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By Satyanarayana V. Lokam

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**Example text**

We conclude that Volr (A) ≥ ( vr )r . On the other hand, consider the r-dimensional subspace V = v1 , . . , vr . Clearly, Rigr (A) ≤ dist(ai , V ). 3 Bounded Coeﬃcient Complexity of Linear Transformations 51 and imagining continuing the selection for one more step, we note that dist(ai , V ) ≤ vr+1 ≤ vr . Hence, we have Rigr (A) ≤ vr . It follows that Volr (A) ≥ ( vr )r ≥ (Rigr (A))r proving the lemma. ) is a natural mathematical measure. , numbers of arbitrary value and precision if working over R or C).

Then, (1) rank(A) = r if and only if σ1 (A) ≥ · · · ≥ σr (A) > σr+1 (A) = · · · = σn (A) = 0. (2) A 2F = σ12 (A) + · · · + σn2 (A). (3) A = σ1 (A). The following important inequality [38] is often useful. 4 (Hoﬀman-Wielandt Inequality). Let A and B be matrices in Cn×n . Then, n [σi (A) − σi (B)]2 ≤ A − B 2 F. i=1 Hoﬀman and Wielandt [38] prove their result for eigenvalues of normal matrices using the Birkhoﬀ–von Neumann characterization of doubly stochastic matrices. 3]. 2 43 Variants of Rigidity We will ﬁrst consider two variants of the rigidity function based on the 2 -distance to low-rank matrices.

13), we get, (n − 2s)n/2 ≤ nr + n/2 n/2 2 . 30 Matrix Rigidity Since n k ≤ (ne/k)k , Hence, (n − 2s)n/2 ≤ 2·n/2 (nr + n/2)e n/2 . (n − 2s) ≤ ((1 + 2r)e)2 This gives us, ≤ c · r2 , for some constant c > 0. (n − c · r2 ) s≥ . 2 Since RV (r) = |C| = n · s, we have proved the theorem. 3. 17 can be improved by a constant factor with a more careful choice of the parameters. Indeed, as pointed out by Tom Hayes (private communication), by selecting √ t ≈ ( 2ern)2/3 in the proof above, it can be shown that RV (r) ≥ n2 − O(n5/3 r2/3 ).