0

I know that a stochastic matrix will have 1 as one of its eigenvalues. But do the stochastic matrices all have a stationary probability vector?

Basically, could there be a case where the eigen vector doesn't sum to 1?

user258269
  • 79
  • 2
    I think this is a consequence of Perron-Frobenius theorem. A stochastic matrix has a positive (each entry is positive) eigenvector corresponding to the eigenvalue $1$. So, dividing by the sum of the entries, you get a probability vector which is also an eigenvector. – Augustin Aug 29 '15 at 21:39
  • See here: http://math.stackexchange.com/questions/119891/finite-state-markov-chain-stationary-distribution –  Aug 29 '15 at 21:43

1 Answers1

1

Thank you Augustin and Bryon. It seems that: There can in fact be many eigen vectors for eigenvalue 1 which are not probability vectors. There can also be many probability vectors (aka stationary probability vectors), for the eigenvalue 1.

thank you for your answers.

user258269
  • 79