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How do you prove that the set of vectors satisfying x₃=x₁+x₂ is a subspace of ℝ³?

To prove that the set W={(x1,x2,x3)∈R3∣x3=x1+x2}W = \{(x_1, x_2, x_3) \in \mathbb{R}^3 \mid x_3 = x_1 + x_2\} is a subspace, you must verify three conditions: nonemptiness (containing the zero vector), closure under vector addition, and closure under scalar multiplication.

Conditions

  • The set is defined by the linear equation x3=x1+x2x_3 = x_1 + x_2 in R3\mathbb{R}^3.
  • The field of scalars is R\mathbb{R}.

Reasoning, step by step

  1. Verify nonemptiness by checking if the zero vector (0,0,0)(0,0,0) satisfies the defining equation.
  2. Verify closure under addition by taking two arbitrary vectors x,y∈Wx, y \in W and showing that x+y∈Wx+y \in W.
  3. Verify closure under scalar multiplication by taking an arbitrary vector x∈Wx \in W and scalar c∈Rc \in \mathbb{R} and showing that cx∈Wcx \in W.
  4. Conclude that since all three criteria are met, WW is a subspace.

Example

The script states: 'The vector (0,0,0) satisfies x₃=x₁+x₂, so the set is nonempty. This is the first required condition.' It further shows: 'If x₃=x₁+x₂ and y₃=y₁+y₂, then x₃+y₃=(x₁+y₁)+(x₂+y₂)' and 'Multiplying x₃=x₁+x₂ by any real scalar gives cx₃=cx₁+cx₂'.

Common misconceptions

  • Forgetting to check for nonemptiness or the zero vector.
  • Assuming that satisfying one closure property implies the other.
  • Confusing the subspace criteria with the axioms of a vector space itself.

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