Proving W Is A Subspace Of R3: A Comprehensive Guide

Hey everyone! Today, we're diving into a fundamental concept in linear algebra: subspaces. Specifically, we're going to show that W is indeed a subspace of R3. This is super important because understanding subspaces helps us grasp the structure of vector spaces and how different sets of vectors relate to each other. So, grab your coffee, and let's get started!

What Exactly is a Subspace, Anyway?

Alright, before we jump into the nitty-gritty of proving that W is a subspace, let's make sure we're all on the same page about what a subspace actually is. Think of a subspace as a little vector space living inside a bigger one. More formally, a subspace W of a vector space V (like our R3) must satisfy three key conditions:

1. The Zero Vector: The zero vector (the vector with all components equal to zero) must be in W. This is like the starting point; without it, we can't really call it a vector space.
2. Closure under Addition: If you take any two vectors in W and add them together, the result must also be in W. In other words, adding vectors within W doesn't magically create a vector that's outside of W.
3. Closure under Scalar Multiplication: If you take any vector in W and multiply it by a scalar (a regular number), the resulting vector must also be in W. Scaling a vector within W should keep it within W.

These three conditions are the gatekeepers of the subspace club. If a set of vectors doesn't meet all three, it's not a subspace. These properties ensure that subspaces behave consistently, allowing us to perform operations like addition and scalar multiplication without leaving the confines of the subspace itself. Let's get more in-depth on this!

Let’s start with the basics. In mathematics, a vector space is a collection of objects called vectors, which can be added together and multiplied by numbers, called scalars. R3, or three-dimensional space, is a classic example of a vector space. Its vectors are triples of real numbers, like (1, 2, 3) or (-2, 0, 1). Addition and scalar multiplication are defined in the usual way: to add two vectors, you add their corresponding components, and to multiply a vector by a scalar, you multiply each component of the vector by that scalar. For a subset W to be a subspace of R3, it needs to satisfy three crucial properties. First, it must contain the zero vector. The zero vector in R3 is (0, 0, 0). Second, it must be closed under addition, meaning that if you add any two vectors in W, the result must also be in W. Third, it must be closed under scalar multiplication, meaning that if you multiply any vector in W by a scalar, the result must also be in W. Proving that a set W is a subspace involves showing that these three properties hold. Let’s prove the first key component.

Showing W Contains the Zero Vector

Now, let's get down to brass tacks and start demonstrating that W is indeed a subspace. The first hurdle we need to clear is the zero vector. The zero vector is the vector where all the components are zero (0, 0, 0). To show that the zero vector is in W, we need to verify that it satisfies the defining condition of W. If W is defined by a specific equation or set of rules, we need to plug in the zero vector and see if it holds true. If the zero vector satisfies the condition, it's a good start! If not, then W is not a subspace and we don't need to go further.

For example, if W is the set of all vectors in R3 where the sum of the components is zero, we would check if (0, 0, 0) satisfies this condition: 0 + 0 + 0 = 0. Since it does, the zero vector is in W. If, however, the sum needed to be 1, the zero vector would not be in W and it would not be a subspace. The importance of the zero vector lies in its fundamental role in vector space operations. It serves as the additive identity. In other words, adding the zero vector to any vector in W does not change that vector. This is essential for maintaining the structure and properties of the vector space, ensuring consistency in operations, and serving as a crucial reference point for other vectors within the space. Let's see how this works by doing a problem:

Example:

Let’s say W is defined as the set of all vectors (x, y, z) in R3 such that x + y – z = 0. To prove that the zero vector is in W, we plug in the values of the zero vector (0, 0, 0) into the equation x + y – z = 0.

0 + 0 – 0 = 0.

Since this equation holds true, the zero vector (0, 0, 0) is indeed in W. This satisfies the first condition to prove W is a subspace.

Proving Closure Under Addition

Next up, we need to demonstrate that W is closed under addition. This means that if you pick any two vectors in W and add them together, the resulting vector must also be in W. Let's break this down further and look at how we can prove this. Think of it like this: if you have two vectors, u and v, that both belong to W, their sum (u + v) must also be a member of W. This is a critical property because it ensures that you can perform addition within W without