Problem 18 - Advanced Engineering Mathematics

Question

In Problems \(17-20\), determine whether the given set is a subspace of the indicated vector space. Polynomials \(p\) that are divisible by \(x-2 ; P_{2}\)

Step-by-Step Solution

Answer

The set is a subspace of \(P_2\).

Step 1: Identify the Vector Space

We are given a set of polynomials that are divisible by \(x-2\) within the vector space of all polynomials of degree less than or equal to 2, denoted by \(P_2\). In this context, the vector space \(P_2\) consists of polynomials in the form \(a_0 + a_1x + a_2x^2\) where \(a_0, a_1, a_2\) are real numbers.

Step 2: Check for Zero Polynomial

For a subset to be a subspace, it must contain the zero vector of \(P_2\). The zero vector in this space is the zero polynomial, \(p(x) = 0\). Since \(0\) is divisible by any polynomial, it is in our subset.

Step 3: Check for Closure Under Addition

We need to show the subset is closed under addition. Take two polynomials \(p(x)\) and \(q(x)\) in our subset, meaning \(p(x) = (x-2)r(x)\) and \(q(x) = (x-2)s(x)\) where \(r(x)\) and \(s(x)\) are polynomials of degree 1 or less. We have:\[(p+q)(x) = p(x) + q(x) = (x-2)r(x) + (x-2)s(x) = (x-2)(r(x) + s(x)).\]Since \(r(x) + s(x)\) is also a polynomial of degree 1 or less, \(p(x)+q(x)\) is divisible by \(x-2\) and remains in the subset.

Step 4: Check for Closure Under Scalar Multiplication

Next, check closure under scalar multiplication. Let \(c\) be any real scalar and \(p(x) = (x-2)r(x)\) be in the subset. Then:\[c \, p(x) = c \, (x-2)r(x) = (x-2)(c \, r(x)).\]Since \(c \, r(x)\) is still a polynomial of degree 1 or less, \(c \, p(x)\) is also divisible by \(x-2\), hence remains in the subset.

Step 5: Conclusion

The subset of polynomials in \(P_2\) that are divisible by \(x-2\) satisfies all three conditions: containing the zero polynomial, closure under addition, and closure under scalar multiplication. Therefore, it is a subspace of \(P_2\).

Key Concepts

Closure under addition

Closure under addition is one of the key requirements for a subset to qualify as a subspace of a vector space. In simple terms, this means that when you add any two elements from the subset, the result must also be an element of the same subset. For the problem at hand, we're looking at polynomials in the form of \((x-2)r(x)\), where \(r(x)\) is a polynomial of degree 1 or less.

Consider two such polynomials, \(p(x)\) and \(q(x)\), which can be represented as \(p(x) = (x-2)r(x)\) and \(q(x) = (x-2)s(x)\). When we add them, we get:

• \[(p+q)(x) = p(x) + q(x) = (x-2)r(x) + (x-2)s(x) = (x-2)(r(x) + s(x)).\]

Here, \((r(x) + s(x))\) is a new polynomial, still of degree 1 or less. This proves that the sum \((p+q)(x)\) also remains divisible by \(x-2\), hence closure under addition is satisfied.

Closure under scalar multiplication

Closure under scalar multiplication ensures that when you multiply an element of the subset by a scalar (a real number in this context), the resulting polynomial remains in the subset. Imagine taking a polynomial \(p(x)\) from our subset, where \(p(x) = (x-2) r(x)\).

If you multiply \(p(x)\) by a real number \(c\), you get:

• \[c \, p(x) = c \, (x-2) r(x) = (x-2) (c \, r(x)).\]

In this expression, \((c \, r(x))\) is still a polynomial of degree 1 or less, because multiplying by a scalar does not increase its degree. Since the product \((x-2)(c \, r(x))\) remains divisible by \(x-2\), we confirm closure under scalar multiplication for the subset.

Polynomials divisible by a polynomial

Understanding polynomials divisible by another polynomial is crucial in determining the characteristics of the subset. In this case, we're focusing on polynomials divisible by \(x-2\). A polynomial \(p(x)\) is said to be divisible by another polynomial \(q(x)\) if there exists a polynomial \(r(x)\) such that \(p(x) = q(x)\cdot r(x)\).

For instance, if \(p(x)\) is given by \((x-2)r(x)\), we immediately know \(p(x)\) is divisible by \(x-2\), since we're expressing it exactly in that form. Divisibility ensures that transformations upon \(p(x)\), within the subset, respect this structure, such as when adding or scaling within the set.

Vector space

A vector space is a fundamental concept in linear algebra. It is a collection of objects, called vectors, that can be added together and multiplied by scalars while satisfying specific properties. The entire set of polynomials of degree less than or equal to 2, denoted as \(P_2\), forms such a space.

This space \(P_2\) consists of polynomials in the form \(a_0 + a_1x + a_2x^2\), where \(a_0, a_1,\) and \(a_2\) are real numbers. Subspaces are like smaller, self-contained vector spaces within a larger vector space. For instance, the set of polynomials divisible by \(x-2\) is a candidate for a subspace of \(P_2\). It must preserve the same algebraic operations as \(P_2\), but limited to elements that meet the divisibility condition.

Zero polynomial

The zero polynomial plays a pivotal role in defining subspaces. It is the polynomial where all coefficients are zero, represented as \(p(x) = 0\). In any vector space, the presence of this zero element is essential. It serves as the identity element for addition; meaning any polynomial added to the zero polynomial remains unchanged.

In our subset, the zero polynomial \(p(x) = 0\) is naturally included, as it is divisible by any polynomial, including \(x-2\). This is necessary for the subset to act properly as a subspace of \(P_2\), ensuring that all subset-defining operations can be performed without exception. Thus, any valid subspace must include this key polynomial.