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- This is trivial when . If we can prove it for all , then by induction we have proved it for all . Thus it remains to prove it for . This we prove by induction on .
Base case: , trivial.
Induction case. Represent . If some , then the proof is finished. So assume all
If is linearly dependent, then we can use induction on its linear span to eliminate one nonzero term in , and thus eliminate it in as well.
Else, there exists , such that . Then we can interpolate a full interval of representations:
If for all , then set . Otherwise, let be the smallest such that one of . Then we obtain a convex representation of with nonzero terms. (en)
- For any , represent for some , then , and we use the lemma.
The second part reduces to the first part by "lifting up one dimension", a common trick used to reduce affine geometry to linear algebra, and reduce convex bodies to convex cones.
Explicitly, let , then identify with the subset . This induces an embedding of into .
Any , by first part, has a "lifted" representation where at most of are nonzero. That is, , and , which completes the proof. (en)
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