<p>Suppose we consider $s$-folded Reed-Solomon codes that are based on polynomials over a field $\mathbb{F} = \mathrm{GF}(p^t)$. Then the alphabet of those codes is of size $p^{t \cdot s}$. Hence, in order to be linear, those codes should be closed under multiplication by scalar from the field $\mathbb{F}' = \mathrm{GF}(p^{t \cdot s})$. There is no apparent reason why they should be closed under such operation.</p>

<p>However, it is true that they are closed under multiplication by scalar from $\mathbb{F}$. In other words, the codes are $\mathbb{F}$-linear.</p>

CC BY-SA 3.0
2017-04-09T21:16:25.383Z
2017-04-09T21:16:25.383Z
Or Meir
<p>This is for my understanding. What am I missing?</p>

<linear-algebra><coding-theory>
Why are folded Reed Solomon Codes considered non linear?
Jardine
# Why are folded Reed Solomon Codes considered non linear?

(Asked by: username_0 on 2017-04-09)

This is for my understanding. What am I missing?

**Tags:** linear-algebra, coding-theory

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## 1 answers

### Answer by username_1 on 2017-04-09. Score: 2 | Accepted answer

Suppose we consider $s$-folded Reed-Solomon codes that are based on polynomials over a field $\mathbb{F} = \mathrm{GF}(p^t)$. Then the alphabet of those codes is of size $p^{t \cdot s}$. Hence, in order to be linear, those codes should be closed under multiplication by scalar from the field $\mathbb{F}' = \mathrm{GF}(p^{t \cdot s})$. There is no apparent reason why they should be closed under such operation.

However, it is true that they are closed under multiplication by scalar from $\mathbb{F}$. In other words, the codes are $\mathbb{F}$-linear.


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