Problem 5 - 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 \(S:=\left\\{a_{0}+a_{1} x+a_{2} x^{2}: a_{0}+a_{1}+a_{2}=1\right\\}.\)

Step-by-Step Solution

Answer

The set \(S\) is not closed under vector addition, as the sum of the coefficients of the resulting vector is not equal to 1. Similarly, \(S\) is not closed under scalar multiplication for all real numbers, except for the specific case of \(k = 1\).

Step 1: Check if the set is closed under vector addition

Let \(v, w \in S\) be two vectors. Then, $$v = a_{0} + a_{1}x + a_{2}x^{2} \quad \text{and} \quad w = b_{0} + b_{1}x + b_{2}x^{2},$$ such that \(a_{0} + a_{1} + a_{2} = b_{0} + b_{1} + b_{2} = 1\). Now, let's add the two vectors: $$v + w = (a_{0} + a_{1}x + a_{2}x^{2}) + (b_{0} + b_{1}x + b_{2}x^{2}).$$ Let's combine the terms to form a new vector \(z\): $$z = (a_{0} + b_{0}) + (a_{1} + b_{1})x + (a_{2} + b_{2})x^{2}.$$ Now, we need to check if the sum of coefficients in \(z\) is equal to \(1\): $$(a_{0} + b_{0}) + (a_{1} + b_{1}) + (a_{2} + b_{2}) = (a_{0} + a_{1} + a_{2}) + (b_{0} + b_{1} + b_{2}) = 1 + 1 = 2.$$ Since the sum of the coefficients of \(z\) is not equal to \(1\), \(S\) is not closed under vector addition.

Step 2: Check if the set is closed under scalar multiplication

Let \(v \in S\) and \(k \in \mathbb{R}\) be a scalar. Then, $$v = a_{0} + a_{1}x + a_{2}x^{2},$$ with \(a_{0} + a_{1} + a_{2} = 1\). Now, let's multiply \(v\) by the scalar \(k\): $$kv = ka_{0} + ka_{1}x + ka_{2}x^{2}.$$ Let's form a new vector \(z\): $$z = c_{0} + c_{1}x + c_{2}x^{2},$$ where \(c_{0} = ka_{0}\), \(c_{1} = ka_{1}\), and \(c_{2} = ka_{2}\). Now, let's check if the sum of coefficients of \(z\) is equal to \(1\): $$c_{0} + c_{1} + c_{2} = ka_{0} + ka_{1} + ka_{2} = k(a_{0} + a_{1} + a_{2}) = k \cdot 1 = k.$$ Since the sum of the coefficients of \(z\) is equal to \(k\) and not \(1\), \(S\) is not closed under scalar multiplication, except for the specific case of \(k = 1\). To conclude, the set \(S\) is neither closed under vector addition nor closed under scalar multiplication for all real numbers.

Key Concepts

Vector Space

A vector space is a collection of objects, known as vectors, where two operations—vector addition and scalar multiplication—are defined and satisfy certain conditions. These conditions, or 'axioms', include the ability to add two vectors together to get another vector within the space, and the ability to multiply a vector by a scalar (a number from a defined field, typically the real numbers, \(\mathbb{R}\)), resulting in another vector within the space.

For example, consider the set \( S \), defined in our exercise. To be a vector space, any vectors you pick from \( S \), when added together, must also be part of \( S \). Similarly, multiplying any vector from \( S \), by any real number, must result in a vector that is still within \( S \). However, from our exercise, we learn that the set \( S \) doesn’t satisfy these requirements, hence it is not a vector space under the usual addition and scalar multiplication.

Scalar Multiplication

Scalar multiplication in linear algebra is the operation of scaling a vector by a scalar, that is, multiplying each component of the vector by the same scalar value. This operation is one of the foundational actions in vector spaces and should leave us within the same vector space after it’s performed.

In simpler terms, if you have a vector \( v = a_0 + a_1x + a_2x^2 \) and a scalar \( k \) from the real numbers \( \mathbb{R} \), scalar multiplying \( v \) by \( k \) would result in \( kv = ka_0 + ka_1x + ka_2x^2 \). Yet, according to our exercise, performing this action with vectors from the set \( S \) and any scalar other than 1 does not produce a new vector that meets the defining condition of \( S \), showing that \( S \) is not closed under scalar multiplication.

Vector Addition

Vector addition is the process of combining two vectors to create a new vector by adding the corresponding components from each vector. In a proper vector space, when you add two vectors, the resultant vector should also belong to the same space. This property is referred to as being 'closed under addition.'

In the context of our set \( S \), which consists of polynomials of certain coefficients whose sum equals 1, adding two vectors from \( S \) should result in another polynomial that also has its coefficients adding up to 1. Our exercise, however, shows that adding any two vectors from \( S \) results in coefficients adding up to 2, thereby violating the necessary condition for closure under addition.

Linear Algebra

Linear algebra is a branch of mathematics focusing on vectors, vector spaces, and linear transformations. It deals with lines, planes, and subspaces, and is fundamental in understanding how different mathematical systems interact and transform. The concepts of vector addition and scalar multiplication are at the heart of linear algebra.

The exercise we've discussed exemplifies an application of linear algebra concepts by testing if a set of vectors qualifies as a vector space, emphasizing the crucial role these operations play. Linear algebra helps us navigate multi-dimensional spaces, solve systems of linear equations, and deeply understand the geometry and algebra of vectors.