(a) Show that 1,1-t,(1-t)^2,(1-t)^3 is a basis for 𝒫^(3). (b) Write p(t)=1+t^3 in terms of the b

(a) Show that 1,1-t,(1-t)^2,(1-t)^3 is a basis for 𝒫^(3)....

Key Concepts

Polynomial Vector Space

A polynomial vector space, such as ?^(3), consists of all polynomials with degrees up to a specified value (in this case, degree 3) and is a fundamental example of a finite-dimensional vector space. The structure of a polynomial vector space allows for operations such as addition and scalar multiplication, which satisfy the axioms of a vector space.

Basis

A basis for a vector space is a set of vectors that are both linearly independent and span the entire space. In the context of ?^(3), the set {1, 1 - t, (1 - t)^2, (1 - t)^3} is proposed as a basis. Establishing this involves demonstrating that any polynomial in ?^(3) can be expressed as a unique linear combination of these basis elements.

Linear Independence and Spanning Set

The concepts of linear independence and spanning are crucial in determining whether a given set is a basis. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others, and it spans the space if every element in the vector space can be represented as a linear combination of the set. For the given set of polynomials in ?^(3), checking these properties confirms its status as a basis.

Change of Basis and Coordinate Representation

Once a basis is established, every polynomial in the space can be uniquely represented as a linear combination of the basis elements. This process is known as finding the coordinates of the polynomial with respect to the new basis. In part (b) of the problem, writing the polynomial p(t) = 1 + t^3 in terms of the basis elements is an application of this concept, where one must determine the coefficients corresponding to each basis element.

Transcript

00:01 Okay, for this question, they want us to find a basis of the polynomials of degree thyrr less, when all of the elements in that basis, let's call that basis, have the same degree.

00:21 Okay, so i'm going to make them all have degree three.

00:27 It's actually the only way to do this.

00:29 They can't all have degree two, because we'll never be able to get the cube term.

00:31 So, but i'll make it simple still.

00:35 I'll have plus 1, x cubed plus x squared, x cubed plus x squared, x cubed plus x.

00:44 Okay.

00:45 Why is that? well, last time i didn't even have to bother translating them to a system of equations in r4 after translating their coefficients to appropriate vector entries.

01:02 You can do that on your own time.

01:04 If you want, it's buzz in to see.

01:07 And actually, you're going to notice the correspondence between the algorithms on r4 and these basis vectors right here.

01:15 So they want us to prove their independent.

01:18 Normally we do gaussian elimination when trying to solve for the zero vector.

01:22 But here we can keep our language in terms of polynomials.

01:26 There's going to need to be some coefficient times this, that ends up equalling 0.

01:40 Because what are we doing? we're taking these vectors and we're solving for the zero vector.

01:44 Oh, i've scrolled too much.

01:54 Let's see.

01:56 Scroll, awesome.

01:58 Okay...

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