$\sum_{l=0}^{\infty} A^l $ for sub stochastic matrix

Asked Apr 08 '22 at 17:42

Active Apr 08 '22 at 17:42

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Let $A$ be a sub stochastic matrix (sum of rows is less or equal to 1) and also we know that the sum of at least one row is strictly less than 1.

Ho do I show that $$ \sum_{l=0}^{\infty} A^l = (I - A)^{-1} $$

I think I can do it if all eigenvalues of $A$ are less than 1 and $A$ is diagonalizable but not sure how to show it in general.

asked Apr 08 '22 at 17:42

Tomer

Show that the sum on the left converges. Then multiply it by $(I-A)$. – Stefan Lafon Apr 08 '22 at 17:45
It converges if the eigenvalues of A are less than 1 but how do I show that ? – Tomer Apr 08 '22 at 17:51
What is the matrix norm of such matrix ? – SacAndSac Apr 08 '22 at 17:52
Less or equal to 1 I guess but then the largest eigenvalue can be 1 and we don't have converges. – Tomer Apr 08 '22 at 17:54
The answer to that may be of help : https://math.stackexchange.com/questions/1393301/frobenius-norm-of-product-of-matrix – P. Quinton Apr 08 '22 at 21:52

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