Problem 17 - Advanced Engineering Mathematics

Question

In Problems \(17-20\), determine whether the given set is a subspace of the indicated vector space. Polynomials of the form \(p(x)=c_{3} x^{3}+c_{1} x ; P_{3}\)

Step-by-Step Solution

Answer

Yes, the set is a subspace of \(P_3\).

Step 1: Define the Problem

We need to determine if the set of polynomials of the form \(p(x) = c_3 x^3 + c_1 x\) is a subspace of \(P_3\), the space of polynomials of degree at most 3.

Step 2: Check for the Zero Vector

A subspace must contain the zero vector of the vector space. For \(P_3\), the zero polynomial is \(p(x) = 0\). We express this zero polynomial with coefficients as \(c_3 = 0\) and \(c_1 = 0\). Since substituting these values in our form \(p(x) = c_3 x^3 + c_1 x\) gives \(p(x) = 0\), the zero vector is in the set.

Step 3: Verify Closure Under Addition

If \(p_1(x) = c_{3,1} x^3 + c_{1,1} x\) and \(p_2(x) = c_{3,2} x^3 + c_{1,2} x\) are polynomials in our set, then their sum is \(p_1(x) + p_2(x) = (c_{3,1} + c_{3,2}) x^3 + (c_{1,1} + c_{1,2}) x\). This is also a polynomial of the original form, so the set is closed under addition.

Step 4: Check Closure Under Scalar Multiplication

For any polynomial \(p(x) = c_3 x^3 + c_1 x\) in the set and any scalar \(k\), the scalar multiplication gives \(k \cdot p(x) = k c_3 x^3 + k c_1 x\). This is also a polynomial of the form \(c_3' x^3 + c_1' x\), so the set is closed under scalar multiplication.

Step 5: Conclusion

The set of polynomials \(p(x) = c_3 x^3 + c_1 x\) satisfies all the required conditions: it contains the zero polynomial, is closed under addition, and is closed under scalar multiplication. Therefore, this set is indeed a subspace of \(P_3\).

Key Concepts

Polynomials

Polynomials are mathematical expressions involving a sum of powers of a variable multiplied by coefficients. For instance, any expression like \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 \) is a polynomial where \( n \) is a non-negative integer and each \( a_i \) is a coefficient. In this problem, we focus on a particular form: \( p(x) = c_3 x^3 + c_1 x \), which indicates polynomial terms in powers of 3 and 1, without a constant term or any other powered terms.
This specific form is categorized within polynomials due to its structured format with defined power terms, emphasizing that it belongs to a larger family called vector spaces of polynomials such as \( P_3 \), which consists of all polynomials with a degree of at most 3.

Closure Under Addition

In the context of vector spaces, closure under addition is a fundamental property for determining whether a set is a subspace. This property implies that if you take any two elements (in this case, polynomials) from the set and add them, the result should still be within the set.
For instance, consider two polynomials \( p_1(x) = c_{3,1} x^3 + c_{1,1} x \) and \( p_2(x) = c_{3,2} x^3 + c_{1,2} x \). Adding these gives \( p_1(x) + p_2(x) = (c_{3,1} + c_{3,2}) x^3 + (c_{1,1} + c_{1,2}) x \), which is of the same form \( c_3 x^3 + c_1 x \).

• This shows that the addition of any two polynomials of this form results in another polynomial of the same form.
• Therefore, the set is closed under addition, validating one of the requirements to be a subspace.

Closure Under Scalar Multiplication

When discussing vector spaces, scalar multiplication is another criterion for establishing subspaces. A set is closed under scalar multiplication if multiplying any element by a scalar also results in an element within the set.
For a polynomial \( p(x) = c_3 x^3 + c_1 x \), if we multiply by any scalar \( k \), the resulting polynomial will be \( k \cdot p(x) = k c_3 x^3 + k c_1 x \). Notice that this is still a polynomial of the form \( c_3 x^3 + c_1 x \), with new coefficients \( k c_3 \) and \( k c_1 \).

• This confirms the operation maintains the polynomial within the same structure.
• Hence, the set is closed under scalar multiplication, further qualifying it as a subspace of \( P_3 \).

Zero Vector in Vector Space

An essential part of any subspace is the inclusion of the zero vector from the parent vector space. In the realm of polynomials and vector spaces like \( P_3 \), the zero vector is the zero polynomial \( p(x) = 0 \).
For the set of polynomials \( p(x) = c_3 x^3 + c_1 x \), substituting both \( c_3 = 0 \) and \( c_1 = 0 \) results in \( p(x) = 0 \), which is the zero polynomial.

• This shows that the zero vector is indeed part of our polynomial set.
• Therefore, one basic requirement for the set being a subspace is satisfied.