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This term measures degeneracy. In order to prevent it from being negative, [math]\log N[/math]<ref name=hoel_2013 /> was added. The meaning of degeneracy is: if the current state of the system is known, can it be deduced from the state of the system at the previous moment? If it can be inferred, then the degeneracy of this Markov matrix will be relatively low, that is, non-degeneracy. If it is difficult to deduce, the Markov matrix is degenerate, i.e., degenerate. Why can degeneracy be described by the negative entropy of the average row vector distribution? First of all, when all the row vectors in P are independent of each other, then their average distribution will be very close to a uniform distribution, i.e [math]\bar{P}\approx (\frac{1}{N},\frac{1}{N},\cdots,\frac{1}{N})[/math], resulting in maximum [[Shannon Entropy]], i.e., [math]\log N[/math]. In this case, the Markov transition matrix is a '''Invertible Matrix''', indicating that we can deduce the previous state from the current state. Therefore, this Markov matrix is non-degenerate, and the computed degeneracy is zero.

This term measures degeneracy. In order to prevent it from being negative, [math]\log N[/math]<ref name=hoel_2013 /> was added. The meaning of degeneracy is: if the current state of the system is known, can it be deduced from the state of the system at the previous moment? If it can be inferred, then the degeneracy of this Markov matrix will be relatively low. If it is difficult to deduce, the Markov matrix is degenerate. Why can degeneracy be described by the negative entropy of the average row vector distribution? First of all, when all the row vectors in P are independent of each other, then their average distribution will be very close to a uniform distribution, i.e [math]\bar{P}\approx (\frac{1}{N},\frac{1}{N},\cdots,\frac{1}{N})[/math], resulting in maximum [[Shannon Entropy]], i.e., [math]\log N[/math]. In this case, the Markov transition matrix is a '''Invertible Matrix''', indicating that we can deduce the previous state from the current state. Therefore, this Markov matrix is non-degenerate, and the computed degeneracy is zero.

Conversely, when all row vectors of P are identical, the average vector is also a one-hot vector with minimum [[Entropy]]. In this case, it is challenging to infer the previous state from the current state, leading to a degenerate (or non-reversible) Markov matrix, with a computed degeneracy equal to [math]\log N[/math].

Conversely, when all row vectors of P are identical, the average vector is also a one-hot vector with minimum [[Entropy]]. In this case, it is challenging to infer the previous state from the current state, leading to a degenerate (or non-reversible) Markov matrix, with a computed degeneracy equal to [math]\log N[/math].