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I am working on a problem of SCC graph. The matrix representation of graph will be a square non-negative matrix that is column stochastic, irreducible. I will make it aperiodic by adding a self-loop on each node so that I can use the power method.
I will write a program of power method. Now I want to know whether my matrix always converges.
I know that diagonalizable matrix always converges. But if my matrix is not diagonalizable for some graph then will it converge? So what should I do if I want my matrix to converge always? Or my matrix's property(non-negative, stochastic, irreducible, aperiodic) is enough to converge always?
(My program will take input any SCC graph, I will add self-loop on each node. And write a program of power method. )

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  • $\begingroup$ what does it mean for a matrix to converge? $\endgroup$ – Dan Rust Sep 14 '19 at 10:56
  • $\begingroup$ We can get the dominant eigenvalue of a square matrix if it is Diagonalizable and has s dominant eigenvalue by power method. If the matrix has a dominant eigenvalue then in power method the initial vector converges to a dominant eigenvector value. @DanRust $\endgroup$ – criticalmind Sep 14 '19 at 11:17
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Your matrices are primitive which means that every entry is a positive real number. In particular, it means that we may apply the Perron-Frobenius theorem, one consequence of which is that if $M$ is a primitive matrix with (dominant) PF-eigenvalue $\lambda$, and associated left- and right-PF eigenvectors $\textbf{l},\textbf{r}$, respectively (normalised so that $\mathbf{l}^T\mathbf{r} = 1$), and we have $$\lim_{n \to \infty}M^n/\lambda^n = \mathbf{r}\mathbf{l}^T.$$

Is that the kind of thing you were looking for?

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  • $\begingroup$ Sir,my matrices are primitive but not positive,but some power of it is positive. I know my matrix is primitive and the Perron-Frobenius theorem is applicable here. I am working here with only the right eigenvector. I know that dominant eigenvalue is 1 and all other absolute value of eigenvalue is strictly less than 1 according to PF theorem. But I have seen in some blogs/books that a matrix should be diagonalizable for power method to work. I want to know whether it is true that only diagonalizable matrix converges? What about my matrix? @DanRust $\endgroup$ – criticalmind Sep 14 '19 at 11:44
  • $\begingroup$ Sorry yes, I forgot to mention you can be primitive also if some power is positive. The result that I mentioned is true in general, without assuming diagonalisability. $\endgroup$ – Dan Rust Sep 14 '19 at 12:46

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