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# nov16 - U is free on v v we may deﬁne f f U → F by f v...

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Math 5125 Wednesday, November 16 November 16, Ungraded Homework Exercise 10.4.12 on page 376 Let V be a vector space over the field F and let v , v be nonzero elements of V . Prove that v v = v v in V F V if and only if v = av for some a F . Suppose we do not have v = av for some a F . Then v , v are linearly independent and therefore span a subspace U of V which has dimension two. Write V = U W where W is a subspace of V (every subspace of a vector space has a direct complement; for infinite dimensional vector spaces this requires the axiom of choice). Since
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Unformatted text preview: U is free on v , v , we may deﬁne f , f : U → F by f ( v ) = f ( v ) = 1 and f ( v ) = f ( v ) = 0. We may extend f , f to the whole of V by deﬁning f ( w ) = f ( w ) = 0 for all w ∈ W . Then f ⊗ f ( v ⊗ v ) = 1 and f ( v ⊗ v ) = 0. Applying f ⊗ f to v ⊗ v = v ⊗ v , we deduce that 1 = 0 which is a contradiction and the result is proven....
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