Use Exercise 7.1.32(c) to find the dual basis for: (a) 𝐯1=( 1 1 ), 𝐯2=( 1 -1

Use Exercise 7.1.32(c) to find the dual basis for: (a) 𝐯1=( ...

Key Concepts

Dual Basis

The dual basis is a set of linear functionals associated with a given basis of a finite-dimensional vector space. Each element of the dual basis acts on the vectors of the original basis in such a way that it ‘picks out’ the corresponding coefficient in any vector’s unique linear combination of the basis elements, satisfying the condition f_i(v_j) = ?_ij, where ?_ij is the Kronecker delta.

Basis

A basis for a vector space is a set of linearly independent vectors that span the entire space, meaning any vector in the space can be expressed uniquely as a linear combination of the basis vectors. The concept of a basis is fundamental in linear algebra because it provides a framework for coordinate representation and transformations within the space.

Linear Functionals

A linear functional is a linear map from a vector space to its underlying field (such as the real or complex numbers). In the context of the dual basis, each linear functional is defined by its action on the basis vectors, and the collection of these functionals forms the dual space, which captures many important properties of the original space.

Kronecker Delta

The Kronecker delta is a function defined by ?_ij, which equals 1 if i = j and 0 otherwise. It is used in the definition of the dual basis, ensuring that each dual basis functional returns 1 when applied to its corresponding basis vector and 0 when applied to other basis vectors, thereby providing a precise indicator of component selection.

Matrix Inversion

Matrix inversion is a technique often employed to compute the dual basis of a given set of vectors. By representing the basis vectors as columns in a matrix, one can use the matrix inverse to derive the linear functionals of the dual basis. This approach relies on the fact that the matrix formed is invertible when the basis vectors are linearly independent.

Transcript

00:02 All right, in this problem, we are asked to find the basis dual to the basis that we're given here.

00:11 So we're given the basis v1, v2, and v3.

00:15 And this is a basis in r3, of course.

00:18 And we're supposed to find the dual basis, phi 1, phi 2, and phi 3.

00:23 So the main thing to keep in mind with all of these, the definition of the dual basis, is that each of the fies, say, phi, sub i, apply.

00:35 To the basis vector, say v subj, will be one if i and j are the same and zero otherwise.

00:43 So we can write that as the chronic or delta ij.

00:48 So this is the one relation that drives everything in this problem.

00:53 So let's start by figuring out what phi sub 1 is.

00:58 And we should give it a name in terms of its coefficients.

01:02 So let's just say that five sub 1 applied to an arbitrary.

01:08 Vector xy z and we'll just use a's for this one so in other words a1x plus a 2y plus a 3 z and all we have to do to determine phi 1 is figure out what are the the a's and to do that we'll just set up three equations so again coming back to this relation we know that 5 sub 1 applied to v1 which is 1 minus minus 1, 3.

01:45 Again, i'll just remind you why i'm doing this.

01:47 This is v1.

01:50 Well, if we expand this out, this is a1 times 1, plus a2 times negative 1, in other words, minus a2, plus a3 times 3.

02:07 So plus 3a3.

02:08 I'm just using the red line above.

02:12 And you can buy this relation up here in the little cloud.

02:16 This must come out to 1, since.

02:19 We're applying phi 1 to v1 the ones match.

02:24 Okay and then likewise we apply phi 1 to v2 and we'll just copy v2 from up here.

02:35 0 1 minus 1 well a 1 times 0 is 0.

02:43 A2 times 1 is 0.

02:44 A2 times 1 is just a 2 and a 3 times negative 1 is minus a 3 and this time this must come out to 0 since we're applying phi 1 to v2.

02:55 1 and 2 aren't the same.

02:57 And finally, same idea.

03:01 Well, this input v3 here, which we're told is 0, 3 minus 2.

03:09 Go ahead and expand this all out.

03:11 So again, 0 times a1 is 0.

03:13 3 times a 2 gives us 3a2, and negative 2 times a 3 .3 minus 2 a 3a3, and negative 2 times a 3 minus 2 a 3.

03:24 And again, this must come out to zero since it's 5 -1 of v3.

03:30 Okay, so at this point, we have a system of three equations in three variables.

03:38 And if we solve for those variables, a1, a2, a3, will be done.

03:42 We'll have found phi -1.

03:44 So you can solve this whatever your favorite method is to solve linear equations.

03:51 One way, of course, is just by, let's say, using the second line here to see that a2, must equal a3 and then we can substitute maybe in the third equation if a2 equals a3 this is like saying that 3a2 minus 2 a2 in other words just plain a 2 must be 0 okay but then a 3 must be 0 because a 2 equals a 3 and finally we can go up to the top equation and knowing that a 2 and a 3 are 0 we get a 1 equals 1 so to write this all down, since we know the a's, we can now write down, i'll just come over here, we can now write down this linear functional phi -1.

04:56 It must be defined as follows.

05:00 Phi -1 of x, y, must equal, while a -1 was 1.

05:04 So 1 times x, a2 was 0, so plus 0 -y, and a -3 is 0, so plus 0 -z.

05:12 So all we have to really write is this.

05:16 And that's the first element of our dual basis that we're looking for.

05:23 So the idea is exactly the same for the other two.

05:25 I'll go a little faster for those two, but we'll still do all the details for you.

05:32 Let me just zoom out a little bit.

05:34 So we'll do the exact same thing.

05:38 And actually, to save a little time, here's what we'll do.

05:44 Notice we're trying to find phi ii this time.

05:47 So i'll erase all the extraneous solutions.