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I know that I-Q can not have a zero determinant, so it has an inverse, i.e. $N=(I-Q)^{-1}$ exists. I think I know how to prove part b of this question given that we assume $N^{-1}$ exists, my question is that how can I prove that $N^{-1}$ exists?

Does anybody know how to prove this. Any help would be greatly appreciated.

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    $\begingroup$ $N^{-1}$ exists iff det(N)$\ne$0 iff nul(N) is empty iff the eigenvalues $\lambda_i \ne$ 0 for all $i$. Which of these is easiest to prove given the definition of an absorbing chain? How to prove it depends on the theorems at your disposal. It may be enough (for some standards of proof) that $x_t^T N = x_{t+1}^T$ satisfies conservation of probability $\sum_i x_{it} = 1$, so $\lambda_i>0$ for all $i$. When in doubt, show that any alternative would violate the definition: If $\lambda_i=0$, then the matrix does not describe an absorbing chain. $\endgroup$ Dec 1 '20 at 4:05

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