Proving W Is A Subspace Of R³: A Comprehensive Guide

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Proving W is a Subspace of R³: A Comprehensive Guide

Hey there, math enthusiasts! Today, we're diving into the fascinating world of linear algebra and tackling a fundamental concept: proving that a set, specifically W, is a subspace of R³. For those new to the game, R³ represents the set of all 3-dimensional real vectors. A subspace is essentially a smaller vector space nestled within a larger one, inheriting some of its cool properties. So, how do we show that W is, in fact, a subspace? Let's break it down step by step, making it super easy to grasp, even if you're just starting out.

What is a Subspace, Anyway?

Before we jump into the nitty-gritty, let's nail down the definition. A subspace, denoted by W, of a vector space V (in our case, R³) must satisfy three key conditions, often called the subspace criteria. Think of these as the VIP passes to subspace status:

  1. Zero Vector: The zero vector (the vector with all components equal to zero) must be in W. This is like the foundation of your house; without it, the whole thing crumbles.
  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 means the set is "closed" under addition; you can't escape it!
  3. Closure under Scalar Multiplication: If you take any vector in W and multiply it by a scalar (a real number), the result must also be in W. Again, the set is "closed" under scalar multiplication. You can't multiply your way out of it!

If a set W ticks all these boxes, boom! You've got yourself a subspace. Now, let's get down to the practical part of proving it.

Step-by-Step Guide to Proving W is a Subspace

Let's assume we're given a set W defined in some way. For example, W might be defined as the set of all vectors (x, y, z) in R³ such that x + y - z = 0. Or maybe it's defined as the set of all vectors where the second component is always twice the first. The specific definition of W will change depending on the problem, but the approach to proving it's a subspace remains the same.

To prove that W is a subspace, we must verify the three subspace criteria mentioned above. Let's walk through them:

1. Zero Vector Check

The first thing to do is to determine if the zero vector is in W. The zero vector in R³ is (0, 0, 0). To check if the zero vector is in W, substitute the components of the zero vector into the defining equation or condition for W. If the equation or condition is satisfied, then the zero vector is in W. If not, W is NOT a subspace.

For example, if W is defined by x + y - z = 0, substitute x = 0, y = 0, and z = 0. We get 0 + 0 - 0 = 0, which is true. Therefore, the zero vector is in W. If W was defined by x + y - z = 1, substituting the zero vector would give us 0 + 0 - 0 = 1, which is not true, and W would not be a subspace. So, the first condition is satisfied.

2. Closure under Addition

This is where we test if W is closed under addition. Suppose we have two arbitrary vectors, u and v, that belong to W. We need to show that their sum, u + v, also belongs to W. The process is as follows:

  1. Define Two Arbitrary Vectors in W: Represent u and v with general components (e.g., u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂)). Remember, these vectors must satisfy the conditions that define W.
  2. Add the Vectors: Calculate u + v by adding their corresponding components: u + v = (x₁ + x₂, y₁ + y₂, z₁ + z₂).
  3. Check if the Sum Satisfies the Conditions of W: Substitute the components of u + v into the equation or conditions that define W. If the equation or conditions are satisfied, u + v is also in W. If not, W is NOT a subspace.

For example, if W is defined by x + y - z = 0, then we know that for u = (x₁, y₁, z₁), x₁ + y₁ - z₁ = 0. Similarly, for v = (x₂, y₂, z₂), x₂ + y₂ - z₂ = 0. Now, let's check if the sum, u + v = (x₁ + x₂, y₁ + y₂, z₁ + z₂), also belongs to W. Substitute the components of u + v into the equation: (x₁ + x₂) + (y₁ + y₂) - (z₁ + z₂) = (x₁ + y₁ - z₁) + (x₂ + y₂ - z₂) = 0 + 0 = 0. Since the sum satisfies the equation, u + v is in W, and the second condition is satisfied.

3. Closure under Scalar Multiplication

Finally, we need to check if W is closed under scalar multiplication. Take any vector u in W and any scalar c. We need to prove that cu is also in W.

  1. Define an Arbitrary Vector in W: As before, use general components to represent u (e.g., u = (x, y, z)). Remember that u must satisfy the conditions for W.
  2. Multiply by a Scalar: Multiply u by the scalar c: cu = (cx, cy, cz).
  3. Check if the Result Satisfies the Conditions of W: Substitute the components of cu into the equation or conditions that define W. If the equation or conditions are satisfied, cu is in W. If not, W is NOT a subspace.

For example, using the same W defined by x + y - z = 0. We know that for u = (x, y, z), x + y - z = 0. Now, let's consider cu = (cx, cy, cz). Substitute these components into the defining equation: cx + cy - cz = c(x + y - z) = c(0) = 0. Since the result satisfies the defining equation, cu is in W, meaning the third condition is met.

Example: Let's Do it for a Specific W!

Let's work through an example to solidify our understanding. Suppose W is defined as the set of all vectors (x, y, z) in R³ such that x - 2y + z = 0. Let's demonstrate that W is a subspace.

1. Zero Vector Check:

Substitute (0, 0, 0) into the defining equation: 0 - 2(0) + 0 = 0. The equation holds true! Therefore, the zero vector is in W.

2. Closure under Addition:

Let u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂) be two arbitrary vectors in W. Because they are in W, we know that x₁ - 2y₁ + z₁ = 0 and x₂ - 2y₂ + z₂ = 0. Now, let's add them: u + v = (x₁ + x₂, y₁ + y₂, z₁ + z₂). Let's check if this sum is in W:

(x₁ + x₂) - 2(y₁ + y₂) + (z₁ + z₂) = (x₁ - 2y₁ + z₁) + (x₂ - 2y₂ + z₂) = 0 + 0 = 0.

Since the result is 0, the sum u + v is also in W.

3. Closure under Scalar Multiplication:

Let u = (x, y, z) be an arbitrary vector in W, so x - 2y + z = 0. Let's multiply u by a scalar c: cu = (cx, cy, cz). Now, check if this is in W:

cx - 2(cy) + cz = c(x - 2y + z) = c(0) = 0.

Since the result is 0, cu is also in W.

Conclusion:

Because W satisfies all three conditions, it is, therefore, a subspace of R³!

Common Pitfalls and Tips

Proving a set is a subspace can be a bit tricky. Here are some common pitfalls and helpful tips to keep in mind:

  • Don't assume: Don't assume that a set is a subspace just because it looks like it should be. Always rigorously check the three conditions.
  • Read the definition carefully: Make sure you fully understand the definition of W before you start. This is critical for applying the tests correctly.
  • Use general components: Always use general components (like x, y, z or x₁, y₁, z₁) to represent vectors. This allows you to prove that the conditions hold for all vectors in W, not just specific ones.
  • Be clear and organized: Write out each step clearly and methodically. This will help you avoid errors and make it easier to follow your logic.
  • Practice, practice, practice: The more examples you work through, the more comfortable you'll become with the process. Try different types of sets and definitions for W.

Wrapping Up

And there you have it! You've successfully learned how to prove that W is a subspace of R³. Remember the three key conditions – zero vector, closure under addition, and closure under scalar multiplication – and you'll be well on your way to mastering linear algebra. Keep practicing, and don't be afraid to ask for help if you get stuck. Happy vectoring, guys!