I understand that this is both reflexive and anti-symmetric. However, I don't think it's transitive. I understand that the classic rule is aRb, bRc, aRc. I can't seem to piece that together with this matrix. I've been asked to draw the Hasse diagram if it is, and I feel like it's a trick question. Is there a transitive relation that I am not seeing? I know that transitivity can be tricky, and I feel like I'm missing something.
$$
\begin{matrix}
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
\end{matrix}
$$
(1,1)(1,3)
(2,2)(2,3)(3,3)(3,4)(4,1)(4,2)(4,4)
Best Answer
Hint:
If we let your matrix be $R$, find $R^2$. If the nonzero entries in $R^2$ are in the same positions as those in $R$, then $R$ represents a transitive relation.