Vector Subspace Verification Is S = {(x, Y) ∈ R² | Y = 2x} A Subspace Of R²

Hey guys! Ever wondered if a particular set forms a vector subspace within a larger vector space? Today, we're diving deep into this concept with a specific example. We'll be examining the set S = {(x, y) ∈ R² | y = 2x} and meticulously checking if it qualifies as a vector subspace of R². To do this, we'll go through the three essential conditions that any subset must satisfy to be considered a vector subspace. So, buckle up and let's get started!

Understanding Vector Subspaces

Before we jump into our specific example, let's quickly recap what a vector subspace actually is. In simple terms, a vector subspace is a subset of a vector space that itself forms a vector space under the same operations (addition and scalar multiplication). This means that the subset must be "self-contained" – performing addition or scalar multiplication on its elements should always result in elements that are still within the subset. For a subset to be a vector subspace, it needs to satisfy three crucial conditions, and we will explore them in detail as we analyze our set S. Keeping these conditions in mind will help us determine whether the given set S = {(x, y) ∈ R² | y = 2x} is a vector subspace. Understanding these principles is fundamental to grasping more advanced concepts in linear algebra, making this exploration both practical and insightful.

The Three Conditions for a Vector Subspace

Okay, so what are these magical three conditions? A subset S of a vector space V is a vector subspace if it meets these criteria:

1. Closure under addition: If u and v are in S, then their sum (u + v) must also be in S. This basically means that when you add two vectors within the subset, the resulting vector should still be a member of that subset. Imagine it like this: if S is a room, and u and v are people inside that room, then when they combine (addition), the resulting "combined person" should still be in the room (S).
2. Closure under scalar multiplication: If a is a scalar (a real number) and u is in S, then a*u must also be in S. This condition ensures that if you scale a vector within the subset by any real number, the scaled vector remains within the subset. Think of it as stretching or shrinking a person (u) in our room (S); they should still fit inside the room after the stretching or shrinking.
3. The zero vector: The zero vector (0) must be in S. This is a fundamental requirement. The zero vector acts as the additive identity in a vector space, and its presence is crucial for the subspace to maintain its vector space structure. In our room analogy, this means there must be a "zero person" present in the room (S).

These three conditions are the bedrock of vector subspace identification. If a subset fails even one of these conditions, it cannot be a vector subspace. Let's keep these conditions at the forefront as we dissect our set S = {(x, y) ∈ R² | y = 2x} and determine its fate.

Defining Our Set S: A Line in R²

Now, let's get down to business and analyze our specific set: S = {(x, y) ∈ R² | y = 2x}. What does this set represent? Well, it's the set of all points (x, y) in the real plane (R²) that satisfy the equation y = 2x. If you remember your high school algebra, this is the equation of a straight line passing through the origin (0, 0) with a slope of 2. Visualizing this line is crucial because it gives us a geometric intuition about whether S could be a vector subspace. Think about it: lines through the origin often have the properties we associate with vector subspaces, but we still need to rigorously check the three conditions.

In the context of vector spaces, the points (x, y) are vectors. So, S is essentially a collection of vectors in R², all lying on this specific line. Our mission is to determine whether this collection forms a vector subspace of R². To do this, we must meticulously examine whether S satisfies the three conditions we discussed earlier: closure under addition, closure under scalar multiplication, and the presence of the zero vector. By systematically checking each condition, we can confidently conclude whether S is indeed a vector subspace or not. The geometric interpretation of S as a line through the origin provides a helpful visual aid, but the algebraic verification using the three conditions is the definitive test.

Condition 1: Closure Under Addition

The first condition we need to tackle is closure under addition. This means we need to show that if we take any two vectors in S and add them together, the resulting vector is also in S. Let's pick two arbitrary vectors from S. Since S = {(x, y) ∈ R² | y = 2x}, we can represent these vectors as u = (x₁, y₁) and v = (x₂, y₂), where y₁ = 2x₁ and y₂ = 2x₂. Remember, this condition (y = 2x) is what defines the elements of our set S.

Now, let's add these vectors: u + v = (x₁ + x₂, y₁ + y₂). To prove closure under addition, we need to demonstrate that this resulting vector (u + v) also belongs to S. In other words, we need to show that the y-component of the sum (y₁ + y₂) is equal to twice the x-component (x₁ + x₂). Let's check: y₁ + y₂ = 2x₁ + 2x₂ (since y₁ = 2x₁ and y₂ = 2x₂). We can factor out a 2 from the right side: 2x₁ + 2x₂ = 2(x₁ + x₂). This confirms that y₁ + y₂ = 2(x₁ + x₂), which means the vector (u + v) = (x₁ + x₂, y₁ + y₂) satisfies the condition y = 2x and therefore belongs to S. So, we've successfully shown that S is closed under addition! This is a crucial step in verifying that S is a vector subspace. By meticulously breaking down the vector addition and utilizing the defining condition of S, we've provided a clear and rigorous proof of closure under addition.

Condition 2: Closure Under Scalar Multiplication

Next up, we need to check for closure under scalar multiplication. This condition requires us to prove that if we multiply any vector in S by a scalar (a real number), the resulting vector is still in S. Let's take an arbitrary vector u from S, which we can represent as u = (x, y), where y = 2x (again, this is the defining condition of S). Now, let's multiply this vector by an arbitrary scalar a (a real number): a*u = a(x, y) = (ax, ay).

To show closure under scalar multiplication, we need to demonstrate that the resulting vector (ax, ay) also belongs to S. This means we need to verify that its y-component (ay) is equal to twice its x-component (ax). Let's see: ay = a(2x) (since y = 2x). We can rewrite this as ay = 2(ax). Voila! We've shown that the y-component (ay) is indeed twice the x-component (ax). Therefore, the vector a*u = (ax, ay) satisfies the condition y = 2x and belongs to S. This proves that S is closed under scalar multiplication. We're now two-thirds of the way through proving that S is a vector subspace! The meticulous step-by-step approach, applying the definition of S and scalar multiplication, has allowed us to confidently confirm this crucial condition.

Condition 3: The Zero Vector

Finally, the last condition we need to check is whether the zero vector (0, 0) is an element of S. This might seem like a trivial condition, but it's absolutely essential for S to be a vector subspace. Remember, the zero vector is the additive identity, and its presence is fundamental to the vector space structure.

So, let's see if (0, 0) belongs to S. Our set S is defined as S = {(x, y) ∈ R² | y = 2x}. To check if (0, 0) is in S, we need to substitute x = 0 and y = 0 into the defining equation y = 2x. Does it hold true? Let's plug it in: 0 = 2(0). Yes, 0 = 0! This confirms that the zero vector (0, 0) satisfies the condition y = 2x and is therefore an element of S.

We've successfully verified that the zero vector is indeed present in S. This completes our checklist of the three essential conditions for a vector subspace. With this final confirmation, we can confidently draw our conclusion about whether S is a vector subspace or not. The simplicity of this condition doesn't diminish its importance; it's a cornerstone of the vector subspace definition, and we've now shown that S clears this hurdle with ease.

Conclusion: S is a Vector Subspace!

Alright guys, let's recap! We started with the set S = {(x, y) ∈ R² | y = 2x}, which represents a line passing through the origin in the real plane R². Our mission was to determine if S is a vector subspace of R². To do this, we systematically checked the three essential conditions:

• Closure under addition: We proved that if u and v are in S, then u + v is also in S.
• Closure under scalar multiplication: We showed that if a is a scalar and u is in S, then a*u is also in S.
• The zero vector: We confirmed that the zero vector (0, 0) is an element of S.

Since S satisfies all three conditions, we can confidently conclude that S is a vector subspace of R²! Woohoo! We've successfully navigated through the definition of a vector subspace and applied it to a specific example. This exercise not only reinforces our understanding of vector subspaces but also demonstrates the power of rigorous mathematical verification. So, the next time you encounter a set and wonder if it's a vector subspace, remember these three conditions and you'll be well-equipped to tackle the problem! Understanding these fundamental concepts is key to unlocking more advanced topics in linear algebra and beyond. Keep exploring, keep questioning, and keep learning!