Suppose v1, …, vn is a basis of V and φ1, …, φn is the corresponding dual basis of V^' . Suppose

Suppose v1, …, vn is a basis of V and φ1, …, φn is the...

Key Concepts

Linearity of Functionals

Linearity is a critical property of functionals in the dual space, meaning that they preserve the operations of addition and scalar multiplication. This implies that the value of a functional on any vector (expressed as a linear combination of basis vectors) can be computed as the corresponding linear combination of its values on the basis vectors.

Basis of a Vector Space

A basis of a vector space is a set of vectors that is both linearly independent and spans the entire space. This means that every vector in the vector space can be uniquely expressed as a linear combination of the basis vectors, which provides a framework for coordinates and representations in the space.

Dual Space

The dual space of a vector space V, often denoted by V?, is the set of all linear functionals (linear maps from V to its field of scalars). It is a vector space in its own right and plays an important role in various areas of linear algebra and functional analysis.

Dual Basis

Given a basis in a finite-dimensional vector space, the corresponding dual basis in the dual space is a collection of linear functionals that are defined by their action on the basis vectors: each dual basis element evaluates to 1 on its corresponding basis vector and 0 on all others. This property allows for the unique representation of any linear functional in terms of the dual basis.

Representation of Linear Functionals

The representation of linear functionals in terms of the dual basis is based on the idea that any linear map is completely determined by its values on a basis of the vector space. Thus, any functional ? in the dual space can be uniquely written as a linear combination of the dual basis elements weighted by the functional's evaluations on the original basis. This representation simplifies the study and manipulation of functionals in finite-dimensional spaces.

Transcript

00:01 So we have a basis here of v over k, and we define some linear functionals on k, that is elements of v star.

00:09 I'm sorry, linear functionals on v, not on k.

00:11 Defined basically to be a form of the kronecker delta function, just where phi i of vj is 0 for i not equal to j, but it's 1 for i equal to j.

00:21 They're basically indicator functions.

00:25 So, let's define a big phi in v star to be some linear combination, alpha 1 times phi 1 plus alpha 2 times phi 2, plus etc.

00:39 Alpha n times phi n.

00:41 And we'll try and want to assume that this is equal to 0.

00:48 Alright, let's see if we can do that.

00:50 Well, i'm going to note that phi of v1 is equal to just alpha 1, right? because when you apply v1 to alpha...

01:01 When you apply phi to alpha 1, this guy is going to become 1, this is going to become 0, this is going to become 0, all of them are going to be 0 except for this 1.

01:12 So this is alpha 1.

01:13 We've also assumed that it's equal to 0 on the entire vector space.

01:19 Remember, this 0 is the 0 linear functional, not the 0 scalar.

01:25 So phi of v1 is equal to 0, therefore alpha 1 is equal to 0.

01:29 Likewise, phi of v2 is just going to be equal to alpha 2.

01:32 All of the other terms cancel out, which also must be 0.

01:36 And indeed, in general, phi of vi is equal to alpha i.

01:41 Again, these v's are our basis, and that's 0.

01:45 So if phi is going to be 0 in particular just on all of the basis vectors, much less the entire space, then all of these coefficients must be 0.

01:54 Therefore, these linear functionals are linearly independent.

01:59 So we've proved linear independence.

02:03 Let's go ahead and show that they span the space now.

02:08 We're going to let capital phi be in v star...

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