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32 Proof. Sketch: Each point imposes conditions on the set (IZ )d . If every point takes away one degree of freedom, then we will lose r dimensions. Of course, the conditions imposed by the points might very well be redundant, so we don’t necessarily have equality. We will now do several examples to familiarize you with the abstraction above so that you feel more at ease. Suppose we are looking for polynomials that vanish on the set Z = {[1, 0, 1], [0, 1, 1]}. Clearly no nonzero form of degree 0 can vanish on both of these points, so dim(IZ )0 = 0.

Vr , we can find a vector w such that w · vi = 0 for each i = 1, . . , r − 1, but w · vr = 0. The proof of this is simple: Consider the space U = Span(v1 , . . , vr−1 ). This is an r − 1 dimensional space. So U ⊥ , the set of all vectors perpendicular to U is an m − (r − 1) dimensional space. We may assume r < m. (proof for r = m is left to the reader) Then U ⊥ is at least a two dimensional vector space. Since (v1 , . . , vr ) is independent, vr is not in U so we can choose w in U ⊥ not perpendicular to vr .

Of course, the conditions imposed by the points might very well be redundant, so we don’t necessarily have equality. We will now do several examples to familiarize you with the abstraction above so that you feel more at ease. Suppose we are looking for polynomials that vanish on the set Z = {[1, 0, 1], [0, 1, 1]}. Clearly no nonzero form of degree 0 can vanish on both of these points, so dim(IZ )0 = 0. ) Next we try to tackle forms of degree 1. Forms of degree 1 look like f = AX + BY + CZ. If they vanish at [1, 0, 1] and [0, 1, 1], then we get the system of equations A+C =0 B + C = 0.

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