Proving W Is A Subspace Of R4: A Step-by-Step Guide

Hey guys! Ever wondered how to prove that a given set W is a subspace of R4? It's a fundamental concept in linear algebra, and understanding it opens doors to more advanced topics. In this article, we'll break down the process into easy-to-follow steps, ensuring you grasp the core ideas and can confidently tackle similar problems. So, let's dive in!

Understanding Subspaces

Before we jump into proving that W is a subspace of R4, let's quickly recap what a subspace actually is. Think of it like this: R4 is the entire 4-dimensional space, containing all possible 4-dimensional vectors. A subspace W is a subset of R4 that itself behaves like a vector space. This means it must satisfy three crucial conditions:

1. The Zero Vector: The zero vector (0, 0, 0, 0) must be an element of W. This is the foundation; without the origin, it can't be a true subspace.
2. Closure Under Addition: If you take any two vectors in W and add them together, the resulting vector must also be in W. This ensures that W is "closed" under the operation of addition. In other words, you can't escape W by adding vectors within it.
3. Closure Under Scalar Multiplication: If you take any vector in W and multiply it by any scalar (a real number), the resulting vector must also be in W. This means W is "closed" under scalar multiplication. You can stretch or shrink vectors in W without leaving the subspace.

If a subset W of R4 satisfies all three of these conditions, then we can confidently declare that W is indeed a subspace of R4. Seems simple enough, right? But the devil's in the details. Showing these conditions hold for a specific W often requires careful algebraic manipulation and a clear understanding of the set's definition. In the next section, we'll tackle a concrete example to illustrate the process. Remember, understanding these three conditions is paramount. They are the pillars upon which the entire concept of subspaces rests. They ensure that the subspace behaves consistently as a vector space within the larger space. Without these properties, W would just be a random subset, not a subspace. So, keep these three conditions firmly in mind as we move forward. They are your guiding principles.

Example: Proving a Specific W is a Subspace of R4

Okay, let's get our hands dirty with an example. Suppose W is defined as the set of all vectors (a, b, c, d) in R4 such that a + b = c + d. In mathematical notation:

W = {(a, b, c, d) ∈ R4 | a + b = c + d}

Our mission, should we choose to accept it (and we do!), is to prove that W is a subspace of R4. To do this, we need to verify the three conditions we discussed earlier:

1. The Zero Vector

First, we need to check if the zero vector (0, 0, 0, 0) is in W. Does it satisfy the condition a + b = c + d? Let's see: 0 + 0 = 0 + 0. Yep, it does! Since 0 = 0, the zero vector satisfies the condition, and therefore, the zero vector belongs to W. This is a crucial first step! If the zero vector isn't in W, we can immediately conclude that W is not a subspace, and we can stop right there.

2. Closure Under Addition

Next, we need to show that W is closed under addition. This means that if we take any two vectors in W and add them, the result must also be in W. Let's take two arbitrary vectors in W, say u = (a1, b1, c1, d1) and v = (a2, b2, c2, d2). Since u and v are in W, we know that:

a1 + b1 = c1 + d1 a2 + b2 = c2 + d2

Now, let's add these vectors: u + v = (a1 + a2, b1 + b2, c1 + c2, d1 + d2). To show that u + v is in W, we need to demonstrate that (a1 + a2) + (b1 + b2) = (c1 + c2) + (d1 + d2). Let's see if we can manipulate the equations we already have to prove this:

Start with: a1 + b1 = c1 + d1 And: a2 + b2 = c2 + d2

Add the two equations together: (a1 + b1) + (a2 + b2) = (c1 + d1) + (c2 + d2)

Rearrange the terms: (a1 + a2) + (b1 + b2) = (c1 + c2) + (d1 + d2)

Bingo! We've shown that the sum of the two vectors u and v also satisfies the condition for belonging to W. Therefore, W is closed under addition. This step often involves some algebraic manipulation. The key is to use the defining property of W (in this case, a + b = c + d) to show that the sum of two vectors in W also satisfies that property.

3. Closure Under Scalar Multiplication

Finally, we need to show that W is closed under scalar multiplication. This means that if we take any vector in W and multiply it by any scalar (a real number), the result must also be in W. Let's take an arbitrary vector in W, say u = (a, b, c, d), and an arbitrary scalar k. Since u is in W, we know that a + b = c + d.

Now, let's multiply the vector u by the scalar k: ku = (ka, kb, kc, kd). To show that ku is in W, we need to demonstrate that ka + kb = kc + kd. Let's see if we can use the equation we already have to prove this:

Start with: a + b = c + d

Multiply both sides of the equation by k: k(a + b) = k(c + d)

Distribute the k: ka + kb = kc + kd

Excellent! We've shown that the scalar multiple of the vector u also satisfies the condition for belonging to W. Therefore, W is closed under scalar multiplication. This is usually the easiest of the three conditions to prove. It often involves a simple application of the distributive property.

Conclusion

We've successfully verified all three conditions: the zero vector is in W, W is closed under addition, and W is closed under scalar multiplication. Therefore, we can confidently conclude that W = {(a, b, c, d) ∈ R4 | a + b = c + d} is indeed a subspace of R4.

Proving that a set is a subspace might seem daunting at first, but by breaking it down into these three steps and carefully applying the definitions, you can conquer any subspace problem. Keep practicing, and you'll become a subspace pro in no time!

Key Takeaways:

• A subspace must contain the zero vector.
• A subspace must be closed under addition.
• A subspace must be closed under scalar multiplication.

By understanding and applying these three principles, you can confidently determine whether a given set is a subspace of a vector space. Remember, linear algebra is all about building upon fundamental concepts. Mastering the concept of subspaces is a crucial step in your journey to becoming a linear algebra whiz!

So, there you have it! A comprehensive guide to proving that W is a subspace of R4. Go forth and conquer those linear algebra problems! And remember, practice makes perfect. The more you work with these concepts, the more intuitive they will become. Good luck, and have fun exploring the fascinating world of linear algebra!