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Part CCSIR NET December 2025substitute-the-relation-into-a-general-linear-combination-and-let-independence-of-u-v-x-y-force-every-coefficient

Substitute the relation into a general linear combination and let independence of u v x y force every coefficient

Let V be a real vector space. Suppose that {u,v,w,x,y} ⊆ V is a spanning set of V and that {u,v,x,y} is linearly independent. Which of the following statements are necessarily true?

  1. A.The dimension of V is 4 or 5.✓
  2. B.If 2u−3v+5w=0, then {v,w,x,y} is a basis of V.✓
  3. C.If u−6v+7w=0, then the span of {v,w,x} is a 2-dimensional vector space.
  4. D.If u+w=v+x, then {u,v,w,y} is a basis of V.✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: An injective linear operator on a vector space is surjective

More on this topic

The chapter behind this: Bases, dimension and rank — free to read

From Vector Spaces and Linear Maps › Bases, dimension, rank–nullity

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