Will the diagonal elements of transitive closure matrix of a 'graph' always be 1?

Question

(CAUTION- PLEASE DO NOT TAKE MY QUESTIONS VERY SERIOUSLY.

I received a ban from asking questions,I don't know what to say really,I am just a student trying to learn,not a professional mathematician,so of course the questions could have been,well weird or useless to the mathematical community...I really did not mean any harm or anything except just trying to understand something.I really wish a feature like,telling you exactly what to do with your questions so the ban would get lifted was there.But I guess this is just hard luck and me not taking this community very seriously.I do apologize to the community,and I do understand that asking a question is a privilege not a right here.because,like to stop wasting people's time and not spread wrong ideas?right? I REPEAT,I DID NOT MEAN ANY CONFUSION OR ANYTHING,I AM NO PROFESSIONAL,JUST A LEARNER.)

i was trying to learn this topic from youtube and google and learned that the transitive closure of a graph is the reachability matrix.I looked at this example on geeks for geeks and also some teachers on youtube saying "since A is obviously reachable from A we mark it 1" my question is

1. does this mean the diagonal elements of transitive closure matrix always have to be 1, considering trivially that a vertex is always reachable to itself?

2. how does the formula of transitive closure capture this if yes?

3.does this mean we have to draw loops in the graph of transitive closure?i think that will be no,but anyways how does 1 draw the graph of transitive closure from reachability matrix.

i have also written this question on paper in a neat handwriting and attached it as screenshot to make it clear.( i can't really learn how formulae and graph are written here)

this is the first image

this is the second image

Answer 1

This seems like a disagreement on a choice of convention. If you want to count paths of length $0$, you'll need to add an $A^0$ term onto the beginning of your sum $A^1 + A^2 + A^3 + \cdots $. Each term $A^k$ counts the paths of exactly length $k$.

I find both possible conventions convincing: including paths of length $0$ makes sense if you're thinking about graphs and excluding them makes sense if you're thinking about relations.