Coding theory - Linear codes

How can I prove the following :

1) Let C be a binary linear code in a binary vector space V. Let v be a vector not in C and consider the coset v+C. Show v+C is not a linear code.

2) suppose C is a linear code and that v+C is a coset of C. Suppose a and b are members of v+C. Prove a-b is a member of C.

What I know :

1) A coset of C is x + C = {x + y | y ∈ C}.

2) A linear code is a subspace of V(n, q) which is a union of disjoint cosets of C;

3) A linear code C is a code for which whenever x,y is in C then ax+by is in C for all a,b in Fq , that is C is a linear subspace of (Fq)^n

I understand that I would have to piece together the above three statements to prove both questions.

Any help or guidance is much appreciated.

• looks like homework – NaCl Jun 12 '16 at 9:39
• No, More like preparation for my final exam, it's a question that popped up in a past exam, something I don't want to regret not learning later on. – Keshav Chetty Jun 12 '16 at 9:41

For 1, ask yourself: can $0$ be in $v + V$, if $v \notin V$?
For 2, write $a = v + v_1$ for some $v_1 \in V$. Also $b = v + v_2$ for some $v_2 \in V$. Now what is $a - b$?
For 1, your task is simplified, since you are in the binary vector space domain. Now consider two words $v + k_1$ and $v + k_2$, where $k_1$ and $k_2$ are codewords in $C$. So the sum of the two words becomes $k_1 + k_2 = k'$ where $k' \in C$, and hence does not belong to $v + C$.
For 2, by similar reasoning that the difference of two codewords in $C$ is also a codeword, you can get that $a - b \in C$.
• Exploiting characteristic $2$ does not meaningfully simplify the argument, and somewhat obscures the essence: you could just as easily take the difference of the two words (which you already have done in part 2). – Erick Wong May 4 '17 at 18:48