How many reflexive and antisymmetric relations are there on an n-element set?

Solution

Verified$\textbf{Multiplication principle: }$If one event can occur in $m$ ways AND a second event can occur in $n$ ways, then the number of ways that the two events can occur in sequence is then $m\cdot n$.

A relation $R$ on a set $A$ is $\textbf{antisymmetric}$ if $(b,a)\in R$ and $(a,b) \in R$ implies $a=b$.

A relation $R$ on a set $A$ is $\textbf{reflexive}$ if $(a,a)\in R$ for every element $a\in A$.

Moreover, a relation is not antisymmetric when its corresponding matrix is a matrix that is not symmetric, while the reflexive property requires that all main diagonal elements are 1's.

When $A$ contains $n$ elements, then the relation $R$ is represented by a $n\times n$ matrix.

$\text{\underline{\textbf{Number of symmetric matrices that are not antisymmetric}}}$

When we know all elements on or above the main diagonal of the symmetric matrix, then the other elements of the matrix can be determined due to symmetry.

Since an $n\times n$ matrix contains $n\cdot n=n^2$ elements with $n$ elements on the main diagonal, there are $n^2-n$ elements that are not on the main diagonal.

Since exactly half of the non-main diagonal elements are on either side of the main diagonal, there are $\frac{n^2-n}{2}$ elements above the main diagonal.

For each of the $\frac{n^2-n}{2}$ elements on or above the main diagonal, there are $2$ options: the element is a 0 or a 1.

Use the multiplication principle:

$\underbrace{2\cdot 2\cdot ...\cdot 2}_{(n^2-n)/2\text{ repetitions}}=2^{(n^2-n)/2}$

Thus there are $2^{(n^2-n)/2}$ possible symmetric matrices that are not antisymmetric but reflexive (with 1's on the main diagonal).

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