In coding theory, How do we know a linear code have even weight?
For example, we have a linear code (12,3,6), how to identify each word of it get even weight?
[Math] How do we know a linear code have even weight
coding-theory
Related Solutions
It's sufficient to find a generating matrix.
A linear code is usually defined as a subspace of $F^n$ for some field $F$ (since you're talking about bits, you can take $F = \mathbb{F}_2 = \{0,1\}$). The code $C$ generated by a generating matrix $G$ is the span of the rows of $G$. The span of a set of vectors in $F^n$ is a subspace of $F^n$, so $C$ is a linear code.
Since the discussion in the comments between @GitGud and the OP seems to converging far too slowly to a solution, and since I wrote the answer wherein the statement in question occurred, here goes with a few hints for a different approach which will, along the way force you to learn some useful facts about binary vectors.
Suppose $\mathcal C$ denotes a linear binary code. Partition $\mathcal C$ into two subsets $\mathcal C_0$ and $\mathcal C_1$ consisting respectively of all the codewords of even Hamming weight and all the codewords of odd Hamming weight.
Show that for every linear binary code, $\mathcal C_0$ is a non-empty set. Hint: find a codeword of even weight in $\mathcal C$. (Subhint: $0$ is an even integer).
Explain why if $\mathcal C_1$ is an empty set, then we have proved part of the statement in question.
Digression:
Show that the sum of two binary vectors of even Hamming weight is a vector of even Hamming weight.
Show that the sum of two binary vectors of odd Hamming weight is also a vector of even Hamming weight.
Show that the sum of a binary vector of odd Hamming weight and a binary vector of even Hamming weight is a vector of odd Hamming weight.
End of digression
Suppose that $\mathcal C_1$ is non-empty and let $x$ denote a fixed element of $\mathcal C_1$.
Show that $x + \mathcal C_0 = \{z \colon z = x + y, y \in \mathcal C_0\}$ is a collection of $|\mathcal C_0|$ distinct vectors, all of which have odd Hamming weight. Argue that $x + \mathcal C_0 \subset \mathcal C_1$ and so it must be that $|\mathcal C_1| \geq |\mathcal C_0|$.
Show that $x + \mathcal C_1 = \{u \colon u = x + v, v \in \mathcal C_1\}$ is a collection of $|\mathcal C_1|$ distinct vectors all of which have even Hamming weight. Argue that $x + \mathcal C_1 \subset \mathcal C_0$ and hence $|\mathcal C_1| \leq |\mathcal C_0|$.
Conclude, if you dare, that either $\mathcal C_1 = \emptyset$ and so all the codewords in $\mathcal C$ have even Hamming weight, or that $\mathcal C$ contains codewords of even Hamming weight as well as odd Hamming weight and that $$|\mathcal C_1| = |\mathcal C_0| = \frac 12 |\mathcal C|.$$
Best Answer
In every linear binary code,
either
all the codewords have even weight
or
half the codewords have even weight and half have odd weight
So, if if your question is "How can we tell whether a linear binary code has codewords of even weight or not?" then you have nothing to worry about because all linear binary codes have codewords of even weight.
On the other hand, if your question is "How can we tell whether all the codewords in a linear binary code have even weight?" then some work is needed. One dead giveaway is that if by $[n,k,d]$ is meant a linear binary code of length $n$ with $2^k$ codewords and minimum distance $d$, then the code has codewords of odd weight whenever $d$ is odd. However, when $d$ is even, one cannot say that there are no codewords of odd weight. Consider, for example, the $[5,2,2]$ code with codewords $$00000, 11000, 00111, 11111$$ So, when $d$ is even, we need more information about the code to figure out whether the code has only even-weight codewords.
Here is a $[12,3,6]$ code in which all the codewords have even weight. its generator matrix is given by $$G = \begin{bmatrix}111&111&000&000\\000&000&111&111\\000&111&111&000\end{bmatrix}$$ Can you figure out a relationship between this code and the $[4,3,2]$ code consisting of all even-weight binary vectors of length $4$?