子空间证明疑问：能否通过张成性与线性无关性证明集合为子空间？

Great question—this is a super common mix-up when you’re first getting the hang of linear algebra, so let’s unpack this clearly.

Short Answer #

No, you can’t prove a set is a subspace by showing it’s linearly independent and has a spanning property. These two traits describe a basis for a subspace, not the subspace itself.

Let’s Break Down the Definitions to See Why #

First, let’s align on key terms to avoid confusion:

• A subspace is a subset of a vector space that acts like a vector space on its own. It requires three core properties:
  • It contains the zero vector (or is non-empty—if it’s non-empty and closed under the next two rules, the zero vector is automatically included)
  • It’s closed under addition: if you take any two vectors in the subset, their sum is also in the subset
  • It’s closed under scalar multiplication: if you take any vector in the subset and multiply it by any scalar, the result stays in the subset
• A linearly independent set is one where no vector can be written as a linear combination of the others—you can’t "build" any vector in the set using the rest.
• A spanning set for a space W is a collection of vectors whose linear combinations can create every vector in W. Crucially, the set itself isn’t W—it’s just the "building blocks" that generate W.

The Core Issue: A Basis ≠ A Subspace #

A set that’s both linearly independent and spanning is a basis for some subspace. But the basis isn’t the subspace—it’s the minimal set of vectors that produce the subspace. Let’s use a concrete example to make this obvious:
Take the 2D vector space ℝ². The set {(1,0), (0,1)} is linearly independent (you can’t make (1,0) from (0,1) or vice versa) and spans all of ℝ² (every vector (a,b) can be written as a*(1,0) + b*(0,1)). But this set is not a subspace:

• It doesn’t contain the zero vector (0,0)
• Adding (1,0) and (0,1) gives (1,1), which isn’t in the set (breaks addition closure)
• Multiplying (1,0) by 3 gives (3,0), which also isn’t in the set (breaks scalar multiplication closure)

The subspace here is all of ℝ²—the basis set just generates it, but isn’t the subspace itself.

The Correct Way to Prove a Set is a Subspace #

Stick to the formal definition. You have two equivalent approaches:

1. Verify three conditions:
   • The set is non-empty (the simplest way is to show the zero vector is in it)
   • Closure under addition: For any two vectors u, v in the set, u + v is also in the set
   • Closure under scalar multiplication: For any vector u in the set and any scalar c, c*u is also in the set
2. Verify the set is non-empty and closed under linear combinations: For any scalars c₁, c₂ and vectors u₁, u₂ in the set, c₁*u₁ + c₂*u₂ is in the set (this wraps addition and scalar multiplication into one step)

Quick Recap #

• Linear independence + spanning = a basis (building blocks for a subspace), not the subspace itself.
• To confirm a set is a subspace, you need to check closure properties (and non-emptiness), not whether it’s independent or spanning.

内容的提问来源于stack exchange，提问作者undergrad