Let V be a vector space. Prove that every linear function L: ℝ →V has the form L[x]=x 𝐛, where x

Let V be a vector space. Prove that every linear function L:...

Key Concepts

Basis and Scalar Multiplication

Basis and scalar multiplication highlight how any vector in a finite-dimensional space can be expressed uniquely as a linear combination of basis vectors. In the context of one-dimensional vector spaces, this concept simplifies to the idea that every element is a scalar multiple of a fixed basis element. This principle is key to proving representations of linear functions, where the function's value is determined by the image of the basis element and then extended by linearity.

One-Dimensional Vector Space

A one-dimensional vector space is a space in which every element can be written as a scalar multiple of a single basis vector. In the case of the field of real numbers, this means that every real number is a multiple of the number 1. This property greatly simplifies the analysis of linear functions defined on one-dimensional spaces, as the entire behavior of the function can be understood by its action on the basis element.

Vector Space

A vector space is a collection of objects called vectors, which can be added together and multiplied by scalars (elements of a field) while satisfying specific axioms such as associativity, commutativity, and distributivity. These axioms ensure that vectors behave in a structured and predictable manner, forming the foundation for much of linear algebra.

Linear Transformation

A linear transformation is a function between vector spaces that preserves the operations of vector addition and scalar multiplication. This means that for any vectors and scalars, the transformation distributes over addition and respects the scaling by a scalar. Understanding linear transformations is crucial for exploring maps between vector spaces and the structure of solutions to linear equations.

Transcript

00:02 Okay, so consider a finite dimensional inner product space v, along with a linear functional phi.

00:11 Okay, the dimension of this inner product space v is going to be n.

00:16 Now, the result that we want to show is a classical result.

00:20 We want to show that there exists a unique little v inside this inner product space with a following property.

00:29 So for all vectors, you inside fee, okay, when we apply the linear functional onto you, so any you, this is always equal to the inner product of u and v.

00:47 Okay, so this holds for any u that you can pick inside the inner product space fee.

00:53 So how do we go about showing a result like this? well, note that there are two parts to the argument.

00:59 You need to show that there exists such a v, and then in the second part, we need to show that it is unique.

01:05 So let's do existence first, existence of little v.

01:15 Okay, so the key fact that we're going to use is the fact that because the inner product space v is finite dimensional, there exists a basis that we can use.

01:26 So let e1.

01:28 Up to e .n be an orthomormal basis of your choice, okay? and because this is an orthominal basis, we can write any vector, u, in the following way.

01:41 So we can take the inner product of u along with, you know, like each of the basis vector, e1, and that's the coefficient.

01:51 And we'll just multiply this coefficient along the direction in which each of the basis vectors is pointing to, and then you can just sum them all up.

02:02 Okay.

02:03 So, e, n, right? so i've just done, you know, like a basis expansion of u.

02:12 And now, note that all of these coefficients are in a products, which means that they're scalers.

02:18 And so when we apply a linear functional phi onto you, by linearity, we can pull all these scalar factors out.

02:28 And also by linearity, we can expand it throughout the addition.

02:31 So what that means is we can write this in the following way.

02:47 Okay, so i've done very little here.

02:49 I'm just using the linearity property to break up the addition and to pull out the scalar factors.

02:55 Now, what we want to use is another property of the inner products, which is the homogeneity property.

03:03 And so because of homogeneity, we can pull in this phi factor, this 5e1 all the way to 5en.

03:10 We can pull these into the second argument of the inner product, okay, as long as we take the conjugate.

03:18 So we can write this in the following way.

03:22 E1 bar phi of e1 plus dot, dot, dot, plus e .n, phi, e.

03:33 And so there's a bar on top of all the fies because we have to take the conjugate.

03:39 That's the conjugate homogeneity property of the inner product, and we are putting it into one big inner product just by linearity, right? because this is an addition and the argument, the first argument, is always just you.

03:52 So we can just lump this into a big inner product...

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