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Chapter 6: Problem 27

Which of the codes are linear codes? $$C=\left\\{\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right],\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right]\right\\}$$

Short Answer

Expert verified
The code C is a linear code.

Step by step solution

01

Understand Linear Codes

A code is linear if it forms a vector space. This means that it contains the zero codeword, is closed under vector addition, and scalar multiplication (in this case, addition modulo 2 since we deal with binary vectors in finite fields, where scalar multiplication is essentially multiplying each component by 1, which does not change a vector). We will verify if these properties hold for the given code set.
02

Check for Zero Codeword

Check if the zero vector \([0, 0, 0]\) is in the code. This is a necessary condition for a linear code. In the given set, \([0, 0, 0]\) is indeed present.
03

Test Closure under Addition

Check if the addition of any two vectors from the code \( C \) is also in \( C \). Use vector addition modulo 2:- \([0, 0, 0] + [0, 1, 0] = [0, 1, 0]\) is in \( C \).- \([0, 0, 1] + [0, 1, 0] = [0, 1, 1]\) is in \( C \).- \([0, 1, 0] + [0, 1, 1] = [0, 0, 1]\) is in \( C \).Each addition should result in a vector that is already within the code set, confirming closure under addition.
04

Conclusion on Linearity

Since the code \( C \) contains the zero codeword and is closed under addition, it is a linear code. Scalar multiplication is trivially satisfied in binary codes, so we conclude with these checks.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Space
A vector space is a fundamental concept in linear algebra and mathematics. It's a collection of objects called vectors, where two main operations, vector addition and scalar multiplication, can be performed. For a set to qualify as a vector space, it must satisfy several properties:
  • Closure under addition: Adding any two vectors in the set results in another vector that is also in the set.
  • Closure under scalar multiplication: Multiplying any vector by a scalar (a constant) results in a vector that is still in the set.
  • Contains a zero vector: The set must have a vector that is an additive identity, meaning any vector added to the zero vector remains unchanged.
  • Abides by commutative, associative, and distributive laws for addition and scalar multiplication respectively.
These properties provide a framework within which you can perform linear operations, making vector spaces essential for understanding linear codes in coding theory.
Zero Codeword
The zero codeword is essential in defining a linear code. It acts as the additive identity within a set of vectors or codewords. In the realm of coding theory, the zero codeword is the all-zero binary vector.
  • For the given problem, the zero codeword is \([0, 0, 0]\).
  • Having the zero codeword in your set means that you can add it to any other codeword without changing its value.
  • In a linear code, the zero codeword must be present for the code to qualify as a vector space.
Its presence ensures completeness under addition, ensuring any vector added to the zero codeword remains unchanged, thereby maintaining the code's structural integrity.
Vector Addition
Vector addition is a crucial operation when considering linear codes. It involves combining vectors by adding their corresponding components together. For linear codes, addition must be closed, meaning:
  • When you add any two codewords from the set, the resulting vector should also be part of the set.
  • In binary vectors, you perform vector addition using modulo 2 operations, essentially the same as "bitwise addition," which results in a 0 if the bits are the same and 1 if they are different.
  • In the given example, adding codewords \([0, 0, 1] + [0, 1, 0]\)gives \([0, 1, 1]\), which is also part of the code set.
Ensuring closure under addition is imperative for confirming the code's linearity, as it must consistently produce a vector that fits within the original code set.
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar, which is a real number in most contexts. However, for binary vectors within a linear code, the scalar is typically 0 or 1, meaning:
  • Multiplying a vector by 1 leaves the vector unchanged.
  • Multiplying a vector by 0 converts every component to 0, transforming it into the zero vector \([0, 0, 0]\).
  • This property ensures that binary vector sets satisfy the scalar multiplication requirement of a vector space trivially.
Since binary vectors inherently exhibit these traits with scalars in binary operations, scalar multiplication becomes a straightforward check within the context of linear codes.
Binary Vectors
Binary vectors play a significant role in coding theory due to their simplicity and efficiency. These vectors consist of elements that are strictly 0 or 1 and are typically used in digital and computer systems.
  • Binary vectors rely on operations like bitwise addition, which corresponds to vector addition modulo 2.
  • They provide an efficient means to represent data, especially in communication systems where error detection and correction are necessary.
  • Each position in the binary vector represents a bit, and collectively, they define a particular codeword in coding.
Understanding how binary vectors work lays the groundwork for grasping linear codes, as they allow vectors to be easily manipulated while maintaining strict structural rules that ensure data integrity.

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Most popular questions from this chapter

Find the solution of the differential equation that satisfies the given boundary condition(s). $$y^{\prime}-3 y=0, y(1)=2$$

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Let \(\\{\mathbf{u}, \mathbf{v}, \mathbf{w}\\}\) be a linearly independent set of vectors in a vector space \(V\) (a) \(\mathrm{Is}\\{\mathbf{u}+\mathbf{v}, \mathbf{v}+\mathbf{w}, \mathbf{u}+\mathbf{w}\\}\) linearly independent? Either prove that it is or give a counterexample to show that it is not. (b) \(\mathrm{Is}\\{\mathbf{u}-\mathbf{v}, \mathbf{v}-\mathbf{w}, \mathbf{u}-\mathbf{w}\\}\) linearly independent? Either prove that it is or give a counterexample to show that it is not.

Determine whether the set \(\mathcal{B}\) is a basis for the vector space \(V\) $$\begin{array}{l} V=M_{22} \\ B=\left\\{\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right],\left[\begin{array}{rr} 1 & 3 \\ -3 & 1 \end{array}\right],\left[\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right],\left[\begin{array}{ll} 1 & 2 \\ 3 & 2 \end{array}\right]\right\\} \end{array}$$
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