Question

Which of the following are subspaces of \(\mathbf{P}_{3}\) ? Support your answer. a. \(U=\left\\{f(x) \mid f(x) \in \mathbf{P}_{3}, f(2)=1\right\\}\) b. \(U=\left\\{x g(x) \mid g(x) \in \mathbf{P}_{2}\right\\}\) c. \(U=\left\\{x g(x) \mid g(x) \in \mathbf{P}_{3}\right\\}\) d. \(U=\left\\{x g(x)+(1-x) h(x) \mid g(x)\right.\) and \(\left.h(x) \in \mathbf{P}_{2}\right\\}\) e. \(U=\) The set of all polynomials in \(\mathbf{P}_{3}\) with constant term 0 f. \(U=\left\\{f(x) \mid f(x) \in \mathbf{P}_{3},\right.\) deg \(\left.f(x)=3\right\\}\)

Short answer

Subspaces: b, d, e.

Step-by-step solution

Understanding the problem

We are given subspaces, where each is a set of polynomials expressed in different forms. A subset of a vector space is a subspace if it is closed under addition, closed under scalar multiplication, and contains the zero vector. Here, our vector space is \(\mathbf{P}_3\), the space of all polynomials of degree at most 3.

Evaluating subspace a

Subspace a is \(U=\{f(x) \mid f(x) \in \mathbf{P}_{3}, f(2)=1\}\). It is not a subspace because the zero vector (polynomial 0) is not included since \(f(2)=0\) for \(f(x)=0\).

Evaluating subspace b

Subspace b is \(U=\{xg(x) \mid g(x) \in \mathbf{P}_2\}\). The expression \(xg(x)\) forms polynomials of degree less than or equal to 3, and it is closed under addition and scalar multiplication. The zero polynomial is included by choosing \(g(x)=0\). Therefore, this is a subspace.

Evaluating subspace c

Subspace c is \(U=\{xg(x) \mid g(x) \in \mathbf{P}_3\}\). Any polynomial \(xg(x)\) from multiplying another polynomial of degree 3 will have a degree of 4. However, these polynomials do not belong to \(\mathbf{P}_3\). Therefore, it is not a subspace.

Evaluating subspace d

Subspace d is \(U=\{xg(x) + (1-x)h(x) \mid g(x) \text{ and } h(x) \in \mathbf{P}_2\}\). Since both parts are degree 3 or less and the operations performed stay within this degree, all necessary closure properties and the inclusion of zero polynomial (choose \(g(x)=h(x)=0\)) hold. Thus, it is a subspace.

Evaluating subspace e

Subspace e is the set of all polynomials in \(\mathbf{P}_3\) with constant term 0. This set is closed under addition and scalar multiplication, and the zero polynomial is included. It forms a subspace since removing the constant term does not affect closure properties.

Evaluating subspace f

Subspace f is \(U=\{f(x) \mid f(x) \in \mathbf{P}_{3},\text{ deg}(f(x))=3\}\). It is neither closed under addition nor scalar multiplication. For example, adding two degree 3 polynomials could reduce the degree. Therefore, it's not a subspace.

Key concepts

Subspaces

In linear algebra, a subspace is a special subset of a vector space. To qualify as a subspace, the set must be closed under vector addition and scalar multiplication, and must include the zero vector.

These requirements mean that when you add two vectors (or polynomials, in our case) from the set, the result should still be within the set. Similarly, multiplying a vector by a scalar should not lead you outside the set. Finally, the presence of the zero vector ensures that all linear combinations make sense within the context of the subspace. In the polynomial framework of this problem, we treat polynomials as vectors, and therefore analyze each given set with these rules.

For instance, in subspace example 'b', each polynomial is of the form \( xg(x) \), where \( g(x) \) is a polynomial of degree 2. The form \( xg(x) \) ensures that the added polynomial remains in the space, since the degree of the polynomial doesn’t exceed 3. Moreover, multiplying by a scalar retains the degree, ensuring closure under such operations. Thus, with the inclusion of the zero polynomial, it forms a subspace.

Vector Space

A vector space is essentially a collection of objects called vectors that you can add together and multiply by numbers, which are called scalars in this context. These operations must satisfy certain rules, such as commutativity, associativity of addition, and the existence of an additive identity known as the zero vector.

When you work with polynomials as vectors, as you do in this exercise, the set of polynomials \( \mathbf{P}_3 \) is a vector space consisting of all polynomials of degree at most 3. This includes polynomials with terms like \( x^3, x^2, x, \) and constants.

To determine if something is a subspace of this vector space, like the sets in the provided exercise, we check if these sets satisfy the required properties such as the presence of the zero polynomial, and closure under both addition and scalar multiplication within \( \mathbf{P}_3 \). Each part of the exercise is about testing these properties in different polynomial setups.

Polynomials

Polynomials are mathematical expressions consisting of variables raised to whole number powers and multiplied by coefficients. In the context of vector spaces, they can be very handy because they provide a nice structure for holding operations and rules.

The specific set \( \mathbf{P}_3 \) in the exercise refers to polynomials that have terms up to \( x^3 \), which means these polynomials can be classified as vectors in this space. For example, for subspace example 'e' in the exercise, polynomials within \( \mathbf{P}_3 \) that have a constant term of 0 are considered. The definition of a subspace ensures that any linear combination of these polynomials results in another polynomial with no constant term, ensuring closure.

This combination of rules and definitions based on degrees and terms is what allows us to classify and understand different subsets within this polynomial vector space.