Algebraic Methods in Combinatorics
نویسنده
چکیده
1. (A result of Bourbaki on finite geometries, from Răzvan) Let X be a finite set, and let F be a family of distinct proper subsets of X. Suppose that for every pair of distinct elements in X, there is a unique member of F which contains both elements. Prove that |F| ≥ |X|. Solution: Let X = [n] and F = {A1, . . . , Am}. We need to show that n ≤ m. Define the m × n incidence matrix A over R by putting 1 in the i-th row and j-th column if j ∈ Ai. Consider the product AA, which is an n× n matrix. For i 6= j, its entry at (i, j) is precisely 1. Also, the diagonal entries are strictly larger than 1, because if some element j ∈ X belongs to only one set Ak ∈ F , then the condition implies that every element i ∈ X is also in Ak, contradicting requirement that Ak be proper. Therefore, AA is nonsingular by Lemma 1, hence rank(AA) = n. But rank(AA) ≤ rank(A) ≤ m, so we are done.
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تاریخ انتشار 2008