The concept of a subspace is fundamental in linear algebra, representing a subset of a vector space that itself forms a vector space under the same operations. Determining whether a set is a subspace requires verifying specific properties. This article delves into the process of proving a subset is a subspace of a vector space, outlining the key steps and illustrating them with examples.
Understanding Subspaces
Before embarking on proofs, let's solidify our understanding of subspaces. A subset W of a vector space V is considered a subspace if it satisfies the following three conditions:
- Zero Vector: The zero vector of V must be an element of W.
- Closure under Addition: For any two vectors u and v in W, their sum u + v must also be in W.
- Closure under Scalar Multiplication: For any scalar c and vector u in W, the scalar multiple cu must also be in W.
These conditions ensure that W maintains the essential properties of a vector space within the larger context of V.
Steps for Proving a Subspace
To rigorously prove that a given subset is a subspace, follow these steps:
-
Identify the Vector Space: Begin by explicitly stating the vector space V that the subset belongs to. This clarifies the operations (addition and scalar multiplication) under which we are working.
-
Define the Subset: Precisely define the subset W. This might involve describing it using set-builder notation or providing a specific rule that characterizes its elements.
-
Verify Closure under Addition: Take two arbitrary vectors u and v in W. Demonstrate that their sum u + v also belongs to W. This involves applying the definition of W and the vector space operations.
-
Verify Closure under Scalar Multiplication: Choose an arbitrary scalar c and an arbitrary vector u in W. Show that the scalar multiple cu is also an element of W. Again, utilize the definition of W and vector space operations.
-
Verify the Zero Vector: Show that the zero vector of V is included in W. This might be straightforward or require applying the definition of W.
-
Conclude: After verifying all three conditions, state that since W satisfies the necessary properties, it is a subspace of V.
Illustrative Examples
Let's solidify these steps with a few examples:
Example 1:
- Vector Space: V = ℝ², the set of all 2-dimensional real vectors.
- Subset: W = {(x, y) ∈ ℝ² | x + 2y = 0}
Proof:
- We are working with the vector space ℝ².
- W consists of all vectors in ℝ² whose components satisfy the equation x + 2y = 0.
- Closure under Addition: Let u = (x₁, y₁) and v = (x₂, y₂) be any two vectors in W. This means x₁ + 2y₁ = 0 and x₂ + 2y₂ = 0. Their sum is u + v = (x₁ + x₂, y₁ + y₂). We need to show that this sum also satisfies the condition for W: (x₁ + x₂) + 2(y₁ + y₂) = (x₁ + 2y₁) + (x₂ + 2y₂) = 0 + 0 = 0. Therefore, u + v is in W.
- Closure under Scalar Multiplication: Let c be any scalar and u = (x₁, y₁) be in W. This means x₁ + 2y₁ = 0. The scalar multiple cu = (cx₁, cy₁). We need to show that this satisfies the condition for W: (cx₁) + 2(cy₁) = c(x₁ + 2y₁) = c(0) = 0. Therefore, cu is in W.
- Zero Vector: The zero vector of ℝ² is (0, 0), which satisfies the condition 0 + 2(0) = 0. Therefore, the zero vector is in W.
- Conclusion: Since W satisfies all three conditions, it is a subspace of ℝ².
Example 2:
- Vector Space: V = ℝ³, the set of all 3-dimensional real vectors.
- Subset: W = {(x, y, z) ∈ ℝ³ | x + y + z = 1}
Proof:
- We are working with the vector space ℝ³.
- W consists of all vectors in ℝ³ whose components satisfy the equation x + y + z = 1.
- Closure under Addition: Let u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂) be any two vectors in W. This means x₁ + y₁ + z₁ = 1 and x₂ + y₂ + z₂ = 1. Their sum is u + v = (x₁ + x₂, y₁ + y₂, z₁ + z₂). We need to show that this sum also satisfies the condition for W: (x₁ + x₂) + (y₁ + y₂) + (z₁ + z₂) = (x₁ + y₁ + z₁) + (x₂ + y₂ + z₂) = 1 + 1 = 2. However, this does not equal 1, meaning the sum u + v is not in W.
Since W does not satisfy closure under addition, we can conclude that it is not a subspace of ℝ³.
Conclusion
Proving a subset is a subspace of a vector space requires verifying three fundamental conditions: closure under addition, closure under scalar multiplication, and the presence of the zero vector. These steps ensure that the subset inherits the crucial properties of a vector space, allowing for meaningful mathematical operations and analysis within the larger context of the original vector space. By applying this rigorous framework, we can confidently determine whether a subset forms a subspace, paving the way for further exploration and application within linear algebra.
