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So I recently found about representation of relation through matrix and found this post https://math.stackexchange.com/a/228956/1153005 in which he explains the stuff and finally writes that in summary if you have the transitive matrix $M_R$ then all the non zero entries in it must correspondingly have non zero entries in $M_R^2$ but playing around with this I found one "exception" of sorts.

Let $A={1,2,3}$ and we have a relation $R$ on $A\times A$ such that $R=\{(1,2),(2,3),(1,3)\}$. Now we can by observation say this is transitive but if we do it by the matrix thingy then we have $$M_R=\begin{bmatrix}0&1&1\\0&0&1\\0&0&0\end{bmatrix}\implies M_R^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$ Now why is not every non zero element in $M_R$ not corresponding non zero element in $M_R^2$ even though the relation is transitive, where am I going wrong?

lilychou
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    I have not read the referenced answer very carefully, but I think you are misstating it: it seems to me to say that if $M_R^2$ is non-zero at $(i,j)$ then so is $M_R$. – ancient mathematician Jan 24 '25 at 13:43

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$M^2_R$ shows the number of $2$-step paths. There is exactly one $2$-step path in this relation, between $1$ and $3$, being $(1,2) \to (2,3)$. Note that if you remove $(1,3)$ from the relation that path still exist and you get the same matrix for $M_R^2$. So to be transitive you require a non-zero entry in the same position for $M_R$ and $M_R^2$.

CyclotomicField
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