Posted by: atri | February 2, 2009

Lecture 9: Gilbert-Varshamov Bound

In today’s class we proved the Gilbert-Varshamov bound. The relevant lecture notes from fall 07 is lecture 15.

In class today I made the following claim without proof: given a non-zero vector \mathbf{m}\in\mathbb{F}_q^k and a random k\times n matrix \mathbf{G} over \mathbb{F}_q, the vector \mathbf{m}\cdot\mathbf{G} is uniformly distributed over \mathbb{F}_q^n. I realized later that this might not be so obvious if you have not fiddled with finite fields much. So below is a proof of this statement.

Let the (j,i)th entry in \mathbf{G} (1\le j\le k, 1\le i\le n) be denoted by g_{ji}. Note that as \mathbf{G} is a random k\times n matrix over \mathbb{F}_q, each of the g_{ji} is a uniformly random element from \mathbb{F}_q. Now, note that we would be done if we can show that for every 1\le i\le n, the ith entry in \mathbf{m}\cdot\mathbf{G}  (call it b_i) is a uniformly random element from \mathbb{F}_q. To finish the proof, we prove this latter fact. If we denote \mathbf{m}=(m_1,\dots,m_k), then b_i=\sum_{j=1}^k m_jg_{ji}. (Recall that the g_{ji}‘s are uniformly random elements from \mathbb{F}_q.) The rest of the proof is a generalization the argument we used in class to show that every non-zero codeword in {\sc Had}_r has Hamming weight 2^{r-1} to the general \mathbb{F}_q.  

Note that to show that b_i is uniformly distributed over \mathbb{F}_q, it is sufficient to prove that b_i takes every value in \mathbb{F}_q equally often over all the choices of values that can be assigned to g_{1i},g_{2i},\dots,g_{ki}. Now as \mathbf{m} is non-zero, at least one of the its element is non-zero: with loss of generality assume that m_1\neq 0. Thus, we can write b_i= m_1g_{1i}+\sum_{j=2}^k m_jg_{ji}. Now for every fixed assignment of values to g_{2i},g_{3i},\dots,g_{ki} (note that there are q^{k-1} such assignments), b_i takes a different value for each of the q distinct possible assignments to g_{1i} (this is where we use the assumption that m_{1i}\neq 0). Thus, over all the possible assignments of g_{1i},\dots,g_{ki}, b_i takes each of the values in \mathbb{F}_q exactly q^{k-1} times, which proves our claim.

Please use the comments section for any questions you might have about today’s lecture or the proof above.

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