# Randomized Algorithm Lemma

Hello I am struggling with proving a lemma, it goes as follows:

Suppose we have a vector r = (r1....rn)^T where rj is either 0 or 1 which is selected uniformly at random with probability 1/2. Suppose now we have a n x n matrix called M (which has at least one non zero element).

I need to prove that the Pr[Mr = 0] <= 1/2(basically prove that the probability of one random position of the matrix M is zero is less than or equal 1/2).

Hint: We can assume that M11 ≠ 0. Based on the hypothesis, argue that the probability of the inner product of the first line of the matrix with the vector r being zero, is at most 1/2.

Also with the help of the above lemma give a randomized algorithm that can check if 3 matrices n x n A,B,C satisfy the relation AB = C in O(n^2).

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– D.W.
6 hours ago

To prove the claim

$$P(Mr = 0 ) \leq \frac{1}{2}$$ it is enough to prove it for any row vector $$m = (M_{i,1},...,M_{i,n})$$ that has at least one non-zero coefficient. Thus, we continue by induction on $$n$$.

$$n= 1$$ base case

The base case is trivial since

$$P(mr=0) = P(r=0) = \frac{1}{2}$$ in this case.

Induction case $$n \implies n+1$$.

If $$m$$ has only one nonzero coefficient then it reduces to the base case; otherwise, there are at least 2 non-zero coefficients say $$m_1,m_2$$ and we can condition on the two cases $$r_1 = 0,1$$ to get the result by applying the induction hypothesis.

The second part is given by the following algorithm:

1. Generate a uniform random binary vector $$r$$
2. Multiply $$Br$$ to get $$r_1$$
3. Multiply $$Ar_1$$ to get $$r_2$$
4. Multiply $$Cr$$ to get $$r_3$$
5. Subtract $$r_2-r_3$$ to get $$r_4$$
6. Return $$r_4$$

This algorithm has $$P(r_4 = 0 )>\frac{1}{2} \iff AB = C$$ and has complexity $$\mathcal{O}(n^2)$$.

• Hey thanks for the answer just a quick question. My lemma has the claim P(Mr = 0) <= 1/2 and not P(Mr= 0 ) >= 1/2 is that a typo of yours or not? May 8 at 13:49
• @kostger I think I fixed it. May 9 at 16:01