Sign rank and spectral gaps Sample Clauses

Sign rank and spectral gaps. The lower bound on the sign rank uses Xxxxxxx’x argument [31], who showed how to relate sign rank to spectral norm. He proved that if S is a sign (N × N )-matrix then We would like to apply Xxxxxxx’x theorem to the matrix S in our explicit examples. The spectral norm of S, however, is too large to be useful: if S is ∆ ≤ N/3 regular and x is the vector consisting entirely of 1s then Sx = (2∆ − N )x and so S ≥ N/3. Applying Xxxxxxx’x theorem to S yields that its sign rank is Ω(1), which is not informative. Our solution is based on the observation that Xxxxxxx’x argument actually proves a stronger statement. His proof works as long as the entries of the matrix are not too close to zero, as was already noticed in [32]. We therefore use a variant of the spectral norm of a sign matrix S which we call star norm and denote by12 S ∗ = min{M : Mi,jSi,j ≥ 1 for all i, j}. Three comments seem appropriate: (i) we do not think of the star norm as a norm; (ii) it is always at most the spectral norm, S ∗ ≤ S ; (iii) every M in the above minimum satisfies sign-rank(M ) = sign-rank(S). Theorem 22 (see [32]). Let S be a sign (N × N )-matrix. Then sign-rank(S) ≥ N . For completeness, in § 3.3.2 we provide a short proof of this theorem (which uses the main lemma from [31] as a ‘black box’). To get any improvement using this theorem we must have S ∗ S . It is not a priori obvious that there is a matrix S for which this holds. The following lemma shows that spectral gaps yield such examples.