. Au-préalable, L. On-note-que-les-noyaux-de, and L. Sont, ?n+2D ) En effet, la probabilité que deux sous-espaces aléatoires de dimension d et d ? soient d'intersection de dimension j est d'après le lemme 2 de l'ordre de 2 ?j(n?d?d ? +j) , et donc celle que ces deux sous-espaces soient d'intersection non-nulle est de l'ordre de 2 ?n+d+d ? = O(2 ?n+2D ) Quitte à remplacer ? ++ 2,D (d, d ? ) par son intersection avec la condition ker(L) ? ker(L ? ) = {0} dont la valeur est la même à des termes d'ordre 2 ?n+2D près, nous pouvons donc supposer que cette condition est réalisée, Nous estimons maintenant le nombre ? ++ 2) = n ? r dim(ker(L) ? K) = i dim(ker(L ? ) ? K) = i ? et a, b ? K où ker(L) et ker(L ? ) sont d'intersection zéro. L'intersection I de ker(L) et K est un sous-espace de dimension contenant a : il y a donc

. Choix-pour-un-tel-sous-espace and . De-même, E(d ? , i ? ) choix pour l'intersection I ? de ker(L ? ) et K. Le sous-espace somme des intersections de K avec ker(L) et ker(L ? ) est un sous-espace de dimension i + i ? de K. Le nombre de sous-espaces de dimension n ? r d'intersection I avec ker(L) et I ? avec ker

. Démonstration, Une borne supérieure du nombre cherché est le nombre de sousespaces de dimension n ? r contenant I + I ? . Le nombre de tels sous-espaces est S(n, n ? r) S(n, i + i ? )

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