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MathidiaMathematics Academy
Vector spaces
Lesson 23 of 75
Standard practice

Linear independence

Objective. Detect dependence and construct relations.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

A set of vectors {v₁,…,vₖ} is linearly independent if the only scalars satisfying c₁v₁+⋯+cₖvₖ=0 are c₁=⋯=cₖ=0; otherwise the set is dependent and some vector can be written as a combination of the others. To detect dependence, set the linear combination equal to the zero vector and solve the resulting homogeneous system — a nontrivial solution (not all zero) proves dependence, and substituting those scalars gives an explicit relation. For v₁=(1,2), v₂=(2,4), the equation c₁(1,2)+c₂(2,4)=(0,0) gives c₁+2c₂=0 and 2c₁+4c₂=0, both equivalent to c₁=−2c₂, so nonzero solutions exist (e.g. c₂=1, c₁=−2), proving dependence since v₂=2v₁. Independent sets have only the trivial solution.

Worked example

Determine whether the vectors v₁=(1,2) and v₂=(2,4) are linearly independent, and construct any dependence relation.

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Illustrated card showing 1 … 2
  1. 1Set c₁v₁+c₂v₂=(0,0): c₁(1,2)+c₂(2,4)=(0,0).
  2. 2This gives the equations c₁+2c₂=0 and 2c₁+4c₂=0, which are the same equation scaled by 2.
  3. 3Choose c₂=1, then c₁=−2, a nontrivial solution satisfying both equations.
  4. 4Since a nonzero solution exists, the vectors are linearly dependent, with relation −2v₁+v₂=0, i.e. v₂=2v₁.

Answer: Linearly dependent; v₂ = 2v₁

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Common mistake. Students conclude vectors are independent just because none is visibly a multiple of another by inspection, but three or more vectors can be dependent through a non-obvious combination, so the homogeneous system must actually be solved.

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