ordered basis B = (b,b 2 ) is [2, T In symbols, [ 2 [x] B = This definition requires a specific ordering of the basis since, if the order is not fixed, then someone else might write x = b 2 +2b and say that the coordinate vector of x is [, T Here is the general definition: Coordinate vector Let S be a subspace of R n, let B = (b,b 2,,b s ) be an ordered basis for S, and let x be a vector in S The coordinate vector of x relative to B is [x] B = [α,α 2,,α s ] T, where x = α b +α 2 b 2 + +α s b s

6 6 BASIS 6 Put more simply, the coordinate vector of x relative to B is found by writing x as a linear combination of the (ordered) basis vectors, pulling off the scalar factors, and writing those factors as a column matrix 63 Example Let B = (b,b 2 ) be the ordered basis of R 2 with [ [ 3 2 = (a) Find the coordinate vector [x] B, where x = [9,8] T (b) Sketch the vectors b, b 2, and x, as well as the coordinate system grid determined by B Solution (a) We need to write x as a linear combination of b and b 2 : x = α b +α 2 b 2 [ [ ] [ ] 9 3 = α 8] +α 2 = 2 [ ] 3α +α 2 α +2α 2 Equating components leads to a system with augmented matrix [ ] [ ] [ ] [ ] [ ] Therefore, x = 2b +3b 2 so that [x] B = [2,3] T (b) Here is the sketch:

7 6 BASIS 7 The ordered basis B sets up a new coordinate system (with grid in green) Inthiscoordinatesystemthevectorxhasb -coordinate2andb 2 -coordinate Example Let B = (b,b 2 ) be the ordered basis of Span{b,b 2 } with b =, b 2 = 0 0 (a) Find the coordinate vector [x] B, where x = [,4,3] T (b) Find y, given that [y] B = [9, T Solution (a) We need to write x as a linear combination of b and b 2 : x = α b +α 2 b 2 0 α 4 = α +α 2 = α +α Equating components shows immediately that α = and α 2 = 3 (In this case, there is no need to use the augmented matrix of the system) Therefore, x = b +3b 2 so that [x] B = [,3] T α 2

8 6 BASIS 8 (b) We are given that [y] B = [9, T Therefore, y = 9b +( 2)b 2 = 9 +( 2) 0 = Example Let E = (e,e 2,e 3 ) be the standard ordered basis of R 3 Find the coordinate vector [x] E, where x = [4,3, 5] T Solution Using inspection, we have x = 4 3 = ( 5) 0 0 = 4e +3e 2 +( 5)e Therefore, [x] E = [4,3, 5] T The example illustrates the fact that if E is the standard basis for R n, then for every vector x in R n [x] E = x Put another way, when we specify a vector by giving a list of numbers, like x = [4,3, 5] T, we are really giving the vector s coordinate vector relative to the standard basis 6 Exercises 6 Show that {b,b 2 } is a basis for R 2, where [ ] [ ] = 2

9 6 BASIS Show that {b,b 2,b 3 } is a basis for R 3, where b = 0 2, b 2 = 4, b 3 = Determine whether {b,b 2 } is a basis for S = Span{b,b 2 }, where b = 2 6, b 2 = Determine whether {b,b 2 } is a basis for S = {x R 3 3x +x 2 +4x 3 = 0}, where b = 7, b 2 = Let B = (b,b 2 ) be the ordered basis of R 2 with [ [ ] = 2 (a) Find the coordinate vector [x] B, where x = [,5] T (b) Sketch the vectors b, b 2, and x, as well as the coordinate system grid determined by B 6 6 Let B = (b,b 2 ) be the ordered basis of Span{b,b 2 } with b = 2, b 2 = 3 (a) Find the coordinate vector [x] B, where x = [,7,7] T (b) Find y, given that [y] B = [4, 6] T