Why is an Accepting NFA in NL

Question

I am reading the definition of NL complexity, but I'm having a hard time with a simple example.

Let's say I have an NFA Turing Machine $<M, w>$.

This is an accepting NFA Turing Machine, meaning that $M$ accepts an input string $w$.

Why is this NFA $\in$ NL

Answer 1

This is the solution I came up with.

We know that the problem of determining whether there is a path between two vertices in a directed graph runs in logarithmic space. We can consider A to be a directed graph, and check all final states to see if it is reachable from the start state. Thus A NFA must also be in NL.