Understanding Linear Codes for Homework

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A code is considered linear if it forms a linear subspace with finite elements. To determine if the product of two linear codes, C and C', is also a linear code, one must verify that the resulting code satisfies the three conditions of linear subspaces and maintains a finite number of elements. Each linear code must adhere to these criteria individually for their product to also be a linear code. Understanding the relationship between linear subspaces is crucial in this context. Therefore, studying the properties of linear subspaces is essential for this homework question.
ashnicholls
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How do you know if a code is linear.

I have only been given examples of linear codes, not what makes them linear codes.

I have to find out for homework if the product of two linear codes will be a linear code.

Eg Will C x C' be a linear code if C and C' are linear codes?

Cheers
 
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I am sure I did a search and that never came up, cheers will have a look.
 
Yes that what I know already but don't understand,

How do I relate it to my question?

Cheers
 
Just follow the definition. A code is linear because it is a linear subspace (with finite elements, or vectors). So you need to study linear subspaces (with finite elements).

Say Ci is a linear subspace, for each i = 1, 2 (which means that each Ci satisfies the 3 conditions of linear subspaces and has a finite number of elements). Does that mean that C = C1 x C2 is also a linear subspace (does C satisfy the 3 conditions of linear subspaces and have a finite number of elements, given that each of C1 and C2 does)?
 
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I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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