A) U = (R x R, +_U, *_U)
where (a₁, a₂) +_U (b₁, b₂) = (a₁ + 2b₁, a₂ + 3b₂)
and *_U is the usual scalar multiplication

B) V = (R x R, +_V, *_V)
where (a₁, a₂) +_V (b₁, b₂) = (a₁ + b₁ + 1, a₂ + b₂)
and *_V is the usual scalar multiplication

C) W = (R x R, +_W, *_W)
where (a₁, a₂) +_W (b₁, b₂) = (a₁ + b₁ + 1, a₂ + b₂ + 1)
and r*_W(a₁, a₂) = (ra₁ + r – 1, ra₂ + r – 1)

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Exercise: show that if a vector space (over the real numbers) has at least two vectors, it has infinitely many.

Simple proof: If V has two vectors, one of them is nonzero. The span of such a vector is infinite because there are infinitely many real numbers, and its span is contained in V, so V is infinite.

Sharp-eyed student question: how do we know that aX and bX can’t be the same, for a, b distinct scalars and X a nonzero vector?

After a certain amount of trying to produce a counterexample, a quest I knew was doomed to fail but thought might be illuminating, I produced a two step proof of the infinitude of the span.

Lemma: If for some vector X there is a nonzero real number a such that aX = 0, then for any real number b, bX = 0.

Proof: Suppose X, a are as in the lemma. Since a is nonzero we may divide by it. Then for any scalar b, bX = ((b/a)a)X = (b/a)(aX) = (b/a)0 = 0.

Claim: If X is a nonzero vector and a, b are distinct real numbers, then aX, bX are distinct.

Proof: By the lemma, since 1X is not 0 (this is an axiom of vector spaces), no nonzero real number can give 0 when multiplied by X. Suppose a, b are real numbers that give the same product with X. Then by adding (-b)X to each side of the equality and factoring, we see (a-b)X = (b-b)X = 0, which implies a = b.

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