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We have two linear codes $C_1,C_2$ with vectors from $\mathbb{F}_2^n$ such that the minimum Hamming distance between the elements from $C_i$ is $d_i$. Then for $C_1\cap C_2$ this distance is $d_0$ and for $C_1+C_2$ it's $d_3$.

The problem is to prove that $d$, the minimum distance for the elements from $\{(a+b,a'+b,a+a'+b):a,a'\in C_1, b\in C_2\}$ satisfies $d\geq\min (d_0,2d_1,3d_3)$.

I'm not really sure how much this problem has to do with coding theory because for me it seems to be more related to linear algebra. Anyways, what I thought of so far is that if $a,a'\in C_1\implies a+a'\in C_1$ and hence $a+b,a'+b,a+a'+b\in C_1+C_2$. This gives us $d\geq 3d_3$.

Sadly, I'm still clueless about the rest of the inequalities. Any ideas, hints?

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    Hint: In a linear code, the minimum distance equals the minimum weight. So, what is the weight of the codeword if $a^\prime = b = \mathbf 0$? What if $a=a^\prime=\mathbf 0$? – 2017-02-18
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    Oh, I didn't realize that, thank you. So, for $a'=b=0$ I get $d\geq 2d_1$ as my codewords are of the form $(a,0,a)$. But for $a=a'=0$ we get $(b,b,b)$ which would be equivalent to saying $d\geq 3d_2$ and not $d_0$. – 2017-02-18

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