Consider continuous Markov chain with finite state space. Let’s $P(t)$ demote it’s transition matrix at moment $t$.
We want to prove that for all non-zeroes $$ we have : $det P(t) \ne 0$.
I think this isn’t right. If we consider $P = \xi \xi^{т}$, where $\xi = \{1/\sqrt{2},1/\sqrt{2}\}$ then all powers of this matrix have zero determinant
Maybe I miss something?