Quantum linear code/ Dual Code (CSS) proof

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SUMMARY

This discussion focuses on the proof of properties related to quantum linear codes and their dual codes, specifically addressing the conditions under which a vector \( x \) belongs to the orthogonal complement \( C^\perp \). It establishes that if \( x \in C^\perp \), certain equalities hold true due to the subspace nature of \( C \). The discussion emphasizes the significance of the relationship between \( x \) and elements \( c_0 \in C \) in deriving these equalities, reinforcing the mathematical framework of quantum coding theory.

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steve1763
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What would the proof be for the following identity? I cannot find the proof anywhere
Screenshot 2021-09-05 at 21.52.33.png
 
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How should we know, what your symbols mean? You have to give a bit more information or a link to the paper you are referring too.
 
If ##x\in C^\perp## then it is clear. If not, then there is a ##c_0\in C## such that ##x\cdot c_0 =1##. The you have

##
-1\sum_{c\in C}(-1)^{x\cdot c}=(-1)^{x\cdot c_0}\sum_{c\in C}(-1)^{x\cdot c}=\sum_{c\in C}(-1)^{x\cdot (c-c_0)}=\sum_{c\in C}(-1)^{x\cdot c}
##

The last equality is because ##C## is a subspace.
 
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martinbn said:
If ##x\in C^\perp## then it is clear. If not, then there is a ##c_0\in C## such that ##x\cdot c_0 =1##. The you have

##
-1\sum_{c\in C}(-1)^{x\cdot c}=(-1)^{x\cdot c}\sum_{c\in C}(-1)^{x\cdot c}=\sum_{c\in C}(-1)^{x\cdot (c-c_0)}=\sum_{c\in C}(-1)^{x\cdot c}
##

The last equality is because ##C## is a subspace.
Thank you very much
 

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