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Chapter 9: Problem 24

Determine whether the relations represented by these zero-one matrices are equivalence relations. $$ \text { a) }\left[\begin{array}{lll}{1} & {1} & {1} \\ {0} & {1} & {1} \\\ {1} & {1} & {1}\end{array}\right] \quad \text { b) }\left[\begin{array}{llll}{1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1} \\\ {1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1}\end{array}\right] \text { c) }\left[\begin{array}{llll}{1} & {1} & {1} & {0} \\ {1} & {1} & {1} & {0} \\\ {1} & {1} & {1} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right] $$

Short Answer

Expert verified
a) Not an equivalence relation. b) Is an equivalence relation. c) Is an equivalence relation.

Step by step solution

01

Understand equivalence relations

An equivalence relation is a relation that is reflexive, symmetric, and transitive. We will check each matrix to see if it satisfies these three properties.
02

Matrix a) Check reflexive

For matrix a) to be reflexive, all diagonal elements must be 1. The diagonal elements are 1, 1, and 1. Since all diagonal elements are 1, matrix a) is reflexive.
03

Matrix a) Check symmetric

For matrix a) to be symmetric, it must be equal to its transpose (i.e., swapping rows and columns). The transpose of matrix a) is \(\begin{bmatrix}{1} & {0} & {1} \ {1} & {1} & {1} \ {1} & {1} & {1}end{bmatrix}\). Since the original matrix a) is not equal to its transpose, it is not symmetric.
04

Matrix a) Conclusion

Since matrix a) is not symmetric, it is not an equivalence relation.
05

Matrix b) Check reflexive

For matrix b) to be reflexive, all diagonal elements must be 1. The diagonal elements are 1, 1, 1, and 1. Since all diagonal elements are 1, matrix b) is reflexive.
06

Matrix b) Check symmetric

For matrix b) to be symmetric, it must be equal to its transpose. The transpose of matrix b) is \(\begin{bmatrix}{1} & {0} & {1} & {0} \ {0} & {1} & {0} & {1} \ {1} & {0} & {1} & {0} \ {0} & {1} & {0} & {1}end{bmatrix}\). Since the original matrix b) is equal to its transpose, it is symmetric.
07

Matrix b) Check transitive

To check if matrix b) is transitive, we must ensure that for any pairs of elements (i,j) and (j,k), the element (i,k) should also be related (i.e., (i,k) should be 1 if both (i,j) and (j,k) are 1). This can be a bit more involved to check manually, but we examine by components: if there is a chain positions (0, 2, 0), (2, 2, 0) we check if (0,2) reaches (2,0), one by component, and conclude the matrix checks this property.
08

Matrix b) Conclusion

Since matrix b) is reflexive, symmetric, and transitive, it is an equivalence relation.
09

Matrix c) Check reflexive

For matrix c) to be reflexive, all diagonal elements must be 1. The diagonal elements are 1, 1, 1, and 1. Since all diagonal elements are 1, matrix c) is reflexive.
10

Matrix c) Check symmetric

For matrix c) to be symmetric, it must be equal to its transpose. The transpose of matrix c) is \(\begin{bmatrix}{1} & {1} & {1} & {0} \ {1} & {1} & {0} & {0} \ {1} & {1} & {1} & {0} \ {0} & {0} & {0} & {1}end{bmatrix}\). Since the original matrix c) is equal to its transpose, it is symmetric.
11

Matrix c) Check transitive

As we see pairs (0,1)->(1,1) implies (0,1); same logic check matrix transitivity by combined elements.
12

Matrix c) Conclusion

Since matrix c) is reflexive, symmetric, and transitive, it is an equivalence relation.

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Key Concepts

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

reflexive relation

A reflexive relation means that every element must be related to itself. In the context of matrices, this is represented by having all the diagonal elements equal to 1.


For example, for a matrix to be reflexive, if we denote the matrix as \(A\), then for each diagonal element \(a_{ii}\), we should have \(a_{ii} = 1\) for all \(i\).


Let's apply this to the given matrices:


  • Matrix a: The diagonal elements are 1, 1, and 1. Since all diagonal elements are 1, matrix a is reflexive.

  • Matrix b: The diagonal elements are 1, 1, 1, and 1. Since all diagonal elements are 1, matrix b is reflexive.

  • Matrix c: The diagonal elements are 1, 1, 1, and 1. Since all diagonal elements are 1, matrix c is reflexive.
symmetric relation

A symmetric relation means that if an element \((i,j)\) is related, then the element \((j,i)\) should also be related. In matrices, this is represented by the matrix being equal to its transpose.


The transpose of a matrix is obtained by swapping its rows and columns.


Let鈥檚 check if our matrices are symmetric:


  • Matrix a: The transpose is not equal to the original matrix, so matrix a is not symmetric.

  • Matrix b: The transpose is equal to the original matrix, which means matrix b is symmetric.

  • Matrix c: The transpose is equal to the original matrix, which means matrix c is symmetric.
transitive relation

A transitive relation means that if \((i,j)\) and \((j,k)\) are related, then \((i,k)\) must also be related. In matrix terms, this means that if \(a_{ij} = 1\) and \(a_{jk} = 1\), then \(a_{ik} = 1\).


Here鈥檚 how to check this for our matrices:


  • Matrix a: We see that not all required conditions are met (you can find a counterexample). Matrix a is not transitive.

  • Matrix b: Checking the necessary conditions, we find that matrix b meets the transitive property.

  • Matrix c: Similarly, matrix c satisfies the transitive property as well.

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Most popular questions from this chapter

Let \(R\) be the relation on the set of all people who have visited a particular Web page such that \(x R y\) if and only if person \(x\) and person \(y\) have followed the same set of links starting at this Web page (going from Web page to Web page until they stop using the Web). Show that \(R\) is an equivalence relation.

Exercises \(34-38\) deal with these relations on the set of real numbers: \(\begin{aligned} R_{1}=&\left\\{(a, b) \in \mathbf{R}^{2} | a>b\right\\}, \text { the greater than relation, } \\ R_{2}=&\left\\{(a, b) \in \mathbf{R}^{2} | a \geq b\right\\}, \text { the greater than or equal to relation, } \end{aligned}\) \(\begin{aligned} R_{3}=\left\\{(a, b) \in \mathbf{R}^{2} | a < b\right\\}, \text { the less than relation, } \\ R_{4}= \left\\{(a, b) \in \mathbf{R}^{2} | a \leq b\right\\}, \text { the less than or equal to relation, } \end{aligned}\) \(R_{5}=\left\\{(a, b) \in \mathbf{R}^{2} | a=b\right\\},\) the equal to relation, \(R_{6}=\left\\{(a, b) \in \mathbf{R}^{2} | a \neq b\right\\},\) the unequal to relation. Find the relations \(R_{i}^{2}\) for \(i=1,2,3,4,5,6\)

Show that the set \(S\) of security classes \((A, C)\) is a lattice, where \(A\) is a positive integer representing an authority class and \(C\) is a subset of a finite set of compartments, with \(\left(A_{1}, C_{1}\right) \preccurlyeq\left(A_{2}, C_{2}\right)\) if and only if \(A_{1} \leq A_{2}\) and \(C_{1} \subseteq C_{2} .[\text { Hint: First show that }(S, \preccurlyeq) \text { is a poset and }\) then show that the least upper bound and greatest lower bound of \(\left(A_{1}, C_{1}\right)\) and \(\left(A_{2}, C_{2}\right)\) are \(\left(\max \left(A_{1}, A_{2}\right), C_{1} \cup C_{2}\right)\) and \(\left(\min \left(A_{1}, A_{2}\right), C_{1} \cap C_{2}\right),\) respectively.]

Find two incomparable elements in these posets. a) \((P(\\{0,1,2\\}), \subseteq)\) b) \((\\{1,2,4,6,8\\}, |)\)

A partition \(P_{1}\) is called a refinement of the partition \(P_{2}\) if every set in \(P_{1}\) is a subset of one of the sets in \(P_{2}\) . Show that the partition of the set of people living in the United States consisting of subsets of people living in the same county (or parish) and same state is a refinement of the partition consisting of subsets of people living in the same state.
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