Determining whether a subset of a vector space is a subspace involves checking three crucial conditions. Understanding these conditions is fundamental to linear algebra and crucial for many applications. This guide will walk you through the process clearly and concisely.
Understanding Subspaces
Before diving into the tests, let's clarify what a subspace is. A subspace is a subset of a vector space that is itself a vector space under the same operations (addition and scalar multiplication) as the parent vector space. This means it must be "closed" under these operations. This "closed" property is the key to identifying subspaces.
The Three Subspace Tests
To determine if a subset W of a vector space V is a subspace, you must verify these three conditions:
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The zero vector is in W: The zero vector (the vector where all components are zero) must be an element of the subset W. If the zero vector isn't in W, it cannot be a subspace.
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Closure under addition: For any two vectors u and v in W, their sum (u + v) must also be in W. This ensures that adding vectors within the subset keeps you within the subset.
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Closure under scalar multiplication: For any vector u in W and any scalar c, the scalar multiple (c*u) must also be in W. This ensures that multiplying a vector in the subset by a scalar keeps you within the subset.
Working Through Examples
Let's illustrate with examples. Suppose V is the vector space ℝ² (all two-dimensional real vectors).
Example 1: Is the set W = {(x, y) ∈ ℝ² | x + y = 0} a subspace of ℝ²?
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Zero vector: The zero vector (0, 0) satisfies 0 + 0 = 0, so it's in W.
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Closure under addition: Let u = (x₁, y₁) and v = (x₂, y₂) be in W. This means x₁ + y₁ = 0 and x₂ + y₂ = 0. Their sum is u + v = (x₁ + x₂, y₁ + y₂). Since (x₁ + x₂) + (y₁ + y₂) = (x₁ + y₁) + (x₂ + y₂) = 0 + 0 = 0, the sum is also in W.
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Closure under scalar multiplication: Let u = (x, y) be in W (so x + y = 0) and c be a scalar. Then c*u = (cx, cy). Since (cx) + (cy) = c(x + y) = c(0) = 0, the scalar multiple is also in W.
Conclusion: Since W satisfies all three conditions, it is a subspace of ℝ².
Example 2: Is the set W = {(x, y) ∈ ℝ² | x + y = 1} a subspace of ℝ²?
- Zero vector: The zero vector (0, 0) does not satisfy 0 + 0 = 1, so it's not in W.
Conclusion: Because the zero vector is not in W, W is not a subspace of ℝ². We don't even need to check the other conditions; failure on one condition is sufficient.
Key Takeaways and Further Exploration
Remember, all three conditions must be met for a subset to be a subspace. Failing any one of them is enough to disqualify the subset. Practice working through different examples, varying the vector spaces and subsets, to solidify your understanding of this fundamental concept in linear algebra. Consider exploring more complex vector spaces and subsets to challenge your understanding further.
