By Satyanarayana V. Lokam
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For A ∈ Fm×n , ρA (s) ≥ rank A − dA (s). Proof. Let C be a matrix achieving ρA (s), so rank(A − C) = ρA (s) and each row of C has at most s nonzero entries. Trivially, C + A = C + A − C . Considering the left-hand side, dim( C + A ) = dim A + dim C − dim( A ∩ C ). 6 Paturi–Pudl´ ak Dimensions 37 Considering the right-hand side, dim( C + A − C ) ≤ dim C + dim A − C = dim C + ρA (s). Thus, we have dim A − dim( A ∩ C ) ≤ ρA (s). Since dim A = rank(A) and dA (s) ≥ dim( A ∩ C ), we are done. 29. For any finite-dimensional subspace V , dV (s) + DV (s) ≥ 2 dim V.
Hence, we may assume that Rigr (L ) is no more than n for some small constant > 0. Raz uses this √ to show the existence of a matrix Y such that: (i) Rigr (Y ) = Ω( n) for r = Ω(n), and (ii) ∀j, |Lj (Y )| ≤ O(n log n) for 0 < < 1. To see the crucial role of this matrix, observe that if we fix Y , the circuit B computes a linear transformation on the matrix X and in fact becomes a linear circuit BY . Moreover, it is easy to see that the linear transformation computed by this circuit is given by I ⊗ Y and a lower bound on Rigr (Y ) from (i) implies one on Rignr (I ⊗ Y ).
Let s denote the average of the si , s := (s1 + · · · + sn )/n. Thus, |C| = n · s. There must be at least n/2 rows of V with no more than 2s changes in each of those rows. Fix n/2 such rows and call them “good” rows. ) We claim that Dn/2 (V − C) ≥ (n − 2s)n/2 . 12) To prove this claim, consider the products formed by taking unchanged entries of good rows, one entry per row. They are of the form xji11 xji22 · · · xjitt , where i1 , . . , it are good rows and each jk has at least (n − 2s) possibilities.