Final answer:
The statement is true as the presence of a state with a self-transition probability greater than 0 in an irreducible Markov chain ensures that this state and therefore all other states are aperiodic.
Step-by-step explanation:
The statement is true. In a Markov chain, if there is a state j such that the probability of transitioning from state j to itself (P(j,j)) is greater than 0, it suggests that it's possible to return to state j in one step. This implies that the greatest common divisor of return times to state j is 1, which by definition means that the state j is aperiodic. Since the Markov chain is irreducible, all states communicate with each other, meaning that each state can be reached from every other state in a finite number of steps. Therefore, the property of aperiodicity must extend to all the states in the chain, as they can all reach state j and return to themselves through it, thus they will also have a period of 1, making them aperiodic.