Span and generating sets
Objective. Describe spans and test membership constructively.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
The span of a set of vectors {v₁,…,vₖ} is the set of all linear combinations c₁v₁+⋯+cₖvₖ for scalars cᵢ, forming the smallest subspace containing those vectors. To test whether a vector w lies in the span, set up and solve the equation c₁v₁+⋯+cₖvₖ=w for the unknown scalars; a consistent system means w is in the span, an inconsistent one means it is not. For v₁=(1,0) and v₂=(0,1), span{v₁,v₂}=ℝ² since any (a,b)=a·v₁+b·v₂. For v₁=(1,1) and v₂=(2,2), span{v₁,v₂} is just the line y=x, since v₂=2v₁ adds no new direction. Checking whether generating vectors are multiples of each other reveals whether the span is smaller than expected.
Worked example
Determine whether the vector w=(5,1) lies in the span of v₁=(1,0) and v₂=(1,1).
- 1Set up c₁v₁+c₂v₂=w: c₁(1,0)+c₂(1,1)=(5,1).
- 2This gives the system c₁+c₂=5 and c₂=1.
- 3Substitute c₂=1 into the first equation: c₁+1=5, so c₁=4.
- 4Since c₁=4, c₂=1 is a consistent solution, w=4v₁+1v₂ lies in the span.
Answer: Yes, w = 4v₁ + v₂, so w is in the span
Concept mastery
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Practice
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Start practisingCommon mistake. Students assume any two nonzero vectors in ℝ² automatically span all of ℝ², but if one vector is a scalar multiple of the other, their span is only a line through the origin, not the whole plane.
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