Problem 4 - Differential Equations and Linear Algebra

Question

Determine whether the given set \(S\) of vectors is closed under addition and closed under scalar multiplication. In each case, take the set of scalars to be the set of all real numbers. The set \(S:=\left\\{a_{0}+a_{1} x+a_{2} x^{2}: a_{0}+a_{1}+a_{2}=0\right\\}.\)

Step-by-Step Solution

Answer

The set \(S\) is closed under addition, as verified by the following equation: \((a_{0} + b_{0}) + (a_{1} + b_{1}) + (a_{2} + b_{2}) = (a_{0} + a_{1} + a_{2}) + (b_{0} + b_{1} + b_{2}) = 0 + 0 = 0.\) Additionally, the set \(S\) is closed under scalar multiplication, as shown by this equation: \((k a_{0}) + (k a_{1}) + (k a_{2}) = k(a_{0} + a_{1} + a_{2}) = k \cdot 0 = 0.\) Hence, the set \(S\) is closed under both addition and scalar multiplication.

Step 1: Verify closure under addition

Let \(s_1 = a_{0} + a_{1}x + a_{2}x^{2}\) and \(s_2 = b_{0} + b_{1}x + b_{2}x^{2}\) be two vectors in the set \(S\). By definition of the set, we have \(a_{0} + a_{1} + a_{2} = 0\) and \(b_{0} + b_{1} + b_{2} = 0\). Add these two vectors: $$ s_1 + s_2 = (a_{0} + b_{0}) + (a_{1} + b_{1})x + (a_{2} + b_{2})x^{2}. $$ Now, we need to determine if the sum is still in \(S\). To do this, we check if the coefficients satisfy the condition \(a_{0}+a_{1}+a_{2}=0\): $$ (a_{0} + b_{0}) + (a_{1} + b_{1}) + (a_{2} + b_{2}) = (a_{0} + a_{1} + a_{2}) + (b_{0} + b_{1} + b_{2}) = 0 + 0 = 0. $$ Since \((a_{0} + b_{0}) + (a_{1} + b_{1}) + (a_{2} + b_{2}) = 0\), the sum \(s_1 + s_2\) is in the set \(S\). Hence, the set is closed under addition.

Step 2: Verify closure under scalar multiplication

Let \(s = a_{0} + a_{1}x + a_{2}x^{2}\) be a vector in the set \(S\) and let \(k\) be any scalar in \(\mathbb{R}\). We proceed to multiply the vector by the scalar: $$ k \cdot s = k(a_{0} + a_{1}x + a_{2}x^{2}) = (k a_{0}) + (k a_{1})x + (k a_{2})x^{2}. $$ Now, we need to check the condition \((k a_{0}) + (k a_{1}) + (k a_{2}) = 0\): $$ (k a_{0}) + (k a_{1}) + (k a_{2}) = k(a_{0} + a_{1} + a_{2}) = k \cdot 0 = 0. $$ Since \((k a_{0}) + (k a_{1}) + (k a_{2}) = 0\), the scalar multiplication \(k \cdot s\) is in the set \(S\). Hence, the set is closed under scalar multiplication. #Conclusion# The set \(S\) is closed under addition and scalar multiplication.

Key Concepts

Closure Under Addition

In vector spaces, closure under addition is a fundamental property. It means that if you take any two vectors from a set and add them together, their sum must also be part of that set. Let's explore this using our given set of vectors. The set \( S \) is defined as \( S = \{ a_{0} + a_{1}x + a_{2}x^2 : a_{0} + a_{1} + a_{2} = 0 \} \). To verify closure under addition, we consider two arbitrary vectors: \( s_1 = a_{0} + a_{1}x + a_{2}x^2 \) and \( s_2 = b_{0} + b_{1}x + b_{2}x^2 \), both belonging to \( S \). Given the condition \( a_{0} + a_{1} + a_{2} = 0 \) and \( b_{0} + b_{1} + b_{2} = 0 \), we test if \( s_1 + s_2 = (a_{0} + b_{0}) + (a_{1} + b_{1})x + (a_{2} + b_{2})x^2 \) stays within the set. Since:

• \(a_{0} + a_{1} + a_{2} = 0\)
• \(b_{0} + b_{1} + b_{2} = 0\)

Adding these gives: \[(a_{0} + b_{0}) + (a_{1} + b_{1}) + (a_{2} + b_{2}) = 0 + 0 = 0\]Thus, the sum \( s_1 + s_2 \) satisfies the condition for membership in the set \( S \). Hence, the set \( S \) is indeed closed under addition, confirming this crucial vector space property.

Closure Under Scalar Multiplication

Closure under scalar multiplication is another essential property of vector spaces. This property ensures that if you multiply any vector in a set by a scalar, the resulting vector remains within the same set. Let’s examine how this works with our set \( S \), defined as \( S = \{ a_{0} + a_{1}x + a_{2}x^2 : a_{0} + a_{1} + a_{2} = 0 \} \).
To check this, take a vector \( s = a_{0} + a_{1}x + a_{2}x^2 \) from \( S \) and a scalar \( k \) in \( \mathbb{R} \). Multiply the vector by the scalar to get:\[k \cdot s = k(a_{0} + a_{1}x + a_{2}x^2) = (ka_{0}) + (ka_{1})x + (ka_{2})x^2\]We need to confirm that the new vector \( (ka_{0}) + (ka_{1}) + (ka_{2}) = 0 \). Using the initial condition:

• \(a_{0} + a_{1} + a_{2} = 0\)

Multiplying by \( k \) gives:\[(ka_{0}) + (ka_{1}) + (ka_{2}) = k(a_{0} + a_{1} + a_{2}) = k \cdot 0 = 0\]This condition holds true, confirming that multiplying by any real scalar does not violate the the set's defining condition. As a result, set \( S \) is closed under scalar multiplication, maintaining its integrity within the vector space framework.

Polynomial Vector Spaces

Polynomial vector spaces are special kinds of vector spaces where the elements are polynomial functions. These polynomials have coefficients from a specific field, such as the real numbers, and they are subject to certain conditions that form the vector space. For the set \( S \), defined as polynomials of the form \( a_{0} + a_{1}x + a_{2}x^2 \) with the constraint \( a_{0} + a_{1} + a_{2} = 0 \), let's delve deeper into how it functions as a vector space.
Polynomials from \( S \) are essentially functions mapping each real number \( x \) to a value according to the equation. This set is structured like a vector space because it allows operations like addition and scalar multiplication, integral to the concept of vector spaces. Key characteristics:

• Closure properties: We've discussed closure under addition and scalar multiplication, ensuring that \( S \) forms a valid vector space.
• Zero vector: Every vector space contains a zero vector. In set \( S \), if \( a_{0} = 0, a_{1} = 0, a_{2} = 0 \), then the polynomial becomes \( 0 \), satisfying the condition to belong to \( S \).

These conditions collectively form what mathematicians recognize as a polynomial vector space. By maintaining such properties, \( S \) can participate in larger mathematical exercises, revealing a broad spectrum of algebraic and geometric insights.