By Satyanarayana V. Lokam

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**Example text**

13, Volr (Y ) ≥ (α n)r for r = 3 n. Hence, Vol 3 n2 (I ⊗ Y ) ≥ √ 2 (α n) 3 n . 17, √ size(C ) ≥ log Vol 3 n2 (I ⊗ Y ) ≥ 3 n2 log(α n) ≥ c4 n2 log n, for a constant c4 . By choosing the parameters, we can ensure that c4 > c3 . It follows that size(C) ≥ size(C ) − c3 n2 log n ≥ cn2 log n. 20. Let L1 , . . , Lk be linear forms on Cn as above and 2 L ∈ Cn ×k be the matrix with Lj as its columns. Fix parameters , n×n (also viewed as a 1 , c1 and c2 . Then, there exists a matrix Y ∈ C 2 n vector in C ) such that (1) For each j, 1 ≤ j ≤ k, |Lj (Y )| ≤ c1 Rig n2 (L √ ) 2 ln k + 4.

This by itself, however, cannot be directly applied since we will have to do this for all multiplication gates which might be as many as the size of B itself. Instead, Raz does the following: Let θ be the maximum (in absolute value) among all the scalars |Lj (Y )| to be multiplied. Replace the scalar multiplication by Lj (Y ) at the inputs to the third part with a multiplication by Lj (Y )/θ, which is now bounded by 1 in absolute value. Now, (eﬀectively) multiply each of the outputs (there are only n2 of them) by the scalar θ using O(log θ) additions and a multiplication by a scalar of value at most 1.

16. Any synchronous linear circuit with coeﬃcients bounded by c computing the linear transformation x → Ax must have at least 2e ln det A/c2 edges. 5. 17. Let C be a linear circuit with coeﬃcients bounded by θ ≥ 1, computing a linear transformation x → A∗ x. Then, for 1 ≤ r ≤ n, (1) size(C) = Ω(log2θ Volr (A)). (2) size(C) = log2θ msvr (A) − O(n). Proof. Let s := size(C). Sort the gates of C in a topological order and let gi denote the ith gate in this order where g−n+1 , . . , g0 are the input nodes and g1 , .

### Complexity Lower Bounds using Linear Algebra (Foundations and Trends in Theoretical Computer Science) by Satyanarayana V. Lokam

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